Samacheer Kalvi Class 11 Maths Solutions Chapter 3 Trigonometry Exercise 3.3

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.3

Question 1. Find the values of
(i) sin 480°
(ii) sin (-1110°)
(iii) cos 300°
(iv) tan (1050°)
(v) cot 660°
(vi) tan \( \left(\frac{19 \pi}{3}\right) \)
(vii) sin \( \left(\frac{-11 \pi}{3}\right) \)
Answer:
(i) To find \( \sin 480^\circ \):
\( \sin 480^\circ = \sin (360^\circ + 120^\circ) \)
Since \( \sin (360^\circ + \theta) = \sin \theta \),
\( \sin 480^\circ = \sin 120^\circ \)
We can write \( 120^\circ \) as \( 90^\circ + 30^\circ \).
So, \( \sin 120^\circ = \sin (90^\circ + 30^\circ) \)
Since \( \sin (90^\circ + \theta) = \cos \theta \),
\( \sin 120^\circ = \cos 30^\circ \)
The value of \( \cos 30^\circ \) is \( \frac{\sqrt{3}}{2} \).
Thus, \( \sin 480^\circ = \frac{\sqrt{3}}{2} \).
(ii) To find \( \sin (-1110^\circ) \):
Since \( \sin (-\theta) = -\sin \theta \),
\( \sin (-1110^\circ) = -\sin (1110^\circ) \)
Now, we find the equivalent angle for \( 1110^\circ \). We know \( 3 \times 360^\circ = 1080^\circ \).
So, \( 1110^\circ = 3 \times 360^\circ + 30^\circ \).
\( -\sin (1110^\circ) = -\sin (3 \times 360^\circ + 30^\circ) \)
Since \( \sin (n \times 360^\circ + \theta) = \sin \theta \),
\( -\sin (3 \times 360^\circ + 30^\circ) = -\sin 30^\circ \)
The value of \( \sin 30^\circ \) is \( \frac{1}{2} \).
Thus, \( \sin (-1110^\circ) = -\frac{1}{2} \).
(iii) To find \( \cos 300^\circ \):
We can write \( 300^\circ \) as \( 270^\circ + 30^\circ \).
\( \cos 300^\circ = \cos (270^\circ + 30^\circ) \)
Since \( \cos (270^\circ + \theta) = \sin \theta \),
\( \cos 300^\circ = \sin 30^\circ \)
The value of \( \sin 30^\circ \) is \( \frac{1}{2} \). Alternatively, \( \cos 300^\circ = \cos(360^\circ - 60^\circ) = \cos 60^\circ = \frac{1}{2} \).
Thus, \( \cos 300^\circ = \frac{1}{2} \).
(iv) To find \( \tan (1050^\circ) \):
We find the equivalent angle for \( 1050^\circ \). We know \( 12 \times 90^\circ = 1080^\circ \).
So, \( 1050^\circ = 12 \times 90^\circ - 30^\circ \).
\( \tan (1050^\circ) = \tan (12 \times 90^\circ - 30^\circ) \)
Since \( 1050^\circ \) is in the fourth quadrant (as \( 1080^\circ - 30^\circ \)), and tan is negative in the fourth quadrant.
\( \tan (12 \times 90^\circ - 30^\circ) = \tan (-30^\circ) = -\tan 30^\circ \)
The value of \( \tan 30^\circ \) is \( \frac{1}{\sqrt{3}} \).
Thus, \( \tan (1050^\circ) = -\frac{1}{\sqrt{3}} \).
(v) To find \( \cot 660^\circ \):
We can write \( 660^\circ \) as \( 7 \times 90^\circ + 30^\circ \).
\( \cot 660^\circ = \cot (7 \times 90^\circ + 30^\circ) \)
Since \( 660^\circ \) is in the fourth quadrant (as \( 630^\circ + 30^\circ \)), and cot is negative in the fourth quadrant.
\( \cot (7 \times 90^\circ + 30^\circ) = -\tan 30^\circ \)
The value of \( \tan 30^\circ \) is \( \frac{1}{\sqrt{3}} \).
Thus, \( \cot 660^\circ = -\frac{1}{\sqrt{3}} \).
(vi) To find \( \tan \left(\frac{19 \pi}{3}\right) \):
We can rewrite \( \frac{19 \pi}{3} \) as a sum involving \( 2\pi \) or \( \pi \).
\( \frac{19 \pi}{3} = \frac{18 \pi + \pi}{3} = 6\pi + \frac{\pi}{3} \)
Since \( \tan (2n\pi + \theta) = \tan \theta \),
\( \tan \left(6\pi + \frac{\pi}{3}\right) = \tan \frac{\pi}{3} \)
The value of \( \tan \frac{\pi}{3} \) (which is \( \tan 60^\circ \)) is \( \sqrt{3} \).
Thus, \( \tan \left(\frac{19 \pi}{3}\right) = \sqrt{3} \).
(vii) To find \( \sin \left(\frac{-11 \pi}{3}\right) \):
Since \( \sin (-\theta) = -\sin \theta \),
\( \sin \left(\frac{-11 \pi}{3}\right) = -\sin \left(\frac{11 \pi}{3}\right) \)
We can rewrite \( \frac{11 \pi}{3} \) as a difference from \( 4\pi \).
\( \frac{11 \pi}{3} = \frac{12 \pi - \pi}{3} = 4\pi - \frac{\pi}{3} \)
So, \( -\sin \left(\frac{11 \pi}{3}\right) = -\sin \left(4\pi - \frac{\pi}{3}\right) \)
Since \( \sin (4\pi - \theta) = -\sin \theta \) (angle in the fourth quadrant),
\( -\sin \left(4\pi - \frac{\pi}{3}\right) = - \left(-\sin \frac{\pi}{3}\right) = \sin \frac{\pi}{3} \)
The value of \( \sin \frac{\pi}{3} \) (which is \( \sin 60^\circ \)) is \( \frac{\sqrt{3}}{2} \).
Thus, \( \sin \left(\frac{-11 \pi}{3}\right) = \frac{\sqrt{3}}{2} \).
In simple words: To find the values of trigonometric functions for large angles, first reduce the angle by adding or subtracting multiples of 360 degrees or \( 2\pi \) radians until it's between 0 and 360 degrees. Then, use the properties of trigonometric functions in different quadrants.

🎯 Exam Tip: Remember the sign convention for trigonometric functions in each quadrant and the reduction formulas like \( \sin(90^\circ \pm \theta) \) or \( \sin(180^\circ \pm \theta) \) to simplify angles correctly.

Question 2. \( \left(\frac{5}{7}, \frac{2 \sqrt{6}}{7}\right) \) is a point on the terminal side of an angle \( \theta \) in standard position. Determine the six trigonometric function values of angle \( \theta \).
Answer: Given that point P \( \left(\frac{5}{7}, \frac{2 \sqrt{6}}{7}\right) \) is on the terminal side of an angle \( \theta \) in standard position.
The coordinates of the point P are \( (x, y) = \left(\frac{5}{7}, \frac{2 \sqrt{6}}{7}\right) \).
The distance 'r' from the origin to the point P is given by the formula \( r = \sqrt{x^2 + y^2} \).
\( r = \text{OP} = \sqrt{\left(\frac{5}{7}\right)^2 + \left(\frac{2 \sqrt{6}}{7}\right)^2} \)
\( \implies r = \sqrt{\frac{25}{49} + \frac{4 \times 6}{49}} \)
\( \implies r = \sqrt{\frac{25}{49} + \frac{24}{49}} \)
\( \implies r = \sqrt{\frac{25+24}{49}} \)
\( \implies r = \sqrt{\frac{49}{49}} \)
\( \implies r = \sqrt{1} = 1 \)

Now we can find the six trigonometric function values:
1. Sine of \( \theta \): \( \sin \theta = \frac{y}{r} = \frac{2 \sqrt{6}/7}{1} = \frac{2 \sqrt{6}}{7} \)
2. Cosine of \( \theta \): \( \cos \theta = \frac{x}{r} = \frac{5/7}{1} = \frac{5}{7} \)
3. Tangent of \( \theta \): \( \tan \theta = \frac{y}{x} = \frac{2 \sqrt{6}/7}{5/7} = \frac{2 \sqrt{6}}{5} \)
4. Cosecant of \( \theta \): \( \csc \theta = \frac{r}{y} = \frac{1}{2 \sqrt{6}/7} = \frac{7}{2 \sqrt{6}} = \frac{7 \sqrt{6}}{12} \)
5. Secant of \( \theta \): \( \sec \theta = \frac{r}{x} = \frac{1}{5/7} = \frac{7}{5} \)
6. Cotangent of \( \theta \): \( \cot \theta = \frac{x}{y} = \frac{5/7}{2 \sqrt{6}/7} = \frac{5}{2 \sqrt{6}} = \frac{5 \sqrt{6}}{12} \)
In simple words: First, find the distance from the origin to the given point using the distance formula. This distance is 'r'. Then, use the definitions of sine (y/r), cosine (x/r), and tangent (y/x), along with their reciprocals, to find all six trigonometric values.

🎯 Exam Tip: Always draw a simple diagram to visualize the point and the angle. This helps in correctly determining the quadrant and the signs of the trigonometric functions.

Question 3. Find the values of the other five trigonometric functions of the following
(i) \( \cos \theta = -\frac{1}{2} \), \( \theta \) lies in the III quadrant
(ii) \( \cos \theta = \frac{2}{3} \), \( \theta \) lies in the I quadrant
(iii) \( \sin \theta = -\frac{2}{3} \), \( \theta \) lies in the IV quadrant
(iv) \( \tan \theta = -2 \), \( \theta \) lies in the II quadrant
Answer:
(i) Given \( \cos \theta = -\frac{1}{2} \), and \( \theta \) lies in the III quadrant.
We know the identity: \( \sin^2 \theta + \cos^2 \theta = 1 \)
Substitute the value of \( \cos \theta \):
\( \sin^2 \theta + \left(-\frac{1}{2}\right)^2 = 1 \)
\( \sin^2 \theta + \frac{1}{4} = 1 \)
\( \sin^2 \theta = 1 - \frac{1}{4} \)
\( \sin^2 \theta = \frac{3}{4} \)
\( \sin \theta = \pm \frac{\sqrt{3}}{2} \)
Since \( \theta \) lies in the III quadrant, the sine function is negative.
So, \( \sin \theta = -\frac{\sqrt{3}}{2} \).

Now, we find the other four trigonometric functions:
1. \( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\sqrt{3}/2}{-1/2} = \sqrt{3} \)
2. \( \csc \theta = \frac{1}{\sin \theta} = \frac{1}{-\sqrt{3}/2} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3} \)
3. \( \sec \theta = \frac{1}{\cos \theta} = \frac{1}{-1/2} = -2 \)
4. \( \cot \theta = \frac{1}{\tan \theta} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \)

(ii) Given \( \cos \theta = \frac{2}{3} \), and \( \theta \) lies in the I quadrant.
We know the identity: \( \cos^2 \theta + \sin^2 \theta = 1 \)
Substitute the value of \( \cos \theta \):
\( \left(\frac{2}{3}\right)^2 + \sin^2 \theta = 1 \)
\( \frac{4}{9} + \sin^2 \theta = 1 \)
\( \sin^2 \theta = 1 - \frac{4}{9} \)
\( \sin^2 \theta = \frac{5}{9} \)
\( \sin \theta = \pm \frac{\sqrt{5}}{3} \)
Since \( \theta \) lies in the I quadrant, the sine function is positive.
So, \( \sin \theta = \frac{\sqrt{5}}{3} \).

Now, we find the other four trigonometric functions:
1. \( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sqrt{5}/3}{2/3} = \frac{\sqrt{5}}{2} \)
2. \( \csc \theta = \frac{1}{\sin \theta} = \frac{1}{\sqrt{5}/3} = \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5} \)
3. \( \sec \theta = \frac{1}{\cos \theta} = \frac{1}{2/3} = \frac{3}{2} \)
4. \( \cot \theta = \frac{1}{\tan \theta} = \frac{1}{\sqrt{5}/2} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \)

(iii) Given \( \sin \theta = -\frac{2}{3} \), and \( \theta \) lies in the IV quadrant.
We know the identity: \( \cos^2 \theta + \sin^2 \theta = 1 \)
Substitute the value of \( \sin \theta \):
\( \cos^2 \theta + \left(-\frac{2}{3}\right)^2 = 1 \)
\( \cos^2 \theta + \frac{4}{9} = 1 \)
\( \cos^2 \theta = 1 - \frac{4}{9} \)
\( \cos^2 \theta = \frac{5}{9} \)
\( \cos \theta = \pm \frac{\sqrt{5}}{3} \)
Since \( \theta \) lies in the IV quadrant, the cosine function is positive.
So, \( \cos \theta = \frac{\sqrt{5}}{3} \).

Now, we find the other four trigonometric functions:
1. \( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-2/3}{\sqrt{5}/3} = -\frac{2}{\sqrt{5}} = -\frac{2\sqrt{5}}{5} \)
2. \( \csc \theta = \frac{1}{\sin \theta} = \frac{1}{-2/3} = -\frac{3}{2} \)
3. \( \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\sqrt{5}/3} = \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5} \)
4. \( \cot \theta = \frac{1}{\tan \theta} = \frac{1}{-2/\sqrt{5}} = -\frac{\sqrt{5}}{2} \)

(iv) Given \( \tan \theta = -2 \), and \( \theta \) lies in the II quadrant.
We know the identity: \( \sec^2 \theta = 1 + \tan^2 \theta \)
Substitute the value of \( \tan \theta \):
\( \sec^2 \theta = 1 + (-2)^2 \)
\( \sec^2 \theta = 1 + 4 \)
\( \sec^2 \theta = 5 \)
\( \sec \theta = \pm \sqrt{5} \)
Since \( \theta \) lies in the II quadrant, the secant function is negative.
So, \( \sec \theta = -\sqrt{5} \).

Now, we find the other four trigonometric functions:
1. \( \cos \theta = \frac{1}{\sec \theta} = \frac{1}{-\sqrt{5}} = -\frac{\sqrt{5}}{5} \)
2. \( \sin \theta = \tan \theta \cdot \cos \theta = (-2) \cdot \left(-\frac{\sqrt{5}}{5}\right) = \frac{2\sqrt{5}}{5} \)
3. \( \csc \theta = \frac{1}{\sin \theta} = \frac{1}{2\sqrt{5}/5} = \frac{5}{2\sqrt{5}} = \frac{\sqrt{5}}{2} \)
4. \( \cot \theta = \frac{1}{\tan \theta} = \frac{1}{-2} = -\frac{1}{2} \)
In simple words: Use the Pythagorean identities (\( \sin^2 \theta + \cos^2 \theta = 1 \), \( \sec^2 \theta = 1 + \tan^2 \theta \), \( \csc^2 \theta = 1 + \cot^2 \theta \)) to find one of the missing basic functions. Then, use the given quadrant to decide if the value is positive or negative. Finally, use the reciprocal and quotient identities to find the remaining functions.

🎯 Exam Tip: Always remember the signs of trigonometric functions in each of the four quadrants (ASTC rule: All, Sine, Tan, Cos). This is crucial for selecting the correct sign when taking square roots.

Question 4. Prove that
\( \frac{\cot (180^\circ + \theta) \sin (90^\circ - \theta) \cos (-\theta)}{\sin (270^\circ + \theta) \tan (-\theta) \csc (360^\circ + \theta)} = \cos^2 \theta \cot \theta \)
Answer: Let's simplify the Left Hand Side (LHS) of the equation using trigonometric identities for related angles.
We know the following identities:
1. \( \cot (180^\circ + \theta) = \cot \theta \) (cot is positive in the third quadrant).
2. \( \sin (90^\circ - \theta) = \cos \theta \) (sine in the first quadrant is positive, and it changes to cosine).
3. \( \cos (-\theta) = \cos \theta \) (cosine is an even function).
4. \( \sin (270^\circ + \theta) = -\cos \theta \) (sine in the fourth quadrant is negative, and it changes to cosine).
5. \( \tan (-\theta) = -\tan \theta \) (tangent is an odd function).
6. \( \csc (360^\circ + \theta) = \csc \theta \) (cosecant is positive in the first quadrant, after a full rotation).

Now, substitute these into the LHS:
\( \text{LHS} = \frac{(\cot \theta)(\cos \theta)(\cos \theta)}{(-\cos \theta)(-\tan \theta)(\csc \theta)} \)
Multiply the terms in the numerator and denominator:
\( \text{LHS} = \frac{\cot \theta \cos^2 \theta}{\cos \theta \tan \theta \csc \theta} \)
We also know that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \csc \theta = \frac{1}{\sin \theta} \).
Let's simplify the denominator: \( \cos \theta \tan \theta \csc \theta = \cos \theta \left(\frac{\sin \theta}{\cos \theta}\right) \left(\frac{1}{\sin \theta}\right) \)
The \( \cos \theta \) in the numerator and denominator cancel out, and the \( \sin \theta \) terms also cancel out.
So, \( \cos \theta \tan \theta \csc \theta = 1 \).

Substitute this back into the LHS:
\( \text{LHS} = \frac{\cot \theta \cos^2 \theta}{1} \)
\( \text{LHS} = \cot \theta \cos^2 \theta \)
This is equal to the Right Hand Side (RHS).
Therefore, the identity is proven.
In simple words: We changed each trigonometric function of an angle like \( (180^\circ + \theta) \) into a simpler form using rules about how they behave in different parts of a circle. After replacing all terms and simplifying, we found that both sides of the equation became the same, which means the proof is complete.

🎯 Exam Tip: To prove trigonometric identities, always start with the more complex side (usually LHS) and simplify it using fundamental identities, quadrant rules, and reciprocal relationships until it matches the other side. Keep the target (RHS) in mind during simplification.

Question 5. Find all the angles between 0° and 360° which satisfy the equation \( \sin^2 \theta = \frac{3}{4} \).
Answer: We are given the equation \( \sin^2 \theta = \frac{3}{4} \).
To solve for \( \sin \theta \), take the square root of both sides:
\( \sin \theta = \pm \sqrt{\frac{3}{4}} \)
\( \sin \theta = \pm \frac{\sqrt{3}}{2} \)

This gives us two cases:
**Case 1: \( \sin \theta = \frac{\sqrt{3}}{2} \)**
We know that \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).
Since sine is positive in the first and second quadrants, the angles are:
1. First quadrant: \( \theta = 60^\circ \).
2. Second quadrant: \( \theta = 180^\circ - 60^\circ = 120^\circ \).

**Case 2: \( \sin \theta = -\frac{\sqrt{3}}{2} \)**
Since sine is negative in the third and fourth quadrants, and the reference angle is \( 60^\circ \), the angles are:
1. Third quadrant: \( \theta = 180^\circ + 60^\circ = 240^\circ \).
2. Fourth quadrant: \( \theta = 360^\circ - 60^\circ = 300^\circ \).

So, the angles between 0° and 360° that satisfy the equation are \( 60^\circ, 120^\circ, 240^\circ, \) and \( 300^\circ \).
In simple words: We first find the basic value of \( \sin \theta \) by taking the square root. Since sine can be positive or negative, we find the angles in all four parts of the circle where \( \sin \theta \) has that value. This gives us four possible angles.

🎯 Exam Tip: When solving trigonometric equations involving squares, remember to consider both positive and negative square roots. Then, identify all possible quadrants where the trigonometric function has that sign and find the corresponding angles within the given range (usually 0° to 360° or 0 to \( 2\pi \)).

Question 6. Show that
\( \sin^2 \frac{\pi}{18} + \sin^2 \frac{\pi}{9} + \sin^2 \frac{7\pi}{18} + \sin^2 \frac{4\pi}{9} = 2 \)
Answer: First, let's convert the given angles from radians to degrees for easier understanding.
We know that \( \pi \) radians = \( 180^\circ \).
\( \frac{\pi}{18} = \frac{180^\circ}{18} = 10^\circ \)
\( \frac{\pi}{9} = \frac{180^\circ}{9} = 20^\circ \)
\( \frac{7\pi}{18} = \frac{7 \times 180^\circ}{18} = 7 \times 10^\circ = 70^\circ \)
\( \frac{4\pi}{9} = \frac{4 \times 180^\circ}{9} = 4 \times 20^\circ = 80^\circ \)

Now, substitute these degree values into the expression:
\( \text{LHS} = \sin^2 10^\circ + \sin^2 20^\circ + \sin^2 70^\circ + \sin^2 80^\circ \)

We know the complementary angle identity: \( \sin (90^\circ - \theta) = \cos \theta \).
Let's apply this to the last two terms:
\( \sin 70^\circ = \sin (90^\circ - 20^\circ) = \cos 20^\circ \)
\( \sin 80^\circ = \sin (90^\circ - 10^\circ) = \cos 10^\circ \)

Substitute these back into the expression:
\( \text{LHS} = \sin^2 10^\circ + \sin^2 20^\circ + (\cos 20^\circ)^2 + (\cos 10^\circ)^2 \)
\( \text{LHS} = \sin^2 10^\circ + \sin^2 20^\circ + \cos^2 20^\circ + \cos^2 10^\circ \)

Now, rearrange the terms to group them using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( \text{LHS} = (\sin^2 10^\circ + \cos^2 10^\circ) + (\sin^2 20^\circ + \cos^2 20^\circ) \)
\( \text{LHS} = 1 + 1 \)
\( \text{LHS} = 2 \)
This is equal to the Right Hand Side (RHS).
Therefore, the equation is proven.
In simple words: We changed the radian angles to degrees. Then, we used a rule that says \( \sin(\text{90 degrees} - \theta) \) is the same as \( \cos \theta \). By doing this, we turned some \( \sin^2 \) terms into \( \cos^2 \) terms. When we put them together, we could use the basic rule \( \sin^2 \theta + \cos^2 \theta = 1 \) twice, which gave us \( 1 + 1 = 2 \).

🎯 Exam Tip: When dealing with sums of squares of trigonometric functions, especially with angles like \( \theta \) and \( (90^\circ - \theta) \), look for opportunities to apply the complementary angle identities (like \( \sin(90^\circ - \theta) = \cos \theta \)) and the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \). This often simplifies expressions quickly.