OP Malhotra Class 11 Maths Solutions Chapter 5 Compound and Multiple Angles Exercise 5 (C)

Question 1. Evaluate:
(i) 2 sin 15° cos 15°
(ii) 1 – 2 sin² 22.5°
(iii) 2 cos² 157.5° – 1
(iv) cos² \( \frac { \pi }{ 12 } \) – sin² \( \frac { \pi }{ 12 } \)
(v) \( \frac { 1 }{ 2 } \) – sin² \( \frac { 7\pi }{ 12 } \)
(vi) cos \( \frac { \pi }{ 8 } \) sin \( \frac { \pi }{ 8 } \)
(vii) \( \frac { 2 \tan 22.5^\circ }{ 1 - \tan^2 22.5^\circ } \)
(viii) 8 cos³ \( \frac { \pi }{ 9 } \) – 6 cos \( \frac { \pi }{ 9 } \)
Answer:
(i) We know that \( 2 \sin \theta \cos \theta = \sin 2\theta \).
So, \( 2 \sin 15^\circ \cos 15^\circ = \sin (2 \times 15^\circ) \)
\( \implies \sin 30^\circ \)
\( \implies \frac{1}{2} \)
(ii) We use the identity \( \cos 2\theta = 1 - 2 \sin^2 \theta \).
So, \( 1 - 2 \sin^2 22.5^\circ = \cos (2 \times 22.5^\circ) \)
\( \implies \cos 45^\circ \)
\( \implies \frac{1}{\sqrt{2}} \)
(iii) We use the identity \( \cos 2\theta = 2 \cos^2 \theta - 1 \).
So, \( 2 \cos^2 157.5^\circ - 1 = \cos (2 \times 157.5^\circ) \)
\( \implies \cos 315^\circ \)
\( \implies \cos (360^\circ - 45^\circ) \) Since \( \cos (360^\circ - \theta) = \cos \theta \).
\( \implies \cos 45^\circ \)
\( \implies \frac{1}{\sqrt{2}} \)
(iv) We use the identity \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \).
So, \( \cos^2 \frac{\pi}{12} - \sin^2 \frac{\pi}{12} = \cos (2 \times \frac{\pi}{12}) \)
\( \implies \cos \frac{\pi}{6} \)
\( \implies \frac{\sqrt{3}}{2} \)
(v) First, rewrite the expression: \( \frac{1}{2} - \sin^2 \frac{7\pi}{12} = \frac{1 - 2 \sin^2 \frac{7\pi}{12}}{2} \).
Using the identity \( \cos 2\theta = 1 - 2 \sin^2 \theta \), the numerator becomes \( \cos (2 \times \frac{7\pi}{12}) \).
So, \( \frac{\cos (\frac{7\pi}{6})}{2} \)
We know \( \cos \frac{7\pi}{6} = \cos (\pi + \frac{\pi}{6}) = -\cos \frac{\pi}{6} \).
\( \implies \frac{-\cos \frac{\pi}{6}}{2} \)
\( \implies \frac{-\sqrt{3}/2}{2} \)
\( \implies -\frac{\sqrt{3}}{4} \)
(vi) We rearrange the expression: \( \cos \frac{\pi}{8} \sin \frac{\pi}{8} = \frac{1}{2} (2 \sin \frac{\pi}{8} \cos \frac{\pi}{8}) \).
Using the identity \( 2 \sin \theta \cos \theta = \sin 2\theta \), this becomes \( \frac{1}{2} \sin (2 \times \frac{\pi}{8}) \).
\( \implies \frac{1}{2} \sin \frac{\pi}{4} \)
\( \implies \frac{1}{2} \times \frac{1}{\sqrt{2}} \)
\( \implies \frac{1}{2\sqrt{2}} \)
(vii) We use the identity \( \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \).
So, \( \frac{2 \tan 22.5^\circ}{1 - \tan^2 22.5^\circ} = \tan (2 \times 22.5^\circ) \)
\( \implies \tan 45^\circ \)
\( \implies 1 \)
(viii) We use the identity \( \cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta \).
So, \( 8 \cos^3 \frac{\pi}{9} - 6 \cos \frac{\pi}{9} = 2 (4 \cos^3 \frac{\pi}{9} - 3 \cos \frac{\pi}{9}) \)
\( \implies 2 \cos (3 \times \frac{\pi}{9}) \)
\( \implies 2 \cos \frac{\pi}{3} \)
\( \implies 2 \times \frac{1}{2} \)
\( \implies 1 \)
In simple words: For each part, we used specific trigonometric identities that simplify expressions involving sine and cosine of angles. These identities help us change the expression into a simpler form, like a single sine or cosine of a double angle, which we can then evaluate. This helps calculate values for unusual angles.

🎯 Exam Tip: Remember common trigonometric identities for double angles, such as \( \sin 2\theta \), \( \cos 2\theta \), and \( \tan 2\theta \), as they are essential for simplifying expressions and evaluating values for compound angles.

Question 2. Find the values of sin 2\( \theta \), cos 2\( \theta \), and tan 2\( \theta \), given :
(i) sin \( \theta = \frac{3}{5} \), \( \theta \) in Quadrant I.
(ii) sin \( \theta = \frac{3}{5} \), \( \theta \) in Quadrant II.
(iii) sin \( \theta = -\frac{1}{2} \), \( \theta \) in Quadrant IV.
(iv) tan \( \theta = -\frac{1}{5} \), \( \theta \) in Quadrant II.
Answer:
(i) Given \( \sin \theta = \frac{3}{5} \) and \( \theta \) is in Quadrant I. In Quadrant I, \( \cos \theta \) is positive.
We find \( \cos \theta = \sqrt{1 - \sin^2 \theta} \)
\( \implies \sqrt{1 - (\frac{3}{5})^2} \)
\( \implies \sqrt{1 - \frac{9}{25}} \)
\( \implies \sqrt{\frac{16}{25}} \)
\( \implies \frac{4}{5} \)
Now, we find \( \sin 2\theta = 2 \sin \theta \cos \theta \)
\( \implies 2 \times \frac{3}{5} \times \frac{4}{5} \)
\( \implies \frac{24}{25} \)
Next, we find \( \cos 2\theta = 1 - 2 \sin^2 \theta \)
\( \implies 1 - 2 (\frac{3}{5})^2 \)
\( \implies 1 - 2 \times \frac{9}{25} \)
\( \implies 1 - \frac{18}{25} \)
\( \implies \frac{7}{25} \)
Finally, we find \( \tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} \)
\( \implies \frac{24/25}{7/25} \)
\( \implies \frac{24}{7} \)
(ii) Given \( \sin \theta = \frac{3}{5} \) and \( \theta \) is in Quadrant II. In Quadrant II, \( \cos \theta \) is negative.
We find \( \cos \theta = -\sqrt{1 - \sin^2 \theta} \)
\( \implies -\sqrt{1 - (\frac{3}{5})^2} \)
\( \implies -\sqrt{1 - \frac{9}{25}} \)
\( \implies -\sqrt{\frac{16}{25}} \)
\( \implies -\frac{4}{5} \)
Now, we find \( \sin 2\theta = 2 \sin \theta \cos \theta \)
\( \implies 2 \times \frac{3}{5} \times (-\frac{4}{5}) \)
\( \implies -\frac{24}{25} \)
Next, we find \( \cos 2\theta = 1 - 2 \sin^2 \theta \)
\( \implies 1 - 2 (\frac{3}{5})^2 \)
\( \implies 1 - 2 \times \frac{9}{25} \)
\( \implies 1 - \frac{18}{25} \)
\( \implies \frac{7}{25} \)
Finally, we find \( \tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} \)
\( \implies \frac{-24/25}{7/25} \)
\( \implies -\frac{24}{7} \)
(iii) Given \( \sin \theta = -\frac{1}{2} \) and \( \theta \) is in Quadrant IV. In Quadrant IV, \( \cos \theta \) is positive.
We find \( \cos \theta = \sqrt{1 - \sin^2 \theta} \)
\( \implies \sqrt{1 - (-\frac{1}{2})^2} \)
\( \implies \sqrt{1 - \frac{1}{4}} \)
\( \implies \sqrt{\frac{3}{4}} \)
\( \implies \frac{\sqrt{3}}{2} \)
Now, we find \( \sin 2\theta = 2 \sin \theta \cos \theta \)
\( \implies 2 \times (-\frac{1}{2}) \times \frac{\sqrt{3}}{2} \)
\( \implies -\frac{\sqrt{3}}{2} \)
Next, we find \( \cos 2\theta = 1 - 2 \sin^2 \theta \)
\( \implies 1 - 2 (-\frac{1}{2})^2 \)
\( \implies 1 - 2 \times \frac{1}{4} \)
\( \implies 1 - \frac{1}{2} \)
\( \implies \frac{1}{2} \)
Finally, we find \( \tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} \)
\( \implies \frac{-\sqrt{3}/2}{1/2} \)
\( \implies -\sqrt{3} \)
(iv) Given \( \tan \theta = -\frac{1}{5} \) and \( \theta \) is in Quadrant II. In Quadrant II, \( \sin \theta \) is positive and \( \cos \theta \) is negative.
We find \( \sec \theta = -\sqrt{1 + \tan^2 \theta} \) (negative because \( \cos \theta \) is negative).
\( \implies -\sqrt{1 + (-\frac{1}{5})^2} \)
\( \implies -\sqrt{1 + \frac{1}{25}} \)
\( \implies -\sqrt{\frac{26}{25}} \)
\( \implies -\frac{\sqrt{26}}{5} \)
Then, \( \cos \theta = \frac{1}{\sec \theta} = -\frac{5}{\sqrt{26}} \).
And \( \sin \theta = \tan \theta \cos \theta \)
\( \implies (-\frac{1}{5}) \times (-\frac{5}{\sqrt{26}}) \)
\( \implies \frac{1}{\sqrt{26}} \)
Now, we find \( \sin 2\theta = 2 \sin \theta \cos \theta \)
\( \implies 2 \times \frac{1}{\sqrt{26}} \times (-\frac{5}{\sqrt{26}}) \)
\( \implies \frac{-10}{26} \)
\( \implies -\frac{5}{13} \)
Next, we find \( \cos 2\theta = 1 - 2 \sin^2 \theta \)
\( \implies 1 - 2 (\frac{1}{\sqrt{26}})^2 \)
\( \implies 1 - 2 \times \frac{1}{26} \)
\( \implies 1 - \frac{1}{13} \)
\( \implies \frac{12}{13} \)
Finally, we find \( \tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} \)
\( \implies \frac{-5/13}{12/13} \)
\( \implies -\frac{5}{12} \)
In simple words: To find the double angle values, we first figure out if sine or cosine is positive or negative based on the given quadrant. Then, we find the missing trigonometric ratio (like cosine if sine is given) using the Pythagorean identity. After that, we use the double angle formulas for sine, cosine, and tangent to calculate the final values.

🎯 Exam Tip: Always pay close attention to the quadrant given for the angle \( \theta \), as it determines the sign of \( \cos \theta \) and \( \sin \theta \), which is critical for correct calculations of double angles.

Question 3. ABC is an acute-angled triangle inscribed in a circle of radius 5 cm and centre O. The sine of angle A is equal to \( \frac{3}{5} \). Calculate without using tables :
(i) the length of BC
(ii) sin OBC
(iii) sin BOC
(iv) cos BOC.
Answer:

(i) We know that the angle subtended by an arc at the center is twice the angle subtended by it on the circumference of the circle. So, \( \angle BOC = 2 \times \angle BAC = 2A \).
In \( \triangle OBD \), we have \( \angle BOD = A \) (since OD is perpendicular to BC and bisects \( \angle BOC \)).
Also, \( OB = 5 \) cm (radius).
Now, \( \sin A = \frac{BD}{OB} \)
\( \implies BD = OB \sin A \)
\( \implies BD = 5 \sin A \)
Given \( \sin A = \frac{3}{5} \), so \( BD = 5 \times \frac{3}{5} = 3 \) cm.
Since OD is perpendicular to BC, it bisects BC. So, \( BC = 2 \times BD \).
\( \implies BC = 2 \times 3 = 6 \) cm.
(ii) From the figure, \( \angle B = 90^\circ - A \).
So, \( \sin \angle OBC = \sin (90^\circ - A) = \cos A \).
We find \( \cos A = \sqrt{1 - \sin^2 A} \). Since A is an acute angle, \( \cos A \) is positive.
\( \implies \sqrt{1 - (\frac{3}{5})^2} \)
\( \implies \sqrt{1 - \frac{9}{25}} \)
\( \implies \sqrt{\frac{16}{25}} \)
\( \implies \frac{4}{5} \)
Therefore, \( \sin \angle OBC = \frac{4}{5} \).
(iii) We know \( \angle BOC = 2A \).
So, \( \sin \angle BOC = \sin 2A \).
Using the identity \( \sin 2A = 2 \sin A \cos A \).
\( \implies 2 \times \frac{3}{5} \times \frac{4}{5} \)
\( \implies \frac{24}{25} \)
(iv) We know \( \angle BOC = 2A \).
So, \( \cos \angle BOC = \cos 2A \).
Using the identity \( \cos 2A = 1 - 2 \sin^2 A \).
\( \implies 1 - 2 (\frac{3}{5})^2 \)
\( \implies 1 - 2 \times \frac{9}{25} \)
\( \implies 1 - \frac{18}{25} \)
\( \implies \frac{7}{25} \)
In simple words: We used geometric properties of a circle and triangle, like how angles at the center relate to angles at the edge. Then, we applied trigonometric identities for double angles and the Pythagorean identity to find the lengths and angle values needed.

🎯 Exam Tip: Always sketch the diagram for geometry problems involving circles and triangles. This helps visualize the relationships between angles and sides, making it easier to apply the correct trigonometric formulas and properties.

Question 4. Derive functions of 120° from functions of 60° and check by using relations between functions of supplementary angles.
Answer:
Let's derive the functions for 120° using double angle formulas based on 60°.
For sine:
\( \sin 120^\circ = \sin (2 \times 60^\circ) = 2 \sin 60^\circ \cos 60^\circ \)
\( \implies 2 \times \frac{\sqrt{3}}{2} \times \frac{1}{2} \)
\( \implies \frac{\sqrt{3}}{2} \)
Now, checking with supplementary angles: \( \sin 120^\circ = \sin (180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2} \). This matches.
For cosine:
\( \cos 120^\circ = \cos (2 \times 60^\circ) = 1 - 2 \sin^2 60^\circ \)
\( \implies 1 - 2 (\frac{\sqrt{3}}{2})^2 \)
\( \implies 1 - 2 \times \frac{3}{4} \)
\( \implies 1 - \frac{3}{2} \)
\( \implies -\frac{1}{2} \)
Now, checking with supplementary angles: \( \cos 120^\circ = \cos (180^\circ - 60^\circ) = -\cos 60^\circ = -\frac{1}{2} \). This also matches.
For tangent:
\( \tan 120^\circ = \tan (2 \times 60^\circ) = \frac{2 \tan 60^\circ}{1 - \tan^2 60^\circ} \)
\( \implies \frac{2 \times \sqrt{3}}{1 - (\sqrt{3})^2} \)
\( \implies \frac{2\sqrt{3}}{1 - 3} \)
\( \implies \frac{2\sqrt{3}}{-2} \)
\( \implies -\sqrt{3} \)
Now, checking with supplementary angles: \( \tan 120^\circ = \tan (180^\circ - 60^\circ) = -\tan 60^\circ = -\sqrt{3} \). This matches.
For cotangent:
\( \cot 120^\circ = \frac{\cos 120^\circ}{\sin 120^\circ} = \frac{-1/2}{\sqrt{3}/2} \)
\( \implies -\frac{1}{\sqrt{3}} \)
Now, checking with supplementary angles: \( \cot 120^\circ = \cot (180^\circ - 60^\circ) = -\cot 60^\circ = -\frac{1}{\sqrt{3}} \). This matches.
For secant:
\( \sec 120^\circ = \frac{1}{\cos 120^\circ} = \frac{1}{-1/2} \)
\( \implies -2 \)
Now, checking with supplementary angles: \( \sec 120^\circ = \sec (180^\circ - 60^\circ) = -\sec 60^\circ = -2 \). This matches.
For cosecant:
\( \csc 120^\circ = \frac{1}{\sin 120^\circ} = \frac{1}{\sqrt{3}/2} \)
\( \implies \frac{2}{\sqrt{3}} \)
Now, checking with supplementary angles: \( \csc 120^\circ = \csc (180^\circ - 60^\circ) = \csc 60^\circ = \frac{2}{\sqrt{3}} \). This matches.
In simple words: We used double angle formulas to find the sine, cosine, and tangent of 120 degrees based on the 60-degree values. Then, we double-checked these results using the rules for supplementary angles (angles that add up to 180 degrees), confirming that all the derivations were correct.

🎯 Exam Tip: When deriving trigonometric functions for angles like 120°, always apply double angle formulas systematically and then use supplementary angle relations as a verification step to ensure accuracy.

Question 5. If 10 = b, find an expression for cos \( \theta \) in terms of a and b. hence find a relation between a and b not involving \( \theta \).
Answer:
Given \( \sin \theta = a \) and \( \sin 2\theta = b \).
We know the double angle formula \( \sin 2\theta = 2 \sin \theta \cos \theta \).
Substitute the given values into the formula:
\( b = 2 (a) \cos \theta \)
Now, we want to express \( \cos \theta \) in terms of a and b:
\( \cos \theta = \frac{b}{2a} \)
To find a relation between a and b that does not involve \( \theta \), we use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \).
Substitute the expressions for \( \sin \theta \) and \( \cos \theta \):
\( (a)^2 + (\frac{b}{2a})^2 = 1 \)
\( \implies a^2 + \frac{b^2}{4a^2} = 1 \)
To eliminate the fraction, multiply the entire equation by \( 4a^2 \):
\( 4a^2 \times a^2 + 4a^2 \times \frac{b^2}{4a^2} = 4a^2 \times 1 \)
\( \implies 4a^4 + b^2 = 4a^2 \)
This is the required relationship between a and b.
In simple words: We started with the given sine values and used the double angle formula for sine to find cosine in terms of 'a' and 'b'. Then, we used the basic sine-squared plus cosine-squared equals one rule to get rid of the angle \( \theta \) and find a link between 'a' and 'b' only.

🎯 Exam Tip: When asked to eliminate an angle from an expression, always consider using fundamental trigonometric identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) or other Pythagorean identities, as they directly relate different trigonometric functions of the same angle.

Question 6. (i) Given that tan A = \( \frac{1}{5} \), find the values of tan 2A, tan 4A and tan (45° – 4A).
(ii) If A is an obtuse angle whose sine is \( \frac{5}{13} \) and B is an acute angle whose tangent is \( \frac{3}{4} \), without using tables find the values of
(a) sin 2B, (b) tan (A – B).
Answer:
(i) Given \( \tan A = \frac{1}{5} \).
First, find \( \tan 2A \). We use the formula \( \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} \).
\( \tan 2A = \frac{2 \times \frac{1}{5}}{1 - (\frac{1}{5})^2} \)
\( \implies \frac{\frac{2}{5}}{1 - \frac{1}{25}} \)
\( \implies \frac{\frac{2}{5}}{\frac{24}{25}} \)
\( \implies \frac{2}{5} \times \frac{25}{24} \)
\( \implies \frac{5}{12} \)
Next, find \( \tan 4A \). We use \( \tan 4A = \tan (2 \times 2A) = \frac{2 \tan 2A}{1 - \tan^2 2A} \).
\( \tan 4A = \frac{2 \times \frac{5}{12}}{1 - (\frac{5}{12})^2} \)
\( \implies \frac{\frac{10}{12}}{1 - \frac{25}{144}} \)
\( \implies \frac{\frac{5}{6}}{\frac{119}{144}} \)
\( \implies \frac{5}{6} \times \frac{144}{119} \)
\( \implies \frac{5 \times 24}{119} \)
\( \implies \frac{120}{119} \)
Finally, find \( \tan (45^\circ - 4A) \). We use the formula \( \tan (X - Y) = \frac{\tan X - \tan Y}{1 + \tan X \tan Y} \).
\( \tan (45^\circ - 4A) = \frac{\tan 45^\circ - \tan 4A}{1 + \tan 45^\circ \tan 4A} \)
Since \( \tan 45^\circ = 1 \):
\( \implies \frac{1 - \frac{120}{119}}{1 + 1 \times \frac{120}{119}} \)
\( \implies \frac{\frac{119 - 120}{119}}{\frac{119 + 120}{119}} \)
\( \implies \frac{-1/119}{239/119} \)
\( \implies -\frac{1}{239} \)
(ii) Given \( \sin A = \frac{5}{13} \) and A is an obtuse angle (Quadrant II).
Given \( \tan B = \frac{3}{4} \) and B is an acute angle (Quadrant I).
For angle B (acute, Q1):
We find \( \sec B = \sqrt{1 + \tan^2 B} \) (positive because B is acute).
\( \implies \sqrt{1 + (\frac{3}{4})^2} \)
\( \implies \sqrt{1 + \frac{9}{16}} \)
\( \implies \sqrt{\frac{25}{16}} \)
\( \implies \frac{5}{4} \)
So, \( \cos B = \frac{1}{\sec B} = \frac{4}{5} \).
And \( \sin B = \tan B \cos B = \frac{3}{4} \times \frac{4}{5} = \frac{3}{5} \).
(a) Now, find \( \sin 2B \). We use \( \sin 2B = 2 \sin B \cos B \).
\( \implies 2 \times \frac{3}{5} \times \frac{4}{5} \)
\( \implies \frac{24}{25} \)
For angle A (obtuse, Q2):
Since A is obtuse, \( \cos A \) is negative and \( \tan A \) is negative.
We find \( \cos A = -\sqrt{1 - \sin^2 A} \)
\( \implies -\sqrt{1 - (\frac{5}{13})^2} \)
\( \implies -\sqrt{1 - \frac{25}{169}} \)
\( \implies -\sqrt{\frac{144}{169}} \)
\( \implies -\frac{12}{13} \)
Then, \( \tan A = \frac{\sin A}{\cos A} = \frac{5/13}{-12/13} = -\frac{5}{12} \).
(b) Now, find \( \tan (A - B) \). We use the formula \( \tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \).
\( \implies \frac{-\frac{5}{12} - \frac{3}{4}}{1 + (-\frac{5}{12}) \times (\frac{3}{4})} \)
\( \implies \frac{\frac{-5 - 9}{12}}{1 - \frac{15}{48}} \)
\( \implies \frac{\frac{-14}{12}}{\frac{48 - 15}{48}} \)
\( \implies \frac{-\frac{7}{6}}{\frac{33}{48}} \)
\( \implies -\frac{7}{6} \times \frac{48}{33} \)
\( \implies -\frac{7 \times 8}{33} \)
\( \implies -\frac{56}{33} \)
In simple words: For part (i), we repeatedly used the double angle formula for tangent to find tan 2A and tan 4A, then applied the tangent subtraction formula. For part (ii), we first found all the sine, cosine, and tangent values for angles A and B, remembering to consider their quadrants for correct signs. Then, we applied the required double angle and subtraction formulas.

🎯 Exam Tip: Always remember to check the quadrant of the angle to determine the correct sign for trigonometric ratios, especially when calculating missing values (like \( \cos A \) from \( \sin A \)) or applying formulas like tangent subtraction.

Question 7. Express
(i) cos 6\( \alpha \) in terms of functions of 3\( \alpha \) ;
(ii) sin 10\( \theta \) in terms of functions of 5\( \theta \) ;
(iii) tan 8\( \alpha \) in terms of functions of 4\( \alpha \) ;
(iv) cos² 2\( \theta \) in terms of cos 4\( \theta \) ;
(v) tan² 4\( \Phi \) in terms of cos 8\( \Phi \);
(vi) sin² \( \frac{5\theta}{2} \) in terms of cos 5\( \theta \);
(vii) cos 20\( \theta \) in terms of sin 5\( \theta \).
Answer:
(i) To express \( \cos 6\alpha \) in terms of functions of \( 3\alpha \):
We use the double angle identity \( \cos 2X = 2 \cos^2 X - 1 \). Let \( X = 3\alpha \).
\( \cos 6\alpha = \cos (2 \times 3\alpha) \)
\( \implies 2 \cos^2 3\alpha - 1 \)
(ii) To express \( \sin 10\theta \) in terms of functions of \( 5\theta \):
We use the double angle identity \( \sin 2X = 2 \sin X \cos X \). Let \( X = 5\theta \).
\( \sin 10\theta = \sin (2 \times 5\theta) \)
\( \implies 2 \sin 5\theta \cos 5\theta \)
(iii) To express \( \tan 8\alpha \) in terms of functions of \( 4\alpha \):
We use the double angle identity \( \tan 2X = \frac{2 \tan X}{1 - \tan^2 X} \). Let \( X = 4\alpha \).
\( \tan 8\alpha = \tan (2 \times 4\alpha) \)
\( \implies \frac{2 \tan 4\alpha}{1 - \tan^2 4\alpha} \)
(iv) To express \( \cos^2 2\theta \) in terms of \( \cos 4\theta \):
We use the half-angle power reduction identity \( \cos^2 X = \frac{1 + \cos 2X}{2} \). Let \( X = 2\theta \).
\( \cos^2 2\theta = \frac{1 + \cos (2 \times 2\theta)}{2} \)
\( \implies \frac{1 + \cos 4\theta}{2} \)
This also means \( \cos 2\theta = \pm \sqrt{\frac{1 + \cos 4\theta}{2}} \).
(v) To express \( \tan^2 4\Phi \) in terms of \( \cos 8\Phi \):
We know \( \tan^2 X = \frac{\sin^2 X}{\cos^2 X} \).
Using \( \sin^2 X = \frac{1 - \cos 2X}{2} \) and \( \cos^2 X = \frac{1 + \cos 2X}{2} \). Let \( X = 4\Phi \).
\( \tan^2 4\Phi = \frac{\frac{1 - \cos (2 \times 4\Phi)}{2}}{\frac{1 + \cos (2 \times 4\Phi)}{2}} \)
\( \implies \frac{1 - \cos 8\Phi}{1 + \cos 8\Phi} \)
(vi) To express \( \sin^2 \frac{5\theta}{2} \) in terms of \( \cos 5\theta \):
We use the half-angle power reduction identity \( \sin^2 X = \frac{1 - \cos 2X}{2} \). Let \( X = \frac{5\theta}{2} \).
\( \sin^2 \frac{5\theta}{2} = \frac{1 - \cos (2 \times \frac{5\theta}{2})}{2} \)
\( \implies \frac{1 - \cos 5\theta}{2} \)
(vii) To express \( \cos 20\theta \) in terms of \( \sin 5\theta \):
We use the identity \( \cos 4X = 1 - 8 \sin^2 X (1 - \sin^2 X) \). Let \( X = 5\theta \).
\( \cos 20\theta = \cos (4 \times 5\theta) \)
\( \implies 1 - 8 \sin^2 5\theta (1 - \sin^2 5\theta) \)
\( \implies 1 - 8 \sin^2 5\theta + 8 \sin^4 5\theta \)
In simple words: For each part, we used specific trigonometric identities that relate an angle to its multiple or half-angle. This helps us rewrite complex trigonometric expressions in simpler forms based on a given base angle. Understanding these identities is key to solving many trigonometry problems.

🎯 Exam Tip: Mastering the various forms of double and half-angle identities (e.g., for \( \sin 2A \), \( \cos 2A \), \( \tan 2A \), \( \sin^2 A \), \( \cos^2 A \)) is crucial. Practice recognizing which form to apply to simplify expressions and achieve the desired representation.

Question 8. Using formulas, find the exact value of (i) sin 15° (ii) sin 292 \( \frac{1^\circ}{2} \).
Answer:
(i) To find \( \sin 15^\circ \):
We use the half-angle formula \( \sin^2 X = \frac{1 - \cos 2X}{2} \). Let \( X = 15^\circ \).
\( \sin^2 15^\circ = \frac{1 - \cos (2 \times 15^\circ)}{2} \)
\( \implies \frac{1 - \cos 30^\circ}{2} \)
\( \implies \frac{1 - \frac{\sqrt{3}}{2}}{2} \)
\( \implies \frac{\frac{2 - \sqrt{3}}{2}}{2} \)
\( \implies \frac{2 - \sqrt{3}}{4} \)
Since \( 15^\circ \) is in Quadrant I, \( \sin 15^\circ \) is positive.
\( \sin 15^\circ = \sqrt{\frac{2 - \sqrt{3}}{4}} \)
\( \implies \frac{\sqrt{2 - \sqrt{3}}}{2} \)
(ii) To find \( \sin 292 \frac{1^\circ}{2} \):
Let \( X = 292 \frac{1^\circ}{2} \). We use the half-angle formula \( \sin^2 X = \frac{1 - \cos 2X}{2} \).
\( \sin^2 (292 \frac{1^\circ}{2}) = \frac{1 - \cos (2 \times 292 \frac{1^\circ}{2})}{2} \)
\( \implies \frac{1 - \cos 585^\circ}{2} \)
Now, we simplify \( \cos 585^\circ \).
\( \cos 585^\circ = \cos (360^\circ + 225^\circ) = \cos 225^\circ \)
\( \implies \cos (180^\circ + 45^\circ) = -\cos 45^\circ \)
\( \implies -\frac{1}{\sqrt{2}} \)
Substitute this back into the formula:
\( \sin^2 (292 \frac{1^\circ}{2}) = \frac{1 - (-\frac{1}{\sqrt{2}})}{2} \)
\( \implies \frac{1 + \frac{1}{\sqrt{2}}}{2} \)
\( \implies \frac{\frac{\sqrt{2} + 1}{\sqrt{2}}}{2} \)
\( \implies \frac{\sqrt{2} + 1}{2\sqrt{2}} \)
To rationalize the denominator, multiply by \( \frac{\sqrt{2}}{\sqrt{2}} \):
\( \implies \frac{(\sqrt{2} + 1)\sqrt{2}}{2\sqrt{2}\sqrt{2}} \)
\( \implies \frac{2 + \sqrt{2}}{4} \)
Now, we need to find the sign of \( \sin 292 \frac{1^\circ}{2} \).
The angle \( 292 \frac{1^\circ}{2} \) lies in the 4th Quadrant (between 270° and 360°).
In the 4th Quadrant, sine is negative.
So, \( \sin (292 \frac{1^\circ}{2}) = -\sqrt{\frac{2 + \sqrt{2}}{4}} \)
\( \implies -\frac{\sqrt{2 + \sqrt{2}}}{2} \)
In simple words: We used the half-angle sine formula for both parts. For sin 15°, we calculated the value directly as it's in the first quadrant. For sin 292.5°, we first found the value of cos 585° by breaking it down into known angles and then applied the formula. We also had to consider that 292.5° is in the fourth quadrant, where the sine value is negative.

🎯 Exam Tip: When using half-angle formulas, always determine the quadrant of the original angle to correctly assign the positive or negative sign to the final square root. Also, simplify large angles by using periodicity and quadrant rules (e.g., \( \cos(360^\circ + \theta) \), \( \cos(180^\circ + \theta) \)).

Question 9. In the triangle ABC, in which C is the right angle, prove that: sin 2A = \( \frac { 2ab }{ c² } \), cos 2A = \( \frac { b² – a² }{ c² } \), sin \( \frac { A }{ 2 } = \sqrt{\frac{c-b}{2 c}} \), cos \( \frac { A }{ 2 } = \sqrt{\frac{c+b}{2 c}} \).

Answer: Let's consider a right-angled triangle ACB, where angle C is 90 degrees. From this triangle, we know the following: \( \sin A = \frac { BC }{ AB } = \frac { a }{ c } \) \( \cos A = \frac { AC }{ AB } = \frac { b }{ c } \) Now we will prove each part:
\( \sin 2A = 2 \sin A \cos A \)
\( \implies = 2 \left( \frac { a }{ c } \right) \left( \frac { b }{ c } \right) = \frac { 2ab }{ c² } \) (This matches the first part to prove.)
\( \cos 2A = \cos² A - \sin² A \)
\( \implies = \left( \frac { b }{ c } \right)² - \left( \frac { a }{ c } \right)² = \frac { b² - a² }{ c² } \) (This matches the second part to prove.)
We also know that \( \sin² \theta = \frac { 1 - \cos 2\theta }{ 2 } \).
So, \( \sin² \frac { A }{ 2 } = \frac { 1 - \cos A }{ 2 } \)
\( \implies = \frac { 1 - \frac { b }{ c } }{ 2 } = \frac { \frac { c - b }{ c } }{ 2 } = \frac { c - b }{ 2c } \)
Therefore, \( \sin \frac { A }{ 2 } = \sqrt{\frac{c-b}{2 c}} \) (This matches the third part to prove.)
Similarly, we know that \( \cos² \theta = \frac { 1 + \cos 2\theta }{ 2 } \).
So, \( \cos² \frac { A }{ 2 } = \frac { 1 + \cos A }{ 2 } \)
\( \implies = \frac { 1 + \frac { b }{ c } }{ 2 } = \frac { \frac { c + b }{ c } }{ 2 } = \frac { c + b }{ 2c } \)
Therefore, \( \cos \frac { A }{ 2 } = \sqrt{\frac{c+b}{2 c}} \) (This matches the fourth part to prove.) This shows that all the given trigonometric identities hold true for a right-angled triangle.In simple words: We used basic definitions of sine and cosine in a right-angled triangle. Then, we applied the double angle formulas and half-angle formulas to show that the given expressions are correct.

🎯 Exam Tip: Remember the basic trigonometric ratios for a right triangle and the double/half-angle formulas. When proving identities, start with one side and transform it until it matches the other side, using known formulas.

Question 10. If \( \cos \alpha = \frac { 3 }{ 5 } \), \( \cos \beta = \frac { 4 }{ 5 } \), find the value of \( \cos \frac { \alpha-\beta }{ 2 } \), assuming \( \alpha \) and \( \beta \) to be acute angles.
Answer: Given that \( \cos \alpha = \frac { 3 }{ 5 } \) and \( \cos \beta = \frac { 4 }{ 5 } \). Since \( \alpha \) and \( \beta \) are acute angles (meaning they are in the first quadrant), their sine values will also be positive.
\( \sin \alpha = \sqrt{1 - \cos² \alpha} = \sqrt{1 - \left(\frac{3}{5}\right)²} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \)
\( \sin \beta = \sqrt{1 - \cos² \beta} = \sqrt{1 - \left(\frac{4}{5}\right)²} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \)
We use the half-angle formula for cosine, \( \cos² \theta = \frac { 1 + \cos 2\theta }{ 2 } \).
So, \( \cos² \left( \frac { \alpha - \beta }{ 2 } \right) = \frac { 1 + \cos (\alpha - \beta) }{ 2 } \)
We know the formula for \( \cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \).
\( \implies \cos (\alpha - \beta) = \left( \frac { 3 }{ 5 } \right) \left( \frac { 4 }{ 5 } \right) + \left( \frac { 4 }{ 5 } \right) \left( \frac { 3 }{ 5 } \right) \)
\( \implies = \frac { 12 }{ 25 } + \frac { 12 }{ 25 } = \frac { 24 }{ 25 } \)
Now substitute this back into the half-angle formula:
\( \cos² \left( \frac { \alpha - \beta }{ 2 } \right) = \frac { 1 + \frac { 24 }{ 25 } }{ 2 } = \frac { \frac { 25 + 24 }{ 25 } }{ 2 } = \frac { \frac { 49 }{ 25 } }{ 2 } = \frac { 49 }{ 50 } \)
Since \( \alpha \) and \( \beta \) are acute, \( 0 < \alpha < \frac{\pi}{2} \) and \( 0 < \beta < \frac{\pi}{2} \).
This means \( -\frac{\pi}{2} < \alpha - \beta < \frac{\pi}{2} \).
So, \( -\frac{\pi}{4} < \frac { \alpha - \beta }{ 2 } < \frac{\pi}{4} \). This range is in the first or fourth quadrant where cosine is positive.
\( \cos \left( \frac { \alpha - \beta }{ 2 } \right) = \sqrt{\frac{49}{50}} = \frac{\sqrt{49}}{\sqrt{50}} = \frac{7}{5\sqrt{2}} \)
To rationalize the denominator, multiply by \( \frac{\sqrt{2}}{\sqrt{2}} \):
\( \cos \left( \frac { \alpha - \beta }{ 2 } \right) = \frac{7\sqrt{2}}{5 \times 2} = \frac{7\sqrt{2}}{10} \)In simple words: First, we found the sine values using the given cosine values. Then, we used the angle subtraction formula for cosine and plugged those numbers in. Finally, we used a half-angle formula for cosine to get the final answer.

🎯 Exam Tip: Always determine the quadrant of angles and half-angles to correctly choose the sign of the square root when applying half-angle formulas. Remember the fundamental trigonometric identities like \( \sin² \theta + \cos² \theta = 1 \).

Question 11. Given that \( \cos \frac { A }{ 2 } = \frac { 12 }{ 13 } \), calculate without the use of tables, the values of \( \sin A \), \( \cos A \) and \( \tan A \).
Answer: We are given \( \cos \frac { A }{ 2 } = \frac { 12 }{ 13 } \).
To find \( \sin \frac { A }{ 2 } \), we use the identity \( \sin² \theta + \cos² \theta = 1 \).
\( \sin² \frac { A }{ 2 } = 1 - \cos² \frac { A }{ 2 } = 1 - \left( \frac { 12 }{ 13 } \right)² = 1 - \frac { 144 }{ 169 } = \frac { 169 - 144 }{ 169 } = \frac { 25 }{ 169 } \)
So, \( \sin \frac { A }{ 2 } = \pm \sqrt{\frac{25}{169}} = \pm \frac{5}{13} \). We usually take the positive root unless specified otherwise, assuming A/2 is in the first or second quadrant where sine is positive. Let's assume \( \frac{A}{2} \) is an acute angle so \( \sin \frac { A }{ 2 } = \frac{5}{13} \).
Now we find \( \sin A \) using the double angle formula \( \sin A = 2 \sin \frac { A }{ 2 } \cos \frac { A }{ 2 } \).
\( \sin A = 2 \times \frac { 5 }{ 13 } \times \frac { 12 }{ 13 } = \frac { 120 }{ 169 } \)
Next, we find \( \cos A \) using the double angle formula \( \cos A = 2 \cos² \frac { A }{ 2 } - 1 \).
\( \cos A = 2 \left( \frac { 12 }{ 13 } \right)² - 1 = 2 \times \frac { 144 }{ 169 } - 1 = \frac { 288 }{ 169 } - \frac { 169 }{ 169 } = \frac { 288 - 169 }{ 169 } = \frac { 119 }{ 169 } \)
Finally, we find \( \tan A \) using the identity \( \tan A = \frac { \sin A }{ \cos A } \).
\( \tan A = \frac { \frac { 120 }{ 169 } }{ \frac { 119 }{ 169 } } = \frac { 120 }{ 119 } \)
Thus, \( \sin A = \frac{120}{169} \), \( \cos A = \frac{119}{169} \), and \( \tan A = \frac{120}{119} \).In simple words: We used the given half-angle cosine to find the half-angle sine. Then, we applied the double angle formulas for sine and cosine to get the full angle values. Lastly, we divided sine A by cosine A to find tan A.

🎯 Exam Tip: When given a half-angle trigonometric ratio, use identities like \( \sin² \theta + \cos² \theta = 1 \) and double angle formulas (e.g., \( \sin 2\theta = 2 \sin \theta \cos \theta \), \( \cos 2\theta = 2 \cos² \theta - 1 \) or \( 1 - 2 \sin² \theta \)) to find the full angle ratios. Pay attention to the sign of the square root based on the angle's quadrant.

Question 12. Given that \( \tan x = \frac { 12 }{ 5 } \), \( \cos y = \frac { -3 }{ 5 } \), and the angles \( x \) and \( y \) are in the same quadrants, calculate without the use of tables the value of
(i) \( \sin(x + y) \).
(ii) \( \cos \frac { y }{ 2 } \).
Answer: We are given \( \tan x = \frac { 12 }{ 5 } \) and \( \cos y = \frac { -3 }{ 5 } \). The angles \( x \) and \( y \) are in the same quadrant.
Since \( \tan x \) is positive and \( \cos y \) is negative, both angles \( x \) and \( y \) must lie in the third quadrant.
In the third quadrant, both \( \sin \) and \( \cos \) are negative.
For angle \( x \):
We have \( \tan x = \frac { 12 }{ 5 } \). We can find \( \sec x \) using \( \sec² x = 1 + \tan² x \).
\( \sec² x = 1 + \left( \frac { 12 }{ 5 } \right)² = 1 + \frac { 144 }{ 25 } = \frac { 25 + 144 }{ 25 } = \frac { 169 }{ 25 } \)
\( \sec x = \pm \sqrt{\frac{169}{25}} = \pm \frac{13}{5} \). Since \( x \) is in the third quadrant, \( \sec x \) is negative.
So, \( \sec x = -\frac{13}{5} \).
Then \( \cos x = \frac { 1 }{ \sec x } = \frac { 1 }{ -\frac { 13 }{ 5 } } = -\frac{5}{13} \).
Now, \( \sin x = \tan x \cos x = \left( \frac { 12 }{ 5 } \right) \left( -\frac { 5 }{ 13 } \right) = -\frac{12}{13} \). (Alternatively, \( \sin x = -\sqrt{1-\cos²x} \)).
For angle \( y \):
We have \( \cos y = -\frac{3}{5} \). Since \( y \) is in the third quadrant, \( \sin y \) is negative.
\( \sin y = -\sqrt{1 - \cos² y} = -\sqrt{1 - \left( -\frac { 3 }{ 5 } \right)²} = -\sqrt{1 - \frac { 9 }{ 25 } } = -\sqrt{\frac{16}{25}} = -\frac{4}{5} \).
(i) Now we calculate \( \sin(x + y) \) using the sum formula:
\( \sin(x + y) = \sin x \cos y + \cos x \sin y \)
\( \implies = \left( -\frac { 12 }{ 13 } \right) \left( -\frac { 3 }{ 5 } \right) + \left( -\frac { 5 }{ 13 } \right) \left( -\frac { 4 }{ 5 } \right) \)
\( \implies = \frac { 36 }{ 65 } + \frac { 20 }{ 65 } = \frac { 56 }{ 65 } \)
(ii) To calculate \( \cos \frac { y }{ 2 } \), we use the half-angle formula \( \cos² \frac { y }{ 2 } = \frac { 1 + \cos y }{ 2 } \).
\( \cos² \frac { y }{ 2 } = \frac { 1 + \left( -\frac { 3 }{ 5 } \right) }{ 2 } = \frac { \frac { 5 - 3 }{ 5 } }{ 2 } = \frac { \frac { 2 }{ 5 } }{ 2 } = \frac { 2 }{ 10 } = \frac { 1 }{ 5 } \)
Since \( y \) is in the third quadrant, \( \pi < y < \frac{3\pi}{2} \).
Dividing by 2, we get \( \frac{\pi}{2} < \frac { y }{ 2 } < \frac{3\pi}{4} \). This means \( \frac { y }{ 2 } \) is in the second quadrant.
In the second quadrant, \( \cos \) is negative.
So, \( \cos \frac { y }{ 2 } = -\sqrt{\frac{1}{5}} = -\frac{1}{\sqrt{5}} \)
Rationalizing the denominator: \( \cos \frac { y }{ 2 } = -\frac{\sqrt{5}}{5} \)In simple words: First, we figured out that both angles are in the third quarter of the circle. Then, we found the sine and cosine of both angles. For part (i), we added the angles' sines and cosines together using a special formula. For part (ii), we used a half-angle formula for cosine, remembering that cosine is negative in the second quarter where half of y lies.

🎯 Exam Tip: When given that angles are in the same quadrant, always identify that specific quadrant first to determine the correct signs of sine, cosine, and tangent. Remember to check the quadrant of the half-angle as well when applying half-angle formulas.

Question 13. Given \( \tan \sin² \beta = \sin \alpha \cos \alpha \), show that \( \cos 2\beta = 2 \cos² \left[ \frac { \pi }{ 4 } + \alpha \right] \).
Answer: We are given \( \sin² \beta = \sin \alpha \cos \alpha \).
We know the identity \( \sin² \theta = \frac { 1 - \cos 2\theta }{ 2 } \).
So, \( \sin² \beta = \frac { 1 - \cos 2\beta }{ 2 } \).
Substituting this into the given equation:
\( \frac { 1 - \cos 2\beta }{ 2 } = \sin \alpha \cos \alpha \)
From this, we get \( 1 - \cos 2\beta = 2 \sin \alpha \cos \alpha \).
We know that \( 2 \sin \alpha \cos \alpha = \sin 2\alpha \).
So, \( 1 - \cos 2\beta = \sin 2\alpha \)
\( \implies \cos 2\beta = 1 - \sin 2\alpha \) ...(1)
Now let's look at the right-hand side of the expression we need to prove: \( 2 \cos² \left[ \frac { \pi }{ 4 } + \alpha \right] \).
We also know that \( \cos² \theta = \frac { 1 + \cos 2\theta }{ 2 } \).
So, \( 2 \cos² \left[ \frac { \pi }{ 4 } + \alpha \right] = 2 \left[ \frac { 1 + \cos \left( 2 \left( \frac { \pi }{ 4 } + \alpha \right) \right) }{ 2 } \right] \)
\( \implies = 1 + \cos \left( \frac { \pi }{ 2 } + 2\alpha \right) \)
We know that \( \cos \left( \frac { \pi }{ 2 } + \theta \right) = -\sin \theta \).
So, \( 1 + \cos \left( \frac { \pi }{ 2 } + 2\alpha \right) = 1 - \sin 2\alpha \) ...(2)
Comparing equation (1) and equation (2), we see that both sides are equal.
Therefore, \( \cos 2\beta = 2 \cos² \left[ \frac { \pi }{ 4 } + \alpha \right] \) is proven.In simple words: We started by changing the given equation using a formula for sine squared. This helped us get an expression for cos 2 beta. Then, we worked with the expression we needed to prove, using another formula for cosine squared. We showed that both expressions simplify to the same thing, which proves they are equal.

🎯 Exam Tip: This problem requires careful application of several trigonometric identities: \( \sin² \theta = \frac { 1 - \cos 2\theta }{ 2 } \), \( \sin 2\theta = 2 \sin \theta \cos \theta \), \( \cos² \theta = \frac { 1 + \cos 2\theta }{ 2 } \), and angle sum/subtraction formulas like \( \cos (\frac{\pi}{2} + \theta) = -\sin \theta \). Break down the problem and solve each part systematically.

Question 14. Derive formulas for the following in terms of functions of \( 2\theta \) and then of \( \theta \).
(i) \( \sin 4\theta \)
(ii) \( \cos 4\theta \)
(iii) \( \tan 4\theta \)
Answer: Let's derive the formulas as requested.
(i) For \( \sin 4\theta \):
In terms of \( 2\theta \): We can write \( \sin 4\theta = \sin (2 \times 2\theta) \). Using the double angle formula \( \sin 2x = 2 \sin x \cos x \), where \( x = 2\theta \):
\( \sin 4\theta = 2 \sin 2\theta \cos 2\theta \)
In terms of \( \theta \): Now we substitute \( \sin 2\theta = 2 \sin \theta \cos \theta \) and \( \cos 2\theta = \cos² \theta - \sin² \theta \):
\( \sin 4\theta = 2 (2 \sin \theta \cos \theta) (\cos² \theta - \sin² \theta) \)
\( \implies = 4 \sin \theta \cos \theta (\cos² \theta - \sin² \theta) \)
(ii) For \( \cos 4\theta \):
In terms of \( 2\theta \): We can write \( \cos 4\theta = \cos (2 \times 2\theta) \). Using the double angle formula \( \cos 2x = 2 \cos² x - 1 \), where \( x = 2\theta \):
\( \cos 4\theta = 2 \cos² 2\theta - 1 \)
In terms of \( \theta \): Now we substitute \( \cos 2\theta = 2 \cos² \theta - 1 \):
\( \cos 4\theta = 2 (2 \cos² \theta - 1)² - 1 \)
\( \implies = 2 (4 \cos^4 \theta - 4 \cos² \theta + 1) - 1 \)
\( \implies = 8 \cos^4 \theta - 8 \cos² \theta + 2 - 1 \)
\( \implies = 8 \cos^4 \theta - 8 \cos² \theta + 1 \)
(iii) For \( \tan 4\theta \):
In terms of \( 2\theta \): We can write \( \tan 4\theta = \tan (2 \times 2\theta) \). Using the double angle formula \( \tan 2x = \frac { 2 \tan x }{ 1 - \tan² x } \), where \( x = 2\theta \):
\( \tan 4\theta = \frac { 2 \tan 2\theta }{ 1 - \tan² 2\theta } \)
In terms of \( \theta \): Now we substitute \( \tan 2\theta = \frac { 2 \tan \theta }{ 1 - \tan² \theta } \):
\( \tan 4\theta = \frac { 2 \left( \frac { 2 \tan \theta }{ 1 - \tan² \theta } \right) }{ 1 - \left( \frac { 2 \tan \theta }{ 1 - \tan² \theta } \right)² } \)
\( \implies = \frac { \frac { 4 \tan \theta }{ 1 - \tan² \theta } }{ 1 - \frac { 4 \tan² \theta }{ (1 - \tan² \theta)² } } \)
\( \implies = \frac { \frac { 4 \tan \theta }{ 1 - \tan² \theta } }{ \frac { (1 - \tan² \theta)² - 4 \tan² \theta }{ (1 - \tan² \theta)² } } \)
\( \implies = \frac { 4 \tan \theta (1 - \tan² \theta) }{ (1 - \tan² \theta)² - 4 \tan² \theta } \)
\( \implies = \frac { 4 \tan \theta (1 - \tan² \theta) }{ 1 - 2 \tan² \theta + \tan^4 \theta - 4 \tan² \theta } \)
\( \implies = \frac { 4 \tan \theta (1 - \tan² \theta) }{ \tan^4 \theta - 6 \tan² \theta + 1 } \)In simple words: We used the double angle formulas repeatedly. First, we wrote \(4\theta\) as \(2 \times 2\theta\) to find expressions in terms of \(2\theta\). Then, we replaced the \(2\theta\) terms with their formulas in terms of \( \theta \) to get the final expressions.

🎯 Exam Tip: Mastering double angle formulas is crucial for these types of derivations. Remember that \( \cos 2\theta \) has three forms (using \( \cos² \theta \), \( \sin² \theta \), or both), and choose the most suitable one for the derivation. For tangent, be careful with algebraic simplification.

Question 15. If \( \sin \alpha = \frac { 3 }{ 5 } \), find the value of
(i) \( \sin 3\alpha \).
(ii) \( \cos 3\alpha \).
(iii) \( \tan 3\alpha \).
Answer: We are given \( \sin \alpha = \frac { 3 }{ 5 } \). We need to find \( \cos \alpha \) and \( \tan \alpha \) first.
Using \( \cos² \alpha = 1 - \sin² \alpha \):
\( \cos² \alpha = 1 - \left( \frac { 3 }{ 5 } \right)² = 1 - \frac { 9 }{ 25 } = \frac { 16 }{ 25 } \)
\( \cos \alpha = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5} \).
This problem implies we need to consider different quadrants for \( \alpha \). Let's evaluate for \( \alpha \) in the first quadrant first (where all ratios are positive) and then discuss other cases.
Case 1: \( \alpha \) is in the first quadrant (so \( \cos \alpha = \frac{4}{5} \) and \( \tan \alpha = \frac{3/5}{4/5} = \frac{3}{4} \)).
(i) Find \( \sin 3\alpha \):
Using the triple angle formula \( \sin 3\alpha = 3 \sin \alpha - 4 \sin³ \alpha \):
\( \sin 3\alpha = 3 \left( \frac { 3 }{ 5 } \right) - 4 \left( \frac { 3 }{ 5 } \right)³ \)
\( \implies = \frac { 9 }{ 5 } - 4 \left( \frac { 27 }{ 125 } \right) \)
\( \implies = \frac { 9 }{ 5 } - \frac { 108 }{ 125 } \)
\( \implies = \frac { 9 \times 25 }{ 125 } - \frac { 108 }{ 125 } = \frac { 225 - 108 }{ 125 } = \frac { 117 }{ 125 } \)
(ii) Find \( \cos 3\alpha \):
Using the triple angle formula \( \cos 3\alpha = 4 \cos³ \alpha - 3 \cos \alpha \):
For \( \alpha \) in the first quadrant, \( \cos \alpha = \frac{4}{5} \):
\( \cos 3\alpha = 4 \left( \frac { 4 }{ 5 } \right)³ - 3 \left( \frac { 4 }{ 5 } \right) \)
\( \implies = 4 \left( \frac { 64 }{ 125 } \right) - \frac { 12 }{ 5 } \)
\( \implies = \frac { 256 }{ 125 } - \frac { 12 \times 25 }{ 125 } = \frac { 256 - 300 }{ 125 } = -\frac { 44 }{ 125 } \)
Case 2: If \( \alpha \) is in the second quadrant, then \( \sin \alpha = \frac{3}{5} \) (positive), but \( \cos \alpha = -\frac{4}{5} \) (negative).
\( \cos 3\alpha = 4 \left( -\frac { 4 }{ 5 } \right)³ - 3 \left( -\frac { 4 }{ 5 } \right) \)
\( \implies = 4 \left( -\frac { 64 }{ 125 } \right) + \frac { 12 }{ 5 } \)
\( \implies = -\frac { 256 }{ 125 } + \frac { 300 }{ 125 } = \frac { 44 }{ 125 } \)
(iii) Find \( \tan 3\alpha \):
Using the triple angle formula \( \tan 3\alpha = \frac { 3 \tan \alpha - \tan³ \alpha }{ 1 - 3 \tan² \alpha } \):
Case 1: For \( \alpha \) in the first quadrant, \( \tan \alpha = \frac{3}{4} \):
\( \tan 3\alpha = \frac { 3 \left( \frac { 3 }{ 4 } \right) - \left( \frac { 3 }{ 4 } \right)³ }{ 1 - 3 \left( \frac { 3 }{ 4 } \right)² } \)
\( \implies = \frac { \frac { 9 }{ 4 } - \frac { 27 }{ 64 } }{ 1 - 3 \left( \frac { 9 }{ 16 } \right) } = \frac { \frac { 9 \times 16 - 27 }{ 64 } }{ 1 - \frac { 27 }{ 16 } } \)
\( \implies = \frac { \frac { 144 - 27 }{ 64 } }{ \frac { 16 - 27 }{ 16 } } = \frac { \frac { 117 }{ 64 } }{ -\frac { 11 }{ 16 } } \)
\( \implies = \frac { 117 }{ 64 } \times \left( -\frac { 16 }{ 11 } \right) = -\frac { 117 }{ 4 \times 11 } = -\frac { 117 }{ 44 } \)
Case 2: If \( \alpha \) is in the second quadrant, then \( \tan \alpha = -\frac{3}{4} \):
\( \tan 3\alpha = \frac { 3 \left( -\frac { 3 }{ 4 } \right) - \left( -\frac { 3 }{ 4 } \right)³ }{ 1 - 3 \left( -\frac { 3 }{ 4 } \right)² } \)
\( \implies = \frac { -\frac { 9 }{ 4 } - \left( -\frac { 27 }{ 64 } \right) }{ 1 - 3 \left( \frac { 9 }{ 16 } \right) } = \frac { -\frac { 9 }{ 4 } + \frac { 27 }{ 64 } }{ 1 - \frac { 27 }{ 16 } } \)
\( \implies = \frac { \frac { -9 \times 16 + 27 }{ 64 } }{ \frac { 16 - 27 }{ 16 } } = \frac { \frac { -144 + 27 }{ 64 } }{ -\frac { 11 }{ 16 } } \)
\( \implies = \frac { -117 }{ 64 } \times \left( -\frac { 16 }{ 11 } \right) = \frac { 117 }{ 4 \times 11 } = \frac { 117 }{ 44 } \)
So, depending on the quadrant of \( \alpha \), the values change for cosine and tangent.
If \( \alpha \) is in the first quadrant: \( \sin 3\alpha = \frac{117}{125} \), \( \cos 3\alpha = -\frac{44}{125} \), \( \tan 3\alpha = -\frac{117}{44} \).
If \( \alpha \) is in the second quadrant: \( \sin 3\alpha = \frac{117}{125} \), \( \cos 3\alpha = \frac{44}{125} \), \( \tan 3\alpha = \frac{117}{44} \).In simple words: First, we found the cosine and tangent of \( \alpha \). Then, we used the special triple angle formulas for sine, cosine, and tangent to calculate their values. We showed how the signs change depending on which quarter of the circle \( \alpha \) is in.

🎯 Exam Tip: When using triple angle formulas, always determine the quadrant of the original angle \( \alpha \) to correctly establish the signs of \( \cos \alpha \) and \( \tan \alpha \). The values of \( \sin 3\alpha \), \( \cos 3\alpha \), and \( \tan 3\alpha \) can change sign based on the quadrant of \( \alpha \). Double-check your algebraic calculations, especially with fractions and powers.

Question 16. If \( 2 \cos \theta = x + \frac { 1 }{ x } \), prove that \( 2 \cos 3\theta = x³ + \frac { 1 }{ x³ } \).
Answer: We are given the equation \( 2 \cos \theta = x + \frac { 1 }{ x } \) ...(1)
We need to prove \( 2 \cos 3\theta = x³ + \frac { 1 }{ x³ } \).
Let's consider the right-hand side of the equation we need to prove: \( x³ + \frac { 1 }{ x³ } \).
We know the algebraic identity \( a³ + b³ = (a + b)(a² - ab + b²) \) or \( (a+b)³ - 3ab(a+b) \).
Using the second form with \( a=x \) and \( b=\frac{1}{x} \):
\( x³ + \frac { 1 }{ x³ } = \left( x + \frac { 1 }{ x } \right)³ - 3 \left( x \right) \left( \frac { 1 }{ x } \right) \left( x + \frac { 1 }{ x } \right) \)
\( \implies = \left( x + \frac { 1 }{ x } \right)³ - 3 \left( x + \frac { 1 }{ x } \right) \)
Now, substitute the given value from equation (1), \( x + \frac { 1 }{ x } = 2 \cos \theta \), into this expression:
\( x³ + \frac { 1 }{ x³ } = (2 \cos \theta)³ - 3 (2 \cos \theta) \)
\( \implies = 8 \cos³ \theta - 6 \cos \theta \)
Now, let's consider the triple angle formula for cosine, which is \( \cos 3\theta = 4 \cos³ \theta - 3 \cos \theta \).
Multiply this formula by 2:
\( 2 \cos 3\theta = 2 (4 \cos³ \theta - 3 \cos \theta) \)
\( \implies = 8 \cos³ \theta - 6 \cos \theta \)
We can see that both the expression for \( x³ + \frac { 1 }{ x³ } \) and \( 2 \cos 3\theta \) simplify to the same algebraic form \( 8 \cos³ \theta - 6 \cos \theta \).
Therefore, we have proven that \( 2 \cos 3\theta = x³ + \frac { 1 }{ x³ } \).In simple words: We took the first given equation and used it in an algebraic formula for cubes. This transformed the cube expression into a form involving cosine. Then, we compared it with the triple angle formula for cosine and found that they are exactly the same, which proved the statement.

🎯 Exam Tip: This question combines algebraic identities with trigonometric formulas. Recognize that \( x³ + \frac{1}{x³} \) is related to \( x + \frac{1}{x} \) by an algebraic identity. Also, know the triple angle formula for cosine: \( \cos 3\theta = 4 \cos³ \theta - 3 \cos \theta \).

Question 17. Calculate without using tables \( \tan 20° \tan 40° \tan 80° \).
Answer: We need to calculate \( \tan 20° \tan 40° \tan 80° \).
We can write \( 40° \) as \( 60° - 20° \) and \( 80° \) as \( 60° + 20° \).
So, the expression becomes \( \tan 20° \tan (60° - 20°) \tan (60° + 20°) \).
We know the identity \( \tan \theta \tan (60° - \theta) \tan (60° + \theta) = \tan 3\theta \).
In this case, \( \theta = 20° \).
Applying the identity:
\( \tan 20° \tan (60° - 20°) \tan (60° + 20°) = \tan (3 \times 20°) \)
\( \implies = \tan 60° \)
We know that \( \tan 60° = \sqrt{3} \).
Therefore, \( \tan 20° \tan 40° \tan 80° = \sqrt{3} \).In simple words: We changed the angles \( 40° \) and \( 80° \) to be related to \( 20° \) and \( 60° \). Then, we used a special trigonometric rule that simplifies \( \tan \theta \tan (60 - \theta) \tan (60 + \theta) \) into \( \tan 3\theta \). With \( \theta = 20° \), this became \( \tan 60° \), which is \( \sqrt{3} \).

🎯 Exam Tip: Recognize the pattern \( \tan \theta \tan (60° - \theta) \tan (60° + \theta) = \tan 3\theta \). This identity is very useful for simplifying products of tangent values at angles that are arithmetic progressions with a common difference of 20 degrees around 60 degrees. Always know the standard trigonometric values for common angles like 30°, 45°, 60°.