GSEB Class 11 Maths Solutions Chapter 3 Trigonometric Functions Exercise 3.3

Prove That:

Question 1. \( \sin^2 \frac{\pi}{6} + \cos^2 \frac{\pi}{3} - \tan^2 \frac{\pi}{4} = - \frac{1}{2} \)
Answer: We begin by taking the Left Hand Side (L.H.S.) of the equation:
L.H.S. = \( \sin^2 \frac{\pi}{6} + \cos^2 \frac{\pi}{3} - \tan^2 \frac{\pi}{4} \)
We can rewrite this expression using squares:
= \( (\sin \frac{\pi}{6})^2 + (\cos \frac{\pi}{3})^2 - (\tan \frac{\pi}{4})^2 \)
Now, we substitute the known values for these trigonometric functions:
[: \( \sin \frac{\pi}{6} = \frac{1}{2}, \cos \frac{\pi}{3} = \frac{1}{2}, \tan \frac{\pi}{4} = 1 \)]
= \( (\frac{1}{2})^2 + (\frac{1}{2})^2 - 1^2 \)
We then calculate the squares and perform the addition and subtraction:
= \( \frac{1}{4} + \frac{1}{4} - 1 \)
= \( \frac{2}{4} - 1 \)
= \( \frac{1}{2} - 1 \)
= \( - \frac{1}{2} \)
This matches the Right Hand Side (R.H.S.) of the equation, so the proof is complete.
In simple words: We started with the left side of the equation and replaced the sine, cosine, and tangent values with their actual numbers. Then, we squared those numbers and did the math. The result we got was exactly the same as the right side, which shows the equation is true.

Exam Tip: Always state the L.H.S. and R.H.S. clearly. Use standard values for common angles like \( \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3} \) and their multiples. Show each step of calculation for full marks.

Question 2. \( 2 \sin^2 \frac{\pi}{6} + \operatorname{cosec}^2 \frac{7\pi}{6} \cos^2 \frac{\pi}{3} = \frac{3}{2} \)
Answer: We begin with the Left Hand Side (L.H.S.) of the given equation:
L.H.S. = \( 2 \sin^2 \frac{\pi}{6} + \operatorname{cosec}^2 \frac{7\pi}{6} \cos^2 \frac{\pi}{3} \)
We can rewrite the squared terms and express \( \operatorname{cosec} \frac{7\pi}{6} \) in a simpler form:
= \( 2(\sin \frac{\pi}{6})^2 + (\operatorname{cosec} (\pi + \frac{\pi}{6}))^2 (\cos \frac{\pi}{3})^2 \)
Using the identity \( \operatorname{cosec} (\pi + \theta) = -\operatorname{cosec} \theta \), we get:
[: \( \operatorname{cosec} (\pi + \theta) = -\operatorname{cosec} \theta \)]
= \( 2(\frac{1}{2})^2 + (-\operatorname{cosec} \frac{\pi}{6})^2 (\frac{1}{2})^2 \)
Now, we substitute the known values for \( \sin \frac{\pi}{6} \), \( \operatorname{cosec} \frac{\pi}{6} \), and \( \cos \frac{\pi}{3} \):
= \( 2 \times \frac{1}{4} + (-2)^2 \times \frac{1}{4} \)
We perform the calculations:
= \( \frac{1}{2} + 4 \times \frac{1}{4} \)
= \( \frac{1}{2} + 1 \)
= \( \frac{3}{2} \)
This result matches the Right Hand Side (R.H.S.) of the equation, confirming the identity.
In simple words: We took the left side of the equation, simplified the angle \( \frac{7\pi}{6} \), and then put in the values for sine, cosecant, and cosine. After squaring and adding, we ended up with the right side of the equation, proving it correct.

Exam Tip: When dealing with angles greater than \( \frac{\pi}{2} \), always simplify them using quadrant rules (e.g., \( \pi + \theta \), \( 2\pi - \theta \)) before substituting values. Pay careful attention to the signs of trigonometric functions in different quadrants.

Question 3. \( \cot^2 \frac{\pi}{6} + \operatorname{cosec} \frac{5\pi}{6} + 3 \tan^2 \frac{\pi}{6} = 6 \)
Answer: We consider the Left Hand Side (L.H.S.) of the given equation:
L.H.S. = \( \cot^2 \frac{\pi}{6} + \operatorname{cosec} \frac{5\pi}{6} + 3 \tan^2 \frac{\pi}{6} \)
First, we rewrite the squared terms and simplify the angle for cosecant:
= \( (\cot \frac{\pi}{6})^2 + \operatorname{cosec} (\pi - \frac{\pi}{6}) + 3(\tan \frac{\pi}{6})^2 \)
Using the identity \( \operatorname{cosec} (\pi - \theta) = \operatorname{cosec} \theta \), and substituting known values:
= \( (\sqrt{3})^2 + \operatorname{cosec} \frac{\pi}{6} + 3(\frac{1}{\sqrt{3}})^2 \)
Now, we evaluate the terms:
= \( 3 + 2 + 3 \times \frac{1}{3} \)
We perform the multiplication and then the addition:
= \( 3 + 2 + 1 \)
= \( 6 \)
This result is equal to the Right Hand Side (R.H.S.), thus proving the identity.
In simple words: We started with the left side of the equation. We simplified the angle for cosecant and then put in the specific values for cotangent, cosecant, and tangent. After doing all the calculations, we found that the answer was 6, which matches the right side, so the statement is proven.

Exam Tip: Remember to simplify angles using quadrant rules before calculation, especially for functions like cosecant. Knowing the exact values of trigonometric functions for standard angles (e.g., \( \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3} \)) is crucial for solving such problems quickly.

Question 4. \( 2 \sin^2 \frac{3\pi}{4} + 2 \cos^2 \frac{\pi}{4} + 2 \sec^2 \frac{\pi}{3} = 10 \)
Answer: Let's start with the Left Hand Side (L.H.S.) of the equation:
L.H.S. = \( 2 \sin^2 \frac{3\pi}{4} + 2 \cos^2 \frac{\pi}{4} + 2 \sec^2 \frac{\pi}{3} \)
We rewrite the terms using squares and simplify \( \sin \frac{3\pi}{4} \):
= \( 2(\sin (\pi - \frac{\pi}{4}))^2 + 2(\cos \frac{\pi}{4})^2 + 2(\sec \frac{\pi}{3})^2 \)
Using \( \sin (\pi - \theta) = \sin \theta \), and substituting known values:
= \( 2(\sin \frac{\pi}{4})^2 + 2(\frac{1}{\sqrt{2}})^2 + 2(2)^2 \)
Now we substitute the value for \( \sin \frac{\pi}{4} \):
= \( 2(\frac{1}{\sqrt{2}})^2 + 2(\frac{1}{2}) + 2(4) \)
We perform the multiplications and additions:
= \( 2 \times \frac{1}{2} + 1 + 8 \)
= \( 1 + 1 + 8 \)
= \( 10 \)
This result equals the Right Hand Side (R.H.S.), hence the identity is proven.
In simple words: We started with the left side of the equation. We simplified the sine term by using an angle rule and then put in the specific numbers for sine, cosine, and secant. After we did the squaring, multiplying, and adding, we got 10, which matches the right side, so the equation is shown to be true.

Exam Tip: Be careful with the squaring of terms like \( (\frac{1}{\sqrt{2}})^2 \) and \( (2)^2 \). Remember to simplify angles like \( \frac{3\pi}{4} \) to their equivalent acute angle forms before finding their trigonometric values.

Question 5. Find the value of:
(i) \( \sin 75^\circ \)
(ii) \( \tan 15^\circ \)
Answer:
(i) To find the value of \( \sin 75^\circ \), we express \( 75^\circ \) as a sum of standard angles:
\( \sin 75^\circ = \sin (45^\circ + 30^\circ) \)
Using the trigonometric identity for the sine of a sum of two angles, \( \sin (A + B) = \sin A \cos B + \cos A \sin B \):
\( = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \)
Now we substitute the known values of these trigonometric functions:
\( = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} \)
We combine the terms:
\( = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} \)
\( = \frac{\sqrt{3}+1}{2\sqrt{2}} \)
(ii) To find the value of \( \tan 15^\circ \), we express \( 15^\circ \) as a difference of standard angles:
\( \tan 15^\circ = \tan (45^\circ - 30^\circ) \)
Using the trigonometric identity for the tangent of a difference of two angles, \( \tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \):
\( = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \)
Now we substitute the known values for \( \tan 45^\circ \) and \( \tan 30^\circ \):
[: \( \tan 45^\circ = 1, \tan 30^\circ = \frac{1}{\sqrt{3}} \)]
\( = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \times \frac{1}{\sqrt{3}}} \)
We simplify the numerator and the denominator:
\( = \frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}} \)
\( = \frac{\sqrt{3}-1}{\sqrt{3}+1} \)
To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator, \( (\sqrt{3}-1) \):
\( = \frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} \)
\( = \frac{(\sqrt{3}-1)^2}{(\sqrt{3})^2 - 1^2} \)
\( = \frac{3+1-2\sqrt{3}}{3-1} \)
\( = \frac{4-2\sqrt{3}}{2} \)
\( = 2-\sqrt{3} \)
In simple words: For \( \sin 75^\circ \), we broke \( 75^\circ \) into \( 45^\circ + 30^\circ \), used a formula, and put in the known values for those angles to get the answer. For \( \tan 15^\circ \), we broke \( 15^\circ \) into \( 45^\circ - 30^\circ \), used another formula, put in the known values, and then simplified the fraction by getting rid of the square root in the bottom part.

Exam Tip: For such problems, always express the given angle as a sum or difference of standard angles (\( 0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ \)). Remember the addition/subtraction formulas for sine, cosine, and tangent accurately. Rationalize the denominator if a radical appears in the final fraction.

Prove the Following:

Question 6. \( \cos (\frac{\pi}{4} - x) \cos (\frac{\pi}{4} - y) - \sin (\frac{\pi}{4} - x) \sin (\frac{\pi}{4} - y) = \sin (x + y) \)
Answer: We begin with the Left Hand Side (L.H.S.) of the given equation:
L.H.S. = \( \cos (\frac{\pi}{4} - x) \cos (\frac{\pi}{4} - y) - \sin (\frac{\pi}{4} - x) \sin (\frac{\pi}{4} - y) \)
To simplify, we let \( A = \frac{\pi}{4} - x \) and \( B = \frac{\pi}{4} - y \).
Then, the L.H.S. becomes:
L.H.S. = \( \cos A \cos B - \sin A \sin B \)
We recognize this as the formula for \( \cos (A + B) \):
= \( \cos (A + B) \)
Now, we substitute back the values for A and B:
= \( \cos [(\frac{\pi}{4} - x) + (\frac{\pi}{4} - y)] \)
= \( \cos [\frac{\pi}{4} + \frac{\pi}{4} - x - y] \)
= \( \cos [\frac{2\pi}{4} - (x + y)] \)
= \( \cos [\frac{\pi}{2} - (x + y)] \)
Using the identity \( \cos (\frac{\pi}{2} - \theta) = \sin \theta \), we get:
[: \( \cos (\frac{\pi}{2} - \theta) = \sin \theta \)]
= \( \sin (x + y) \)
This equals the Right Hand Side (R.H.S.), completing the proof.
In simple words: We took the left side of the equation and used a simple trick to call the complex angle parts 'A' and 'B'. This made the expression look like a known cosine formula, \( \cos (A+B) \). Then we put the original angle parts back and simplified them. Finally, using a special rule for cosine, we got the right side of the equation, proving it correct.

Exam Tip: When an expression matches a standard trigonometric identity, use substitution (like A and B) to make the steps clearer. Remember complementary angle identities like \( \cos (\frac{\pi}{2} - \theta) = \sin \theta \) for quick simplification.

Question 7. \( \frac{\tan (\frac{\pi}{4}+x)}{\tan (\frac{\pi}{4}-x)}=\frac{(1+\tan x)^{2}}{(1-\tan x)^{2}} \)
Answer: We start with the Left Hand Side (L.H.S.) of the equation:
L.H.S. = \( \frac{\tan (\frac{\pi}{4}+x)}{\tan (\frac{\pi}{4}-x)} \)
We use the tangent sum and difference formulas:
\( \tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)
\( \tan (A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \)
And we know that \( \tan \frac{\pi}{4} = 1 \).
First, let's expand the numerator \( \tan (\frac{\pi}{4}+x) \):
\( \tan (\frac{\pi}{4}+x) = \frac{\tan \frac{\pi}{4} + \tan x}{1 - \tan \frac{\pi}{4} \tan x} = \frac{1 + \tan x}{1 - 1 \cdot \tan x} = \frac{1 + \tan x}{1 - \tan x} \)
Next, we expand the denominator \( \tan (\frac{\pi}{4}-x) \):
\( \tan (\frac{\pi}{4}-x) = \frac{\tan \frac{\pi}{4} - \tan x}{1 + \tan \frac{\pi}{4} \tan x} = \frac{1 - 1 \cdot \tan x}{1 + 1 \cdot \tan x} = \frac{1 - \tan x}{1 + \tan x} \)
Now, we substitute these expanded forms back into the L.H.S.:
L.H.S. = \( \frac{\frac{1 + \tan x}{1 - \tan x}}{\frac{1 - \tan x}{1 + \tan x}} \)
To simplify this complex fraction, we multiply the numerator by the reciprocal of the denominator:
= \( \frac{1 + \tan x}{1 - \tan x} \times \frac{1 + \tan x}{1 - \tan x} \)
= \( \frac{(1 + \tan x)^2}{(1 - \tan x)^2} \)
This matches the Right Hand Side (R.H.S.), completing the proof.
In simple words: We took the left side of the equation and used the specific formulas for tangent of a sum and tangent of a difference. After putting in the value for tangent of \( \frac{\pi}{4} \), we simplified the top and bottom parts of the fraction. Then we divided these simplified parts, which resulted in the right side of the equation, proving it.

Exam Tip: Mastering the sum and difference formulas for tangent is vital. Remember that \( \tan \frac{\pi}{4} = 1 \). When simplifying complex fractions, multiplying by the reciprocal of the denominator is a common and effective technique.

Question 8. \( \frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos (\frac{\pi}{2}+x)} = \cot^2 x \)
Answer: We start with the Left Hand Side (L.H.S.) of the equation:
L.H.S. = \( \frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos (\frac{\pi}{2}+x)} \)
We apply the trigonometric identities for angles in different quadrants:
\( \cos (\pi + x) = -\cos x \) (cosine is negative in the 3rd quadrant)
\( \cos (-x) = \cos x \) (cosine is an even function)
\( \sin (\pi - x) = \sin x \) (sine is positive in the 2nd quadrant)
\( \cos (\frac{\pi}{2} + x) = -\sin x \) (cosine is negative in the 2nd quadrant, and changes to sine)
Now, we substitute these simplified terms back into the L.H.S. expression:
L.H.S. = \( \frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)} \)
We perform the multiplication:
= \( \frac{-\cos^2 x}{-\sin^2 x} \)
The negative signs cancel out:
= \( \frac{\cos^2 x}{\sin^2 x} \)
We recognize this as the identity for cotangent squared:
= \( \cot^2 x \)
This matches the Right Hand Side (R.H.S.), thus the proof is complete.
In simple words: We took the left side of the equation and used special rules for how cosine and sine act with angles like \( \pi+x \) or \( \frac{\pi}{2}+x \). After changing each part to its simpler form, we put them back into the fraction. The minus signs canceled, and the expression became \( \cot^2 x \), which is what we needed to show.

Exam Tip: A strong understanding of trigonometric identities and quadrant rules is essential. Carefully determine the sign and function change for each term like \( \cos (\pi+x) \) and \( \sin (\pi-x) \). Incorrect signs are a common source of errors.

Question 9. \( \cos (\frac{3\pi}{2} + x) \cos (2\pi + x) [\cot (\frac{3\pi}{2} - x) + \cot (2\pi + x)] = 1 \)
Answer: We take the Left Hand Side (L.H.S.) of the equation:
L.H.S. = \( \cos (\frac{3\pi}{2} + x) \cos (2\pi + x) [\cot (\frac{3\pi}{2} - x) + \cot (2\pi + x)] \)
We apply the trigonometric identities for angles in different quadrants and periods:
\( \cos (\frac{3\pi}{2} + x) = \sin x \) (cosine is positive in the 4th quadrant, and changes to sine)
\( \cos (2\pi + x) = \cos x \) (cosine has a period of \( 2\pi \))
\( \cot (\frac{3\pi}{2} - x) = \tan x \) (cotangent is positive in the 3rd quadrant, and changes to tangent)
\( \cot (2\pi + x) = \cot x \) (cotangent has a period of \( 2\pi \))
Now, we substitute these simplified terms back into the L.H.S. expression:
L.H.S. = \( (\sin x) (\cos x) [\tan x + \cot x] \)
We express \( \tan x \) and \( \cot x \) in terms of \( \sin x \) and \( \cos x \):
= \( \sin x \cos x [\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}] \)
We combine the terms inside the bracket using a common denominator:
= \( \sin x \cos x [\frac{\sin^2 x + \cos^2 x}{\cos x \sin x}] \)
Using the fundamental identity \( \sin^2 x + \cos^2 x = 1 \):
[: \( \sin^2 x + \cos^2 x = 1 \)]
= \( \sin x \cos x [\frac{1}{\cos x \sin x}] \)
The \( \sin x \cos x \) terms cancel out:
= \( 1 \)
This equals the Right Hand Side (R.H.S.), thus proving the identity.
In simple words: We started with the left side of the equation. We used rules to simplify each part like \( \cos (\frac{3\pi}{2} + x) \) to its simpler sine or cosine form. Then we put these simple forms back into the expression. Next, we changed tangent and cotangent into sine and cosine fractions, combined them, and finally, everything canceled out to leave 1, which matches the right side.

Exam Tip: Pay attention to both quadrant rules (for \( \frac{3\pi}{2} \pm x \)) and periodicity rules (for \( 2\pi \pm x \)) for simplification. Always convert tangent and cotangent to sine and cosine when adding or subtracting them, as it often leads to simpler forms using \( \sin^2 x + \cos^2 x = 1 \).

Prove the Following:

Question 10. \( \sin (n + 1)x \sin (n + 2)x + \cos (n + 1)x \cos (n + 2)x = \cos x \)
Answer: We begin with the Left Hand Side (L.H.S.) of the equation:
L.H.S. = \( \sin (n + 1)x \sin (n + 2)x + \cos (n + 1)x \cos (n + 2)x \)
We can rearrange the terms to match a standard identity form:
= \( \cos (n + 2)x \cos (n + 1)x + \sin (n + 2)x \sin (n + 1)x \)
Let \( A = (n + 2)x \) and \( B = (n + 1)x \).
The expression then becomes:
L.H.S. = \( \cos A \cos B + \sin A \sin B \)
We recognize this as the formula for \( \cos (A - B) \):
= \( \cos (A - B) \)
Now, we substitute back the values for A and B:
= \( \cos [(n + 2)x - (n + 1)x] \)
We expand the terms inside the bracket:
= \( \cos [nx + 2x - nx - x] \)
The \( nx \) terms cancel out, and we combine the \( x \) terms:
= \( \cos [x] \)
= \( \cos x \)
This matches the Right Hand Side (R.H.S.), completing the proof.
In simple words: We started with the left side of the equation and reorganized it to fit the rule for \( \cos(A-B) \). By setting \( A = (n+2)x \) and \( B = (n+1)x \), we applied the formula. After simplifying the angles inside the cosine, we were left with just \( \cos x \), which matches the right side.

Exam Tip: Recognizing standard trigonometric identities like \( \cos(A-B) = \cos A \cos B + \sin A \sin B \) is key. When dealing with algebraic terms within angles (like \( (n+1)x \)), treat them as single variables to simplify application of the identity.

Question 11. \( \cos (\frac{3\pi}{4} + x) - \cos (\frac{3\pi}{4} - x) = - \sqrt{2} \sin x \)
Answer: We consider the Left Hand Side (L.H.S.) of the given equation:
L.H.S. = \( \cos (\frac{3\pi}{4} + x) - \cos (\frac{3\pi}{4} - x) \)
We use the sum-to-product identity for \( \cos C - \cos D \):
\( \cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2} \)
Here, let \( C = \frac{3\pi}{4} + x \) and \( D = \frac{3\pi}{4} - x \).
First, calculate \( C+D \) and \( C-D \):
\( C+D = (\frac{3\pi}{4} + x) + (\frac{3\pi}{4} - x) = \frac{3\pi}{4} + \frac{3\pi}{4} = \frac{6\pi}{4} = \frac{3\pi}{2} \)
\( C-D = (\frac{3\pi}{4} + x) - (\frac{3\pi}{4} - x) = x + x = 2x \)
Now, apply the formula:
L.H.S. = \( -2 \sin (\frac{\frac{3\pi}{2}}{2}) \sin (\frac{2x}{2}) \)
= \( -2 \sin (\frac{3\pi}{4}) \sin x \)
Next, we find the value of \( \sin \frac{3\pi}{4} \):
\( \sin \frac{3\pi}{4} = \sin (\pi - \frac{\pi}{4}) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \)
Substitute this value back into the expression:
= \( -2 \times \frac{1}{\sqrt{2}} \times \sin x \)
We simplify the coefficient:
= \( - \sqrt{2} \sin x \)
This result matches the Right Hand Side (R.H.S.), proving the identity.
In simple words: We started with the left side, which was a subtraction of two cosine terms. We used a special formula to turn this subtraction into a multiplication of two sine terms. After simplifying the angles in the sine terms, we found the value of \( \sin \frac{3\pi}{4} \) and put it back. This led us straight to the right side of the equation.

Exam Tip: Memorize the sum-to-product and product-to-sum formulas, as they are frequently used in proving trigonometric identities. When simplifying angles like \( \frac{3\pi}{4} \), always think about their equivalent angles in the first quadrant and consider the sign based on the quadrant.

Question 12. \( \sin^2 6x - \sin^2 4x = \sin 2x \sin 8x \)
Answer: We start with the Left Hand Side (L.H.S.) of the equation:
L.H.S. = \( \sin^2 6x - \sin^2 4x \)
We use the trigonometric identity for the difference of squares of sines:
\( \sin^2 A - \sin^2 B = \sin (A + B) \sin (A - B) \)
Here, let \( A = 6x \) and \( B = 4x \).
Substitute these into the formula:
L.H.S. = \( \sin (6x + 4x) \sin (6x - 4x) \)
We perform the addition and subtraction within the arguments of the sine functions:
= \( \sin (10x) \sin (2x) \)
We can rearrange the terms to match the R.H.S.:
= \( \sin 2x \sin 10x \)
This equals the Right Hand Side (R.H.S.), completing the proof.
In simple words: We took the left side of the equation, which showed two sine terms being squared and subtracted. We used a special identity that turns this into a multiplication of two sine terms. After adding and subtracting the angles as per the formula, we got the same expression as the right side, proving the statement.

Exam Tip: The identity \( \sin^2 A - \sin^2 B = \sin (A + B) \sin (A - B) \) is very useful. Remember to clearly identify A and B before applying the formula. Double-check your arithmetic when combining the angles.

Question 13. \( \cos^2 2x - \cos^2 6x = \sin 4x \sin 8x \)
Answer: We begin with the Left Hand Side (L.H.S.) of the equation:
L.H.S. = \( \cos^2 2x - \cos^2 6x \)
We use the identity \( \cos^2 \theta = 1 - \sin^2 \theta \) to convert the cosine squared terms to sine squared:
= \( (1 - \sin^2 2x) - (1 - \sin^2 6x) \)
We open the brackets and simplify:
= \( 1 - \sin^2 2x - 1 + \sin^2 6x \)
= \( \sin^2 6x - \sin^2 2x \)
Now, we apply the trigonometric identity for the difference of squares of sines:
\( \sin^2 A - \sin^2 B = \sin (A + B) \sin (A - B) \)
Here, let \( A = 6x \) and \( B = 2x \).
Substitute these into the formula:
= \( \sin (6x + 2x) \sin (6x - 2x) \)
We perform the addition and subtraction within the arguments of the sine functions:
= \( \sin (8x) \sin (4x) \)
We can rearrange the terms to match the R.H.S.:
= \( \sin 4x \sin 8x \)
This equals the Right Hand Side (R.H.S.), completing the proof.
In simple words: We started with the left side, which showed two cosine terms being squared and subtracted. We first changed the cosine squares to sine squares. Then, we used a special identity for the difference of sine squares. After adding and subtracting the angles as required by the formula, we got the same expression as the right side, proving the statement.

Exam Tip: Remember to use the identity \( \cos^2 \theta = 1 - \sin^2 \theta \) when you need to transform cosine squared terms into sine squared terms, or vice-versa. This conversion is often the first step in problems involving \( \sin^2 A - \sin^2 B \) or \( \cos^2 A - \cos^2 B \).

Question 14. \( \sin 2x + 2 \sin 4x + \sin 6x = 4 \cos^2 x \sin 4x \)
Answer: We consider the Left Hand Side (L.H.S.) of the given equation:
L.H.S. = \( \sin 2x + 2 \sin 4x + \sin 6x \)
We group the sine terms strategically to apply the sum-to-product formula:
= \( (\sin 6x + \sin 2x) + 2 \sin 4x \)
Using the identity \( \sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2} \):
[: \( \sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2} \)]
= \( 2 \sin (\frac{6x+2x}{2}) \cos (\frac{6x-2x}{2}) + 2 \sin 4x \)
We simplify the angles:
= \( 2 \sin (4x) \cos (2x) + 2 \sin 4x \)
Now, we factor out the common term \( 2 \sin 4x \):
= \( 2 \sin 4x [\cos 2x + 1] \)
We use the double angle identity for cosine, \( \cos 2x + 1 = 2 \cos^2 x \):
= \( 2 \sin 4x (2 \cos^2 x) \)
We perform the multiplication:
= \( 4 \sin 4x \cos^2 x \)
This matches the Right Hand Side (R.H.S.), completing the proof.
In simple words: We took the left side of the equation and grouped the sine terms together. We used a formula to change the sum of two sines into a product. Then, we factored out a common part, \( 2 \sin 4x \). Finally, we used another identity to replace \( (\cos 2x + 1) \) with \( 2 \cos^2 x \), which then simplified to the right side of the equation.

Exam Tip: When faced with multiple sine or cosine terms, look for opportunities to group them for sum-to-product or product-to-sum formulas. Also, remember common double-angle identities like \( 1 + \cos 2x = 2 \cos^2 x \) for further simplification.

Question 15. \( \cot 4x(\sin 5x + \sin 3x) = \cot x(\sin 5x - \sin 3x) \)
Answer: We prove this identity by simplifying both the Left Hand Side (L.H.S.) and the Right Hand Side (R.H.S.) to a common expression.
**Left Hand Side (L.H.S.):**
L.H.S. = \( \cot 4x (\sin 5x + \sin 3x) \)
We use the sum-to-product identity for \( \sin C + \sin D \):
\( \sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2} \)
Here, \( C=5x \) and \( D=3x \).
L.H.S. = \( \cot 4x (2 \sin (\frac{5x+3x}{2}) \cos (\frac{5x-3x}{2})) \)
We simplify the angles:
= \( \cot 4x (2 \sin 4x \cos x) \)
Now, we express \( \cot 4x \) as \( \frac{\cos 4x}{\sin 4x} \):
= \( \frac{\cos 4x}{\sin 4x} (2 \sin 4x \cos x) \)
The \( \sin 4x \) terms cancel out:
= \( 2 \cos 4x \cos x \)
**Right Hand Side (R.H.S.):**
R.H.S. = \( \cot x (\sin 5x - \sin 3x) \)
We use the difference-to-product identity for \( \sin C - \sin D \):
\( \sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2} \)
Here, \( C=5x \) and \( D=3x \).
R.H.S. = \( \cot x (2 \cos (\frac{5x+3x}{2}) \sin (\frac{5x-3x}{2})) \)
We simplify the angles:
= \( \cot x (2 \cos 4x \sin x) \)
Now, we express \( \cot x \) as \( \frac{\cos x}{\sin x} \):
= \( \frac{\cos x}{\sin x} (2 \cos 4x \sin x) \)
The \( \sin x \) terms cancel out:
= \( 2 \cos x \cos 4x \)
Since L.H.S. = \( 2 \cos 4x \cos x \) and R.H.S. = \( 2 \cos x \cos 4x \), both sides are equal, and the identity is proven.
In simple words: To prove this, we worked on both the left and right sides separately. On the left, we changed the sum of sines into a product and then used \( \cot 4x = \frac{\cos 4x}{\sin 4x} \) to simplify. On the right, we changed the difference of sines into a product and used \( \cot x = \frac{\cos x}{\sin x} \) to simplify. Both sides simplified to the exact same expression, showing that the original equation is true.

Exam Tip: When an identity is complex, it's often effective to simplify both the L.H.S. and R.H.S. independently until they reach a common expression. Ensure you apply the correct sum-to-product and difference-to-product formulas for sine.

Prove the Following:

Question 16. \( \frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} = - \frac{\sin 2x}{\cos 10x} \)
Answer: We begin with the Left Hand Side (L.H.S.) of the equation:
L.H.S. = \( \frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} \)
We use the sum-to-product identities for the numerator and the denominator:
For the numerator, \( \cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2} \):
Numerator = \( -2 \sin (\frac{9x+5x}{2}) \sin (\frac{9x-5x}{2}) = -2 \sin (7x) \sin (2x) \)
For the denominator, \( \sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2} \):
Denominator = \( 2 \cos (\frac{17x+3x}{2}) \sin (\frac{17x-3x}{2}) = 2 \cos (10x) \sin (7x) \)
Now we substitute these back into the L.H.S. expression:
L.H.S. = \( \frac{-2 \sin 7x \sin 2x}{2 \cos 10x \sin 7x} \)
We cancel out the common terms \( 2 \) and \( \sin 7x \) from the numerator and denominator:
= \( \frac{- \sin 2x}{\cos 10x} \)
This matches the Right Hand Side (R.H.S.), completing the proof.
In simple words: We started with the left side of the equation. For the top part, we used a formula to change the subtraction of two cosines into a multiplication of two sines. For the bottom part, we used a formula to change the subtraction of two sines into a multiplication of a cosine and a sine. Then, we canceled out the parts that were the same on the top and bottom, which left us with the right side of the equation.

Exam Tip: Correctly applying the sum-to-product formulas for both numerator and denominator is crucial. Pay careful attention to the signs, especially when using the \( \cos C - \cos D \) identity, which introduces a negative sign.

Question 17. \( \frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x \)
Answer: We consider the Left Hand Side (L.H.S.) of the equation:
L.H.S. = \( \frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} \)
We use the sum-to-product identities for both the numerator and the denominator:
For the numerator, \( \sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2} \):
Numerator = \( 2 \sin (\frac{5x+3x}{2}) \cos (\frac{5x-3x}{2}) = 2 \sin (4x) \cos (x) \)
For the denominator, \( \cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2} \):
Denominator = \( 2 \cos (\frac{5x+3x}{2}) \cos (\frac{5x-3x}{2}) = 2 \cos (4x) \cos (x) \)
Now we substitute these back into the L.H.S. expression:
L.H.S. = \( \frac{2 \sin 4x \cos x}{2 \cos 4x \cos x} \)
We cancel out the common terms \( 2 \) and \( \cos x \) from the numerator and denominator:
= \( \frac{\sin 4x}{\cos 4x} \)
We recognize this as the identity for \( \tan 4x \):
= \( \tan 4x \)
This matches the Right Hand Side (R.H.S.), completing the proof.
In simple words: We started with the left side of the equation. We used special formulas to change both the sum of sines on top and the sum of cosines on the bottom into products. After simplifying and canceling out common parts, we were left with \( \frac{\sin 4x}{\cos 4x} \), which is the same as \( \tan 4x \), the right side of the equation.

Exam Tip: This type of problem is a classic application of sum-to-product identities. Make sure to apply the correct formula for sine sums and cosine sums. Simplification by cancelling common factors is a critical step.

Question 18. \( \frac{\sin x - \sin y}{\cos x + \cos y} = \tan \frac{x - y}{2} \)
Answer: We take the Left Hand Side (L.H.S.) of the equation:
L.H.S. = \( \frac{\sin x - \sin y}{\cos x + \cos y} \)
We use the sum-to-product identities for both the numerator and the denominator:
For the numerator, \( \sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2} \):
Numerator = \( 2 \cos (\frac{x+y}{2}) \sin (\frac{x-y}{2}) \)
For the denominator, \( \cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2} \):
Denominator = \( 2 \cos (\frac{x+y}{2}) \cos (\frac{x-y}{2}) \)
Now we substitute these back into the L.H.S. expression:
L.H.S. = \( \frac{2 \cos (\frac{x+y}{2}) \sin (\frac{x-y}{2})}{2 \cos (\frac{x+y}{2}) \cos (\frac{x-y}{2})} \)
We cancel out the common terms \( 2 \) and \( \cos (\frac{x+y}{2}) \) from the numerator and denominator:
= \( \frac{\sin (\frac{x-y}{2})}{\cos (\frac{x-y}{2})} \)
We recognize this as the identity for tangent:
= \( \tan (\frac{x-y}{2}) \)
This matches the Right Hand Side (R.H.S.), completing the proof.
In simple words: We began with the left side of the equation. We used formulas to change the difference of sines on top and the sum of cosines on the bottom into products. After simplifying and removing the parts that appeared in both the top and bottom, we were left with \( \frac{\sin (\frac{x-y}{2})}{\cos (\frac{x-y}{2})} \), which is the same as \( \tan (\frac{x-y}{2}) \), matching the right side.

Exam Tip: Be sure to distinguish between \( \sin C + \sin D \) and \( \sin C - \sin D \) formulas. The intermediate angle terms \( \frac{C+D}{2} \) and \( \frac{C-D}{2} \) are critical for correctly applying these identities.

Question 19. \( \frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x \)
Answer: We take the Left Hand Side (L.H.S.) of the equation:
L.H.S. = \( \frac{\sin x + \sin 3x}{\cos x + \cos 3x} \)
We use the sum-to-product identities for both the numerator and the denominator:
For the numerator, \( \sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2} \):
Numerator = \( 2 \sin (\frac{x+3x}{2}) \cos (\frac{x-3x}{2}) = 2 \sin (2x) \cos (-x) \)
For the denominator, \( \cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2} \):
Denominator = \( 2 \cos (\frac{x+3x}{2}) \cos (\frac{x-3x}{2}) = 2 \cos (2x) \cos (-x) \)
Now we substitute these back into the L.H.S. expression:
L.H.S. = \( \frac{2 \sin 2x \cos (-x)}{2 \cos 2x \cos (-x)} \)
We cancel out the common terms \( 2 \) and \( \cos (-x) \) from the numerator and denominator:
= \( \frac{\sin 2x}{\cos 2x} \)
We recognize this as the identity for \( \tan 2x \):
= \( \tan 2x \)
This matches the Right Hand Side (R.H.S.), completing the proof.
In simple words: We started with the left side of the equation. We used special formulas to change the sum of sines on top and the sum of cosines on the bottom into products. After simplifying the angles and canceling out common parts, including \( \cos(-x) \), we were left with \( \frac{\sin 2x}{\cos 2x} \), which is the same as \( \tan 2x \), matching the right side.

Exam Tip: Remember that \( \cos (-x) = \cos x \). This property is important for simplifying terms like \( \cos (\frac{x-3x}{2}) \) where the angle becomes negative. The technique of cancelling common factors after applying sum-to-product identities is fundamental.

Question 20. \( \frac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x} = 2 \sin x \)
Answer: We take the Left Hand Side (L.H.S.) of the equation:
L.H.S. = \( \frac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x} \)
We can factor out \( -1 \) from both the numerator and the denominator to make them resemble standard identities:
= \( \frac{-(\sin 3x - \sin x)}{-(\cos^2 x - \sin^2 x)} \)
The negative signs cancel out:
= \( \frac{\sin 3x - \sin x}{\cos^2 x - \sin^2 x} \)
We use the difference-to-product identity for the numerator, \( \sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2} \):
Numerator = \( 2 \cos (\frac{3x+x}{2}) \sin (\frac{3x-x}{2}) = 2 \cos (2x) \sin (x) \)
For the denominator, we use the double angle identity \( \cos 2x = \cos^2 x - \sin^2 x \):
Denominator = \( \cos 2x \)
Now we substitute these back into the L.H.S. expression:
L.H.S. = \( \frac{2 \cos 2x \sin x}{\cos 2x} \)
We cancel out the common term \( \cos 2x \) from the numerator and denominator:
= \( 2 \sin x \)
This matches the Right Hand Side (R.H.S.), completing the proof.
In simple words: We started with the left side of the equation. We made the numerator and denominator positive by factoring out minus signs. Then, we used a formula to change the difference of sines on top into a product. For the bottom, we recognized it as the formula for \( \cos 2x \). After simplifying and canceling out the \( \cos 2x \) terms, we were left with \( 2 \sin x \), which matches the right side.

Exam Tip: Always be on the lookout for common double-angle identities like \( \cos 2x = \cos^2 x - \sin^2 x \) or \( -(\cos^2 x - \sin^2 x) = - \cos 2x \). Manipulating the signs in the numerator and denominator can help align the expression with known formulas.

Prove the Following:

Question 21. \( \frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x \)
Answer: We begin with the Left Hand Side (L.H.S.) of the equation:
L.H.S. = \( \frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} \)
We rearrange the terms in both the numerator and denominator to group \( \cos 4x \) with \( \cos 2x \) and \( \sin 4x \) with \( \sin 2x \), as these terms can be simplified using sum-to-product identities:
= \( \frac{(\cos 4x + \cos 2x) + \cos 3x}{(\sin 4x + \sin 2x) + \sin 3x} \)
For the grouped cosine terms, we use \( \cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2} \):
\( \cos 4x + \cos 2x = 2 \cos (\frac{4x+2x}{2}) \cos (\frac{4x-2x}{2}) = 2 \cos 3x \cos x \)
For the grouped sine terms, we use \( \sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2} \):
\( \sin 4x + \sin 2x = 2 \sin (\frac{4x+2x}{2}) \cos (\frac{4x-2x}{2}) = 2 \sin 3x \cos x \)
Now, we substitute these simplified expressions back into the L.H.S.:
= \( \frac{2 \cos 3x \cos x + \cos 3x}{2 \sin 3x \cos x + \sin 3x} \)
We observe that \( \cos 3x \) is a common factor in the numerator and \( \sin 3x \) is a common factor in the denominator:
= \( \frac{\cos 3x (2 \cos x + 1)}{\sin 3x (2 \cos x + 1)} \)
We cancel out the common factor \( (2 \cos x + 1) \) from both the numerator and denominator:
= \( \frac{\cos 3x}{\sin 3x} \)
We recognize this as the identity for \( \cot 3x \):
= \( \cot 3x \)
This matches the Right Hand Side (R.H.S.), completing the proof.
In simple words: We started with the left side of the equation. We rearranged the terms in the top and bottom parts so we could use formulas to change sums of sines and cosines into products. After doing that, we noticed a common part in the top and bottom, which we canceled out. This left us with \( \frac{\cos 3x}{\sin 3x} \), which is the same as \( \cot 3x \), matching the right side.

Exam Tip: When dealing with three terms, strategically group two terms that simplify nicely using sum-to-product identities. Look for common factors after applying these identities, as they often lead to cancellations and simplification.

Question 22. \( \cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1 \)
Answer: We begin by using the angle relationship \( 3x = x + 2x \).
We take the cotangent of both sides of this equation:
\( \cot(3x) = \cot(x + 2x) \)
We apply the cotangent addition formula: \( \cot(A+B) = \frac{\cot A \cot B - 1}{\cot A + \cot B} \)
So, \( \cot 3x = \frac{\cot x \cot 2x - 1}{\cot x + \cot 2x} \)
Now, we cross-multiply:
\( \cot 3x (\cot x + \cot 2x) = \cot x \cot 2x - 1 \)
We distribute \( \cot 3x \) on the left side:
\( \cot x \cot 3x + \cot 2x \cot 3x = \cot x \cot 2x - 1 \)
To get the desired form, we move all terms involving cotangents to one side and the constant to the other. We rearrange the terms to match the equation we need to prove:
\( \cot x \cot 2x - \cot x \cot 3x - \cot 2x \cot 3x = 1 \)
Rearranging the order of terms to exactly match the question:
\( \cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1 \)
This proves the identity.
In simple words: We started with the simple idea that \( 3x \) is the same as \( x + 2x \). Then we used the cotangent formula for adding angles to expand \( \cot(x + 2x) \). After cross-multiplying and moving all the cotangent terms to one side, we ended up with 1 on the other side, which proves the original equation.

Exam Tip: For identities involving three different angles that sum up (or relate) to each other, consider starting from the sum/difference of angles. The identity \( \cot(A+B) \) is particularly useful here. Algebraic manipulation after applying the trigonometric identity is crucial.

Question 23. \( \tan 4x = \frac{4 \tan x(1-\tan^2 x)}{1-6 \tan^2 x+\tan^4 x} \)
Answer: We start with the Left Hand Side (L.H.S.) of the equation:
L.H.S. = \( \tan 4x \)
We can write \( \tan 4x \) as \( \tan 2(2x) \).
Using the double angle formula for tangent, \( \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} \), where \( A = 2x \):
= \( \frac{2 \tan 2x}{1 - \tan^2 2x} \)
Now, we substitute the formula for \( \tan 2x \) again, where \( A = x \):
\( \tan 2x = \frac{2 \tan x}{1 - \tan^2 x} \)
Substitute this into the expression for L.H.S.:
= \( \frac{2 (\frac{2 \tan x}{1 - \tan^2 x})}{1 - (\frac{2 \tan x}{1 - \tan^2 x})^2} \)
Simplify the numerator and the denominator separately:
Numerator = \( \frac{4 \tan x}{1 - \tan^2 x} \)
Denominator = \( 1 - \frac{4 \tan^2 x}{(1 - \tan^2 x)^2} \)
To combine terms in the denominator, find a common denominator:
Denominator = \( \frac{(1 - \tan^2 x)^2 - 4 \tan^2 x}{(1 - \tan^2 x)^2} \)
Expand \( (1 - \tan^2 x)^2 \):
Denominator = \( \frac{(1 - 2 \tan^2 x + \tan^4 x) - 4 \tan^2 x}{(1 - \tan^2 x)^2} \)
Denominator = \( \frac{1 - 6 \tan^2 x + \tan^4 x}{(1 - \tan^2 x)^2} \)
Now, substitute the simplified numerator and denominator back:
L.H.S. = \( \frac{\frac{4 \tan x}{1 - \tan^2 x}}{\frac{1 - 6 \tan^2 x + \tan^4 x}{(1 - \tan^2 x)^2}} \)
Multiply the numerator by the reciprocal of the denominator:
= \( \frac{4 \tan x}{1 - \tan^2 x} \times \frac{(1 - \tan^2 x)^2}{1 - 6 \tan^2 x + \tan^4 x} \)
Cancel out one factor of \( (1 - \tan^2 x) \):
= \( \frac{4 \tan x (1 - \tan^2 x)}{1 - 6 \tan^2 x + \tan^4 x} \)
This matches the Right Hand Side (R.H.S.), completing the proof.
In simple words: We started with \( \tan 4x \) on the left. We used the double-angle formula for tangent twice. First, we wrote \( \tan 4x \) as \( \tan(2 \times 2x) \). Then, inside that, we wrote \( \tan 2x \) using the formula again. After a lot of algebra to simplify the complex fraction, we ended up with the expression on the right side of the equation.

Exam Tip: This problem requires careful, step-by-step application of the double-angle formula for tangent. Algebraic skills, especially simplifying complex fractions and expanding squared binomials, are crucial here. Avoid combining steps to minimize errors.

Question 24. \( \cos 4x = 1 - 8 \sin^2 x \cos^2 x \)
Answer: We start with the Left Hand Side (L.H.S.) of the equation:
L.H.S. = \( \cos 4x \)
We can write \( \cos 4x \) as \( \cos 2(2x) \).
Using the double angle formula \( \cos 2A = 1 - 2 \sin^2 A \), where \( A = 2x \):
= \( 1 - 2 \sin^2 (2x) \)
Now, we use the double angle formula for sine, \( \sin 2x = 2 \sin x \cos x \):
Substitute this into the expression:
= \( 1 - 2 (2 \sin x \cos x)^2 \)
We square the term inside the parenthesis:
= \( 1 - 2 (4 \sin^2 x \cos^2 x) \)
We perform the multiplication:
= \( 1 - 8 \sin^2 x \cos^2 x \)
This matches the Right Hand Side (R.H.S.), completing the proof.
In simple words: We started with \( \cos 4x \) on the left. We used the double-angle rule to change \( \cos 4x \) to \( 1 - 2\sin^2(2x) \). Then, we replaced \( \sin 2x \) with \( 2 \sin x \cos x \), squared it, and multiplied by 2. This process quickly led us to the expression on the right side of the equation.

Exam Tip: This identity is straightforward if you correctly apply the double-angle formulas for cosine and sine. Choose the cosine double-angle formula that directly leads to sine terms (i.e., \( 1 - 2 \sin^2 A \)) to simplify the path to the desired R.H.S.

Question 25. \( \cos 6x = 32 \cos^6 x - 48 \cos^4 x + 18 \cos^2 x - 1 \)
Answer: We begin with the Left Hand Side (L.H.S.) of the equation:
L.H.S. = \( \cos 6x \)
We can express \( \cos 6x \) as \( \cos 3(2x) \).
Using the triple angle formula \( \cos 3A = 4 \cos^3 A - 3 \cos A \), where \( A = 2x \):
= \( 4 \cos^3 (2x) - 3 \cos (2x) \)
Now, we use the double angle formula \( \cos 2x = 2 \cos^2 x - 1 \):
Substitute this into the expression:
= \( 4 (2 \cos^2 x - 1)^3 - 3 (2 \cos^2 x - 1) \)
Expand the term \( (2 \cos^2 x - 1)^3 \) using the binomial expansion \( (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \), where \( a = 2 \cos^2 x \) and \( b = 1 \):
\( (2 \cos^2 x - 1)^3 = (2 \cos^2 x)^3 - 3(2 \cos^2 x)^2(1) + 3(2 \cos^2 x)(1)^2 - 1^3 \)
\( = 8 \cos^6 x - 3(4 \cos^4 x) + 6 \cos^2 x - 1 \)
\( = 8 \cos^6 x - 12 \cos^4 x + 6 \cos^2 x - 1 \)
Substitute this back into the expression for L.H.S.:
= \( 4 (8 \cos^6 x - 12 \cos^4 x + 6 \cos^2 x - 1) - 3 (2 \cos^2 x - 1) \)
Perform the multiplications:
= \( 32 \cos^6 x - 48 \cos^4 x + 24 \cos^2 x - 4 - 6 \cos^2 x + 3 \)
Combine the like terms (powers of \( \cos x \)):
= \( 32 \cos^6 x - 48 \cos^4 x + (24 - 6) \cos^2 x + (-4 + 3) \)
= \( 32 \cos^6 x - 48 \cos^4 x + 18 \cos^2 x - 1 \)
This matches the Right Hand Side (R.H.S.), completing the proof.
In simple words: We started with \( \cos 6x \) on the left. We first wrote it as \( \cos 3(2x) \) and used the triple-angle formula for cosine. Then, we replaced \( \cos 2x \) with its double-angle formula in terms of \( \cos x \). This involved expanding a cubed term and then carefully multiplying and adding all the parts. After combining all the terms, we got the exact expression shown on the right side of the equation.

Exam Tip: This problem is complex due to the multiple layers of identities and algebraic expansion. Be meticulous with the binomial expansion of \( (2 \cos^2 x - 1)^3 \) and careful with distributing and combining like terms. Any small algebraic error can lead to an incorrect result.