CBSE Class 11 Mathematics Trigonometric Functions Worksheet Set 02

Trigonometric Functions MCQ Questions with Answers Class 11 Mathematics

Q. If y = cos x, then what is the maximum value of y?

Q.1 If tan A = √3. then what is tan 2A?

Q.2 Solve : 2 cos² x+3 sin x = 0

Q.3 Evaluate : sin(40°+θ)cos(10°+θ) - cos(40°+θ)sin(10°+θ)

Q.4 Prove that cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1. 

Q.5 Find the value of sin 150° + cos 300°.

Q.6 If in two circles, arcs of the same length subtend angles 75° and 120º at the centre, find the ratio of their radii.

Q.7 If in two circles, arcs of same length, subtend angles 120o and 150o at the centre, find the ratio of their radii.

Q.8 Write the value of tan 15°.

Q.9 Prove that : (cos x + cos y)² + (sin x - sin y)² = 4 cos² x+y/2

Q.10 Find the value of cos 55° + cos 125° + cos 300°.

Q.11 Find the value of sin 15°.

Q.12 Prove that: (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0. 

Q.13 A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second? 

Q.14 Prove that cos 19° - sin 19° / cos 19° + sin 19° = cot 74°

Q.15 If cot 2A = tan(n - 2)A, then what is A?

Q.16 Solve cos 2θ – cos θ = 0

Q.17 Write the general solution of cos x 1/2

Q.18 Prove that sec 8θ - 1 / sec 4θ - 1 = tan 8θ / tan 2θ

Q.19 Prove that cos2 A + cos² B - 2 cos A cos B cos (A+B) = sin² (A+B)

Q.20  Find the principal solutions of the equation tan x = √3.

Section A: (1 Mark)

Question 1. Find the degree measure for the following radian measure:
(i) \( \frac{7\pi}{12} \)
(ii) \( \frac{1}{4} \)
(iii) -3
Answer:
We convert radians to degrees using the relation: \( \text{Degree measure} = \text{Radian measure} \times \frac{180^\circ}{\pi} \).
(i) \( \frac{7\pi}{12} \text{ radians} = \frac{7\pi}{12} \times \frac{180^\circ}{\pi} = 7 \times 15^\circ = 105^\circ \).
(ii) \( \frac{1}{4} \text{ radian} = \frac{1}{4} \times \frac{180^\circ}{\pi} = \frac{45^\circ}{\pi} \)
Using \( \pi \approx \frac{22}{7} \):
\( = \frac{45 \times 7}{22} = \frac{315^\circ}{22} = 14\frac{7}{22}^\circ \).
We convert the fractional degree into minutes:
\( \frac{7}{22}^\circ = \frac{7 \times 60'}{22} = \frac{210'}{11} = 19\frac{1}{11}' \).
Now, convert the fractional minute into seconds:
\( \frac{1}{11}' = \frac{60''}{11} \approx 5'' \).
Thus, the degree measure is approximately \( 14^\circ 19' 5'' \).
(iii) \( -3 \text{ radians} = -3 \times \frac{180^\circ}{\pi} = -\frac{540^\circ}{\pi} \)
Using \( \pi \approx \frac{22}{7} \):
\( = -\frac{540 \times 7}{22} = -\frac{1890^\circ}{11} = -171\frac{9}{11}^\circ \).
We convert the fractional degree into minutes:
\( \frac{9}{11}^\circ = \frac{9 \times 60'}{11} = \frac{540'}{11} = 49\frac{1}{11}' \).
Convert the fractional minute into seconds:
\( \frac{1}{11}' = \frac{60''}{11} \approx 5'' \).
Thus, the degree measure is approximately \( -171^\circ 49' 5'' \).
In simple words: To change radians to degrees, multiply the value by 180 and divide by pi. Convert any leftover decimal or fraction into minutes and seconds by multiplying by 60.

Exam Tip: Be very careful with the successive divisions when converting fractional degrees into minutes and seconds. Do not round off until the final step.

Question 2. Find the radian measure for the following degree measure:
(i) \( -22^\circ 30' \)
(ii) \( 5^\circ 37' 30'' \)
(iii) \( -270^\circ \)
Answer:
We convert degrees to radians using the relation: \( \text{Radian measure} = \text{Degree measure} \times \frac{\pi}{180^\circ} \).
(i) \( -22^\circ 30' = -\left(22 + \frac{30}{60}\right)^\circ = -22.5^\circ \).
Converting to radians:
\( = -22.5 \times \frac{\pi}{180^\circ} = -\frac{45}{2} \times \frac{\pi}{180} = -\frac{\pi}{8} \text{ radians} \).
(ii) \( 5^\circ 37' 30'' = 5^\circ + \left(37 + \frac{30}{60}\right)' = 5^\circ + 37.5' = 5^\circ + \left(\frac{37.5}{60}\right)^\circ = 5^\circ + 0.625^\circ = 5.625^\circ \).
Converting to radians:
\( = 5.625 \times \frac{\pi}{180^\circ} = \frac{5625}{1000} \times \frac{\pi}{180} = \frac{9}{1.6} \times \frac{\pi}{180} = \frac{9}{160} \times \frac{\pi}{1.8} = \frac{\pi}{32} \text{ radians} \).
(iii) \( -270^\circ \):
Converting to radians:
\( = -270 \times \frac{\pi}{180^\circ} = -\frac{3\pi}{2} \text{ radians} \).
In simple words: To change degrees into radians, first make sure the entire angle is written as a decimal degree. Then, multiply it by pi and divide by 180.

Exam Tip: Keep your final radian answers as simplified fractions containing pi rather than converting them to decimals.

Question 3. The minute hand of a clock is 70cm long. How many centimetres does its tip move in 6 minutes?
Answer:
The minute hand completes one full revolution of \( 360^\circ \) (or \( 2\pi \) radians) in 60 minutes.
In 6 minutes, the angle \( \theta \) traced by the minute hand is:
\( \theta = \frac{6}{60} \times 2\pi = \frac{\pi}{5} \text{ radians} \).
The length of the minute hand represents the radius \( r = 70 \text{ cm} \).
Using the formula for the arc length, \( l = r\theta \):
\( l = 70 \times \frac{\pi}{5} = 14\pi \text{ cm} \).
Using \( \pi \approx \frac{22}{7} \):
\( l \approx 14 \times \frac{22}{7} = 44 \text{ cm} \).
In simple words: First find what fraction of a full circle the minute hand covers in 6 minutes. Then multiply that angle in radians by the length of the hand to find the distance.

Exam Tip: Remember that the arc length formula \( l = r\theta \) only works when the angle \( \theta \) is measured in radians, not degrees.

Question 4. Find the values of :
(i) \( \sin\left(-\frac{21\pi}{4}\right) \)
(ii) \( \cos\left(\frac{83\pi}{6}\right) \)
(iii) \( \tan\left(\frac{35\pi}{6}\right) \)
(iv) \( \cos(1230^\circ) \)
(v) \( \sin(1125^\circ) \)
(vi) \( \tan(2220^\circ) \)
Answer:
(i) \( \sin\left(-\frac{21\pi}{4}\right) = -\sin\left(\frac{21\pi}{4}\right) = -\sin\left(5\pi + \frac{\pi}{4}\right) \)
Since \( \sin(5\pi + \theta) = -\sin\theta \):
\( = -\left(-\sin\frac{\pi}{4}\right) = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} \).
(ii) \( \cos\left(\frac{83\pi}{6}\right) = \cos\left(14\pi - \frac{\pi}{6}\right) \)
Since \( \cos(2n\pi - \theta) = \cos\theta \):
\( = \cos\left(-\frac{\pi}{6}\right) = \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2} \).
(iii) \( \tan\left(\frac{35\pi}{6}\right) = \tan\left(6\pi - \frac{\pi}{6}\right) \)
Since \( \tan(n\pi - \theta) = -\tan\theta \):
\( = -\tan\frac{\pi}{6} = -\frac{1}{\sqrt{3}} \).
(iv) \( \cos(1230^\circ) = \cos(3 \times 360^\circ + 150^\circ) = \cos(150^\circ) \)
\( = \cos(180^\circ - 30^\circ) = -\cos 30^\circ = -\frac{\sqrt{3}}{2} \).
(v) \( \sin(1125^\circ) = \sin(3 \times 360^\circ + 45^\circ) = \sin(45^\circ) = \frac{1}{\sqrt{2}} \).
(vi) \( \tan(2220^\circ) = \tan(6 \times 360^\circ + 60^\circ) = \tan(60^\circ) = \sqrt{3} \).
In simple words: Simplify large angles by removing full multiples of 360 degrees or 2 pi radians, then check the final quadrant to get the correct sign.

Exam Tip: Write down intermediate steps showing the division of large angles into multiples of full rotations to ensure you do not commit simple calculation errors.

Section B: (4 Marks)

Question 5. A horse is tied to a post by a rope. If the horse moves along a circular path always keeping the rope tight and describes 88m when it has traced out \( 72^\circ \) at the centre, find the length of the rope.
Answer:
Let the length of the rope be \( r \), which acts as the radius of the circular path.
The angle traced at the centre is \( \theta = 72^\circ \).
Converting the angle to radians:
\( \theta = 72 \times \frac{\pi}{180} = \frac{2\pi}{5} \text{ radians} \).
The distance described along the path is the arc length \( l = 88 \text{ m} \).
Using the formula \( l = r\theta \):
\( r = \frac{l}{\theta} = \frac{88}{\frac{2\pi}{5}} = \frac{440}{2\pi} = \frac{220}{\pi} \).
Substituting \( \pi \approx \frac{22}{7} \):
\( r = \frac{220 \times 7}{22} = 70 \text{ m} \).
Thus, the length of the rope is \( 70 \text{ m} \).
In simple words: Convert the central angle to radians and divide the arc distance of 88 meters by this angle to find the radius (rope length).

Exam Tip: Clearly state the formula used and highlight your final units to secure full marks in long-answer questions.

Question 6. Prove that \( \cos x + \cos\left(\frac{2\pi}{3} + x\right) + \cos\left(\frac{2\pi}{3} - x\right) = 0 \)
Answer:
Taking the Left Hand Side (L.H.S.):
\( \text{L.H.S.} = \cos x + \cos\left(\frac{2\pi}{3} + x\right) + \cos\left(\frac{2\pi}{3} - x\right) \).
Using the identity \( \cos(A+B) + \cos(A-B) = 2\cos A\cos B \):
\( \cos\left(\frac{2\pi}{3} + x\right) + \cos\left(\frac{2\pi}{3} - x\right) = 2\cos\left(\frac{2\pi}{3}\right)\cos x \).
Substitute \( \cos\left(\frac{2\pi}{3}\right) = \cos\left(\pi - \frac{\pi}{3}\right) = -\cos\frac{\pi}{3} = -\frac{1}{2} \):
\( = 2\left(-\frac{1}{2}\right)\cos x = -\cos x \).
Now, substituting this back into our main expression:
\( \text{L.H.S.} = \cos x + (-\cos x) = 0 = \text{R.H.S.} \).
Hence proved.
In simple words: Group the last two cosine terms together and use a formula to simplify them. The result perfectly cancels out the first term, leaving zero.

Exam Tip: When proving trigonometric identities, always mention the specific identity you are applying at each step of your proof.

Question 7. Find the values of (i) \( \sin(\alpha + \beta) \) (ii) \( \cos(\alpha + \beta) \) (iii) \( \tan(\alpha + \beta) \), if \( \cos\alpha = -\frac{12}{13} \) and \( \cos\beta = -\frac{24}{25} \), where \( \alpha \) lies in the second quadrant and \( \beta \) in the third quadrant.
Answer:
First, we find the values of \( \sin\alpha \) and \( \sin\beta \).
Since \( \alpha \) lies in the second quadrant, \( \sin\alpha \) is positive:
\( \sin\alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - \left(-\frac{12}{13}\right)^2} = \sqrt{1 - \frac{144}{169}} = \frac{5}{13} \).
Since \( \beta \) lies in the third quadrant, \( \sin\beta \) is negative:
\( \sin\beta = -\sqrt{1 - \cos^2\beta} = -\sqrt{1 - \left(-\frac{24}{25}\right)^2} = -\sqrt{1 - \frac{576}{625}} = -\frac{7}{25} \).
Now, we calculate each part:
(i) \( \sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta \)
\( = \left(\frac{5}{13}\right)\left(-\frac{24}{25}\right) + \left(-\frac{12}{13}\right)\left(-\frac{7}{25}\right) = -\frac{120}{325} + \frac{84}{325} = -\frac{36}{325} \).
(ii) \( \cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta \)
\( = \left(-\frac{12}{13}\right)\left(-\frac{24}{25}\right) - \left(\frac{5}{13}\right)\left(-\frac{7}{25}\right) = \frac{288}{325} - \left(-\frac{35}{325}\right) = \frac{288 + 35}{325} = \frac{323}{325} \).
(iii) \( \tan(\alpha + \beta) = \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)} = \frac{-36/325}{323/325} = -\frac{36}{323} \).
In simple words: Find the sine of both angles using the given cosines, keeping the correct signs for each quadrant. Then, plug these values into the standard addition formulas.

Exam Tip: Carefully determine whether sines and cosines are positive or negative based on their quadrants before substituting them into the identities.

Question 8. Prove that \( \tan 70^\circ = \tan 20^\circ + 2\tan 50^\circ \)
Answer:
We can write \( 50^\circ = 70^\circ - 20^\circ \).
Taking the tangent on both sides:
\( \tan 50^\circ = \tan(70^\circ - 20^\circ) \).
Using the formula \( \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A\tan B} \):
\( \tan 50^\circ = \frac{\tan 70^\circ - \tan 20^\circ}{1 + \tan 70^\circ\tan 20^\circ} \).
Since \( \tan 70^\circ = \tan(90^\circ - 20^\circ) = \cot 20^\circ \), we have:
\( \tan 70^\circ \tan 20^\circ = \cot 20^\circ \tan 20^\circ = 1 \).
Substitute this back into our equation:
\( \tan 50^\circ = \frac{\tan 70^\circ - \tan 20^\circ}{1 + 1} = \frac{\tan 70^\circ - \tan 20^\circ}{2} \).
Cross-multiplying by 2:
\( 2\tan 50^\circ = \tan 70^\circ - \tan 20^\circ \).
Rearranging terms:
\( \tan 70^\circ = \tan 20^\circ + 2\tan 50^\circ \).
Hence proved.
In simple words: Express 50 degrees as 70 minus 20. Use the tangent subtraction identity and substitute the fact that tan 70 times tan 20 is equal to 1.

Exam Tip: Recognizing complementary angles like 70 and 20 allows you to quickly simplify terms like \( \tan 70^\circ \tan 20^\circ = 1 \).

Question 9. Solve the following equations:
(i) \( \sin 2\theta + \sin 4\theta + \sin 6\theta = 0 \)
(ii) \( 2\sin^2\theta + \sin^2 2\theta = 2 \)
(iii) \( \sqrt{3}\cos\theta + \sin\theta = \sqrt{2} \)
Answer:
(i) \( \sin 6\theta + \sin 2\theta + \sin 4\theta = 0 \).
Applying \( \sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \) to the first two terms:
\( 2\sin 4\theta\cos 2\theta + \sin 4\theta = 0 \)
\( \sin 4\theta(2\cos 2\theta + 1) = 0 \).
This gives two cases:
Case 1: \( \sin 4\theta = 0 \)

\( \implies 4\theta = n\pi \)

\( \implies \theta = \frac{n\pi}{4} \), where \( n \in \mathbb{Z} \).
Case 2: \( 2\cos 2\theta + 1 = 0 \)

\( \implies \cos 2\theta = -\frac{1}{2} = \cos\frac{2\pi}{3} \)

\( \implies 2\theta = 2k\pi \pm \frac{2\pi}{3} \)

\( \implies \theta = k\pi \pm \frac{\pi}{3} \), where \( k \in \mathbb{Z} \).

(ii) \( 2\sin^2\theta + \sin^2 2\theta = 2 \).
Using \( \sin 2\theta = 2\sin\theta\cos\theta \):
\( 2\sin^2\theta + 4\sin^2\theta\cos^2\theta = 2 \)
Divide the equation by 2:
\( \sin^2\theta + 2\sin^2\theta\cos^2\theta = 1 \)
\( \sin^2\theta(1 + 2\cos^2\theta) = 1 \).
Substitute \( \sin^2\theta = 1 - \cos^2\theta \):
\( (1 - \cos^2\theta)(1 + 2\cos^2\theta) = 1 \)
\( 1 + \cos^2\theta - 2\cos^4\theta = 1 \)
\( \cos^2\theta(1 - 2\cos^2\theta) = 0 \).
This gives two cases:
Case 1: \( \cos^2\theta = 0 \)

\( \implies \cos\theta = 0 \)

\( \implies \theta = (2n+1)\frac{\pi}{2} \), where \( n \in \mathbb{Z} \).
Case 2: \( 1 - 2\cos^2\theta = 0 \)

\( \implies \cos^2\theta = \frac{1}{2} \)

\( \implies \cos\theta = \pm\frac{1}{\sqrt{2}} \)

\( \implies \theta = m\pi \pm \frac{\pi}{4} \), where \( m \in \mathbb{Z} \).

(iii) \( \sqrt{3}\cos\theta + \sin\theta = \sqrt{2} \).
Divide both sides by \( \sqrt{(\sqrt{3})^2 + 1^2} = 2 \):
\( \frac{\sqrt{3}}{2}\cos\theta + \frac{1}{2}\sin\theta = \frac{\sqrt{2}}{2} \)
\( \cos\theta\cos\frac{\pi}{6} + \sin\theta\sin\frac{\pi}{6} = \frac{1}{\sqrt{2}} \).
Using the identity \( \cos(A-B) = \cos A\cos B + \sin A\sin B \):
\( \cos\left(\theta - \frac{\pi}{6}\right) = \cos\frac{\pi}{4} \).
The general solution is:
\( \theta - \frac{\pi}{6} = 2n\pi \pm \frac{\pi}{4} \)

\( \implies \theta = 2n\pi + \frac{\pi}{6} \pm \frac{\pi}{4} \), where \( n \in \mathbb{Z} \).
This yields:
\( \theta = 2n\pi + \frac{5\pi}{12} \) or \( \theta = 2n\pi - \frac{\pi}{12} \).
In simple words: Factorise equations using sum-to-product identities or basic trigonometric substitutions, then write down the general solutions for each resulting factor.

Exam Tip: Never cancel terms like \( \sin\theta \) or \( \cos\theta \) from both sides of an equation as it leads to a loss of valid general solutions.

Question 10. Prove that \( \frac{\sin\theta + \sin 3\theta + \sin 5\theta + \sin 7\theta}{\cos\theta + \cos 3\theta + \cos 5\theta + \cos 7\theta} = \tan 4\theta \)
Answer:
Rearranging the terms in the numerator and denominator:
\( \text{L.H.S.} = \frac{(\sin 7\theta + \sin\theta) + (\sin 5\theta + \sin 3\theta)}{(\cos 7\theta + \cos\theta) + (\cos 5\theta + \cos 3\theta)} \).
Applying the sum-to-product formulas:
\( \sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \),
\( \cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \).
So:
\( \sin 7\theta + \sin\theta = 2\sin 4\theta\cos 3\theta \),
\( \sin 5\theta + \sin 3\theta = 2\sin 4\theta\cos\theta \),
\( \cos 7\theta + \cos\theta = 2\cos 4\theta\cos 3\theta \),
\( \cos 5\theta + \cos 3\theta = 2\cos 4\theta\cos\theta \).
Substitute these back into our expression:
\( \text{L.H.S.} = \frac{2\sin 4\theta\cos 3\theta + 2\sin 4\theta\cos\theta}{2\cos 4\theta\cos 3\theta + 2\cos 4\theta\cos\theta} \)
Factoring out \( 2\sin 4\theta \) from the numerator and \( 2\cos 4\theta \) from the denominator:
\( = \frac{2\sin 4\theta(\cos 3\theta + \cos\theta)}{2\cos 4\theta(\cos 3\theta + \cos\theta)} \)
Cancelling the common factor \( 2(\cos 3\theta + \cos\theta) \):
\( = \frac{\sin 4\theta}{\cos 4\theta} = \tan 4\theta = \text{R.H.S.} \).
Hence proved.
In simple words: Pair the largest and smallest angles together, simplify them using sum-to-product identities, and then factor out the common terms to easily find the result.

Exam Tip: Look for symmetry in angles (like 7 and 1 averaging to 4, and 5 and 3 averaging to 4) to decide which terms to group together.

Section C: (6 Marks)

Question 11. Prove that \( \cos 5x = 16\cos^5 x - 20\cos^3 x + 5\cos x \)
Answer:
We write \( \cos 5x \) using multiple-angle expansion via the Binomial Theorem:
\( \cos 5x = \text{Re}\left[(\cos x + i\sin x)^5\right] \).
Expanding using the binomial formula:
\( \cos 5x = \cos^5 x - 10\cos^3 x\sin^2 x + 5\cos x\sin^4 x \).
Substituting \( \sin^2 x = 1 - \cos^2 x \):
\( \cos 5x = \cos^5 x - 10\cos^3 x(1 - \cos^2 x) + 5\cos x(1 - \cos^2 x)^2 \)
\( = \cos^5 x - 10\cos^3 x + 10\cos^5 x + 5\cos x(1 - 2\cos^2 x + \cos^4 x) \)
\( = 11\cos^5 x - 10\cos^3 x + 5\cos x - 10\cos^3 x + 5\cos^5 x \)
\( = 16\cos^5 x - 20\cos^3 x + 5\cos x \).
L.H.S. = R.H.S. Hence proved.
In simple words: Expand the cosine of a five-fold angle into powers of sine and cosine, and then rewrite all the sines using cosines to get the final formula.

Exam Tip: Ensure that you expand binomial terms like \( (1-\cos^2 x)^2 \) carefully, keeping track of algebraic signs to avoid term mistakes.

Question 12. Show that \( \cos^2\frac{\pi}{8} + \cos^2\frac{3\pi}{8} + \cos^2\frac{5\pi}{8} + \cos^2\frac{7\pi}{8} = 2 \)
Answer:
We know that:
\( \cos\frac{7\pi}{8} = \cos\left(\pi - \frac{\pi}{8}\right) = -\cos\frac{\pi}{8} \)

\( \implies \cos^2\frac{7\pi}{8} = \cos^2\frac{\pi}{8} \).
Similarly:
\( \cos\frac{5\pi}{8} = \cos\left(\pi - \frac{3\pi}{8}\right) = -\cos\frac{3\pi}{8} \)

\( \implies \cos^2\frac{5\pi}{8} = \cos^2\frac{3\pi}{8} \).
Substitute these back into the expression:
\( \text{L.H.S.} = 2\left(\cos^2\frac{\pi}{8} + \cos^2\frac{3\pi}{8}\right) \).
Also, we have:
\( \cos\frac{3\pi}{8} = \cos\left(\frac{\pi}{2} - \frac{\pi}{8}\right) = \sin\frac{\pi}{8} \).
Substitute this into the expression:
\( \text{L.H.S.} = 2\left(\cos^2\frac{\pi}{8} + \sin^2\frac{\pi}{8}\right) \).
Since \( \cos^2\theta + \sin^2\theta = 1 \):
\( \text{L.H.S.} = 2(1) = 2 = \text{R.H.S.} \).
Hence proved.
In simple words: Use the property that cosine squares of supplementary angles are equal. Then, change the remaining term to sine using complementary angle formulas.

Exam Tip: Group supplementary terms together first; this simplifies calculations and keeps the proof elegant and easy to read.

Question 13. Prove that \( \frac{\sin 8\theta\cos\theta - \sin 6\theta\cos 3\theta}{\cos 2\theta\cos\theta - \sin 3\theta\sin 4\theta} = \tan 2\theta \)
Answer:
Multiply the numerator and denominator of L.H.S. by 2:
\( \text{L.H.S.} = \frac{2\sin 8\theta\cos\theta - 2\sin 6\theta\cos 3\theta}{2\cos 2\theta\cos\theta - 2\sin 3\theta\sin 4\theta} \).
Using product-to-sum formulas:
\( 2\sin A\cos B = \sin(A+B) + \sin(A-B) \),
\( 2\cos A\cos B = \cos(A+B) + \cos(A-B) \),
\( 2\sin A\sin B = \cos(A-B) - \cos(A+B) \).
Substitute these formulas into our expression:
\( 2\sin 8\theta\cos\theta = \sin 9\theta + \sin 7\theta \),
\( 2\sin 6\theta\cos 3\theta = \sin 9\theta + \sin 3\theta \),
\( 2\cos 2\theta\cos\theta = \cos 3\theta + \cos\theta \),
\( 2\sin 4\theta\sin 3\theta = \cos\theta - \cos 7\theta \).
Now, rewrite the L.H.S. with these substitutions:
Numerator: \( (\sin 9\theta + \sin 7\theta) - (\sin 9\theta + \sin 3\theta) = \sin 7\theta - \sin 3\theta \).
Denominator: \( (\cos 3\theta + \cos\theta) - (\cos\theta - \cos 7\theta) = \cos 3\theta + \cos 7\theta \).
So:
\( \text{L.H.S.} = \frac{\sin 7\theta - \sin 3\theta}{\cos 7\theta + \cos 3\theta} \).
Using the sum-to-product identities:
\( \sin 7\theta - \sin 3\theta = 2\cos 5\theta\sin 2\theta \),
\( \cos 7\theta + \cos 3\theta = 2\cos 5\theta\cos 2\theta \).
Substitute these back:
\( \text{L.H.S.} = \frac{2\cos 5\theta\sin 2\theta}{2\cos 5\theta\cos 2\theta} = \frac{\sin 2\theta}{\cos 2\theta} = \tan 2\theta = \text{R.H.S.} \).
Hence proved.
In simple words: Multiply the top and bottom by 2 so you can convert products of trig functions into simpler sums. Cancel the identical terms and use a formula to simplify what remains.

Exam Tip: Be cautious with the minus sign in the subtraction identity \( 2\sin A\sin B = \cos(A-B) - \cos(A+B) \) as it is a common point of error.

Question 14. Prove that \( \sin 10^\circ \sin 50^\circ \sin 60^\circ \sin 70^\circ = \frac{\sqrt{3}}{16} \)
Answer:
We know that \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).
Substitute this value into L.H.S.:
\( \text{L.H.S.} = \frac{\sqrt{3}}{2} \left[ \sin 10^\circ \sin 50^\circ \sin 70^\circ \right] \).
Using the identity \( \sin\theta \sin(60^\circ-\theta) \sin(60^\circ+\theta) = \frac{1}{4}\sin 3\theta \).
Let \( \theta = 10^\circ \). The term inside the bracket becomes:
\( \sin 10^\circ \sin(60^\circ-10^\circ) \sin(60^\circ+10^\circ) = \sin 10^\circ \sin 50^\circ \sin 70^\circ \).
So:
\( \sin 10^\circ \sin 50^\circ \sin 70^\circ = \frac{1}{4}\sin(3 \times 10^\circ) = \frac{1}{4}\sin 30^\circ \).
Substitute \( \sin 30^\circ = \frac{1}{2} \):
\( = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \).
Now, substitute this value back into the L.H.S.:
\( \text{L.H.S.} = \frac{\sqrt{3}}{2} \times \frac{1}{8} = \frac{\sqrt{3}}{16} = \text{R.H.S.} \).
Hence proved.
In simple words: Substitute the known value of sin 60 first. Then use a special three-angle product identity with theta equal to 10 to quickly find the rest of the product.

Exam Tip: Utilizing identities like \( \sin\theta\sin(60-\theta)\sin(60+\theta) = \frac{1}{4}\sin 3\theta \) saves time compared to standard product-to-sum expansions.

Question 15. Prove that \( \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \frac{1}{16} \)
Answer:
We know that \( \cos 60^\circ = \frac{1}{2} \).
Substitute this value into L.H.S.:
\( \text{L.H.S.} = \frac{1}{2} \left[ \cos 20^\circ \cos 40^\circ \cos 80^\circ \right] \).
Using the identity \( \cos\theta \cos(60^\circ-\theta) \cos(60^\circ+\theta) = \frac{1}{4}\cos 3\theta \).
Let \( \theta = 20^\circ \). The term inside the bracket becomes:
\( \cos 20^\circ \cos(60^\circ-20^\circ) \cos(60^\circ+20^\circ) = \cos 20^\circ \cos 40^\circ \cos 80^\circ \).
So:
\( \cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{4}\cos(3 \times 20^\circ) = \frac{1}{4}\cos 60^\circ \).
Substitute \( \cos 60^\circ = \frac{1}{2} \):
\( = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \).
Now, substitute this value back into the L.H.S.:
\( \text{L.H.S.} = \frac{1}{2} \times \frac{1}{8} = \frac{1}{16} = \text{R.H.S.} \).
Hence proved.
In simple words: First write cos 60 as 1/2. Then use the special three-cosine product identity with theta as 20 degrees to quickly solve the remaining terms.

Exam Tip: Memorize the three-cosine product formula as it is highly useful for numerical simplification questions in competitive exams.

Question 16. Prove that \( \frac{(\cos\theta - \cos 3\theta)(\sin 8\theta + \sin 2\theta)}{(\sin 5\theta - \sin\theta)(\cos 4\theta - \cos 6\theta)} = 1 \)
Answer:
Using sum-to-product identities:
\( \cos A - \cos B = 2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{B-A}{2}\right) \),
\( \sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \),
\( \sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \).
Applying these to each term of the L.H.S.:
\( \cos\theta - \cos 3\theta = 2\sin 2\theta\sin\theta \),
\( \sin 8\theta + \sin 2\theta = 2\sin 5\theta\cos 3\theta \),
\( \sin 5\theta - \sin\theta = 2\cos 3\theta\sin 2\theta \),
\( \cos 4\theta - \cos 6\theta = 2\sin 5\theta\sin\theta \).
Substitute these back into our main fraction:
\( \text{L.H.S.} = \frac{(2\sin 2\theta\sin\theta)(2\sin 5\theta\cos 3\theta)}{(2\cos 3\theta\sin 2\theta)(2\sin 5\theta\sin\theta)} \).
All terms in the numerator and denominator cancel out perfectly:
\( \text{L.H.S.} = 1 = \text{R.H.S.} \).
Hence proved.
In simple words: Convert each sum or difference term into products of sine and cosine. You will find that the top and bottom terms match perfectly and cancel out to 1.

Exam Tip: Be cautious when simplifying terms like \( \cos 4\theta - \cos 6\theta \) to ensure you do not drop the correct sign in your product.

Question 17. Prove that \( \frac{\sec 8\theta - 1}{\sec 4\theta - 1} = \frac{\tan 8\theta}{\tan 2\theta} \)
Answer:
We convert the secant terms to cosines:
\( \text{L.H.S.} = \frac{\frac{1}{\cos 8\theta} - 1}{\frac{1}{\cos 4\theta} - 1} = \frac{\frac{1 - \cos 8\theta}{\cos 8\theta}}{\frac{1 - \cos 4\theta}{\cos 4\theta}} = \frac{1 - \cos 8\theta}{1 - \cos 4\theta} \times \frac{\cos 4\theta}{\cos 8\theta} \).
Using the half-angle identity \( 1 - \cos 2A = 2\sin^2 A \):
\( 1 - \cos 8\theta = 2\sin^2 4\theta \),
\( 1 - \cos 4\theta = 2\sin^2 2\theta \).
Substitute these back into L.H.S.:
\( \text{L.H.S.} = \frac{2\sin^2 4\theta}{2\sin^2 2\theta} \times \frac{\cos 4\theta}{\cos 8\theta} = \frac{\sin 4\theta\cos 4\theta}{\cos 8\theta} \times \frac{\sin 4\theta}{\sin^2 2\theta} \).
Multiply and divide by 2 inside the expression:
\( = \frac{2\sin 4\theta\cos 4\theta}{2\cos 8\theta} \times \frac{\sin 4\theta}{\sin^2 2\theta} = \frac{\sin 8\theta}{2\cos 8\theta} \times \frac{2\sin 2\theta\cos 2\theta}{\sin^2 2\theta} \).
Simplify the terms:
\( = \tan 8\theta \times \frac{\cos 2\theta}{\sin 2\theta} = \frac{\tan 8\theta}{\tan 2\theta} = \text{R.H.S.} \).
Hence proved.
In simple words: Change all secant terms to cosines. Then, simplify using double angle formulas to turn the sines and cosines back into tangents.

Exam Tip: Expressing secant as cosine is usually the best first step in proofs that involve multiple angles.

Question 18. Prove that \( \cos^2 A + \cos^2(A + 120^\circ) + \cos^2(A - 120^\circ) = \frac{3}{2} \)
Answer:
Using the half-angle identity \( \cos^2\theta = \frac{1 + \cos 2\theta}{2} \):
\( \text{L.H.S.} = \frac{1 + \cos 2A}{2} + \frac{1 + \cos(2A + 240^\circ)}{2} + \frac{1 + \cos(2A - 240^\circ)}{2} \)
\( = \frac{1}{2} \left[ 3 + \cos 2A + \cos(2A + 240^\circ) + \cos(2A - 240^\circ) \right] \).
Using the identity \( \cos(X+Y) + \cos(X-Y) = 2\cos X\cos Y \):
\( \cos(2A + 240^\circ) + \cos(2A - 240^\circ) = 2\cos 2A \cos 240^\circ \).
Since \( \cos 240^\circ = \cos(180^\circ + 60^\circ) = -\cos 60^\circ = -\frac{1}{2} \):
\( 2\cos 2A \cos 240^\circ = 2\cos 2A \left(-\frac{1}{2}\right) = -\cos 2A \).
Substitute this value back into the L.H.S. expression:
\( \text{L.H.S.} = \frac{1}{2} \left[ 3 + \cos 2A - \cos 2A \right] = \frac{3}{2} = \text{R.H.S.} \).
Hence proved.
In simple words: Convert the squared cosine terms to single power cosines using double-angle identities, then use a sum-to-product formula to simplify the remaining terms.

Exam Tip: Be sure to factor out the common fraction \( \frac{1}{2} \) at the beginning to avoid algebraic clutter during steps.

Question 19. Prove that \( \tan 4\theta = \frac{4\tan\theta(1 - \tan^2\theta)}{1 - 6\tan^2\theta + \tan^4\theta} \)
Answer:
We use the double-angle identity \( \tan 2A = \frac{2\tan A}{1 - \tan^2 A} \).
Let \( A = 2\theta \):
\( \tan 4\theta = \frac{2\tan 2\theta}{1 - \tan^2 2\theta} \).
Substitute \( \tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} \):
\( \tan 4\theta = \frac{2\left(\frac{2\tan\theta}{1 - \tan^2\theta}\right)}{1 - \left(\frac{2\tan\theta}{1 - \tan^2\theta}\right)^2} \)
\( = \frac{\frac{4\tan\theta}{1 - \tan^2\theta}}{\frac{(1 - \tan^2\theta)^2 - 4\tan^2\theta}{(1 - \tan^2\theta)^2}} \)
\( = \frac{4\tan\theta}{1 - \tan^2\theta} \times \frac{(1 - \tan^2\theta)^2}{1 - 2\tan^2\theta + \tan^4\theta - 4\tan^2\theta} \)
\( = \frac{4\tan\theta(1 - \tan^2\theta)}{1 - 6\tan^2\theta + \tan^4\theta} = \text{R.H.S.} \).
Hence proved.
In simple words: Write tan 4-theta as tan of 2 times 2-theta, and apply the double angle formula twice. Simplify the resulting fractions to arrive at the solution.

Exam Tip: Carefully expand binomials in the denominator of rational fractions to make sure you do not get incorrect coefficients.

Question 20. If \( 0 \leq x \leq 2\pi \), find \( \sin\frac{x}{2} \), \( \cos\frac{x}{2} \) and \( \tan\frac{x}{2} \), when
(i) \( \tan x = -\frac{4}{3} \), ( x lies in quadrant II
(ii) \( \cos x = -\frac{1}{3} \), ( x ) lies in quadrant III
(iii) \( \sin x = \frac{\sqrt{5}}{3} \), ( x ) lies in quadrant II
Answer:
Since \( 0 \leq x \leq 2\pi \):
If \( x \) is in quadrant II (\( \frac{\pi}{2} < x < \pi \)), then \( \frac{x}{2} \) lies in quadrant I (\( \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2} \)), where all half-angle trigonometric functions are positive.
If \( x \) is in quadrant III (\( \pi < x < \frac{3\pi}{2} \)), then \( \frac{x}{2} \) lies in quadrant II (\( \frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4} \)), where sine is positive, cosine is negative, and tangent is negative.

(i) \( \tan x = -\frac{4}{3} \), \( x \) lies in quadrant II.
Since \( x \) is in quadrant II, \( \cos x \) is negative:
\( \sec^2 x = 1 + \tan^2 x = 1 + \frac{16}{9} = \frac{25}{9} \)

\( \implies \sec x = -\frac{5}{3} \)

\( \implies \cos x = -\frac{3}{5} \).
Since \( \frac{x}{2} \) is in quadrant I, sines and cosines are positive:
\( \sin\frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}} = \sqrt{\frac{1 - (-3/5)}{2}} = \sqrt{\frac{8/5}{2}} = \frac{2}{\sqrt{5}} \).
\( \cos\frac{x}{2} = \sqrt{\frac{1 + \cos x}{2}} = \sqrt{\frac{1 + (-3/5)}{2}} = \sqrt{\frac{2/5}{2}} = \frac{1}{\sqrt{5}} \).
\( \tan\frac{x}{2} = \frac{\sin(x/2)}{\cos(x/2)} = \frac{2/\sqrt{5}}{1/\sqrt{5}} = 2 \).

(ii) \( \cos x = -\frac{1}{3} \), ( x ) lies in quadrant III.
Since \( \frac{x}{2} \) is in quadrant II, \( \sin\frac{x}{2} \) is positive, and \( \cos\frac{x}{2} \) is negative:
\( \sin\frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}} = \sqrt{\frac{1 - (-1/3)}{2}} = \sqrt{\frac{4/3}{2}} = \frac{\sqrt{2}}{\sqrt{3}} \).
\( \cos\frac{x}{2} = -\sqrt{\frac{1 + \cos x}{2}} = -\sqrt{\frac{1 + (-1/3)}{2}} = -\sqrt{\frac{2/3}{2}} = -\frac{1}{\sqrt{3}} \).
\( \tan\frac{x}{2} = \frac{\sin(x/2)}{\cos(x/2)} = \frac{\sqrt{2}/\sqrt{3}}{-1/\sqrt{3}} = -\sqrt{2} \).

(iii) \( \sin x = \frac{\sqrt{5}}{3} \), \( x \) lies in quadrant II.
Since \( x \) is in quadrant II, \( \cos x \) is negative:
\( \cos x = -\sqrt{1 - \sin^2 x} = -\sqrt{1 - \frac{5}{9}} = -\frac{2}{3} \).
Since \( \frac{x}{2} \) is in quadrant I, sines and cosines are positive:
\( \sin\frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}} = \sqrt{\frac{1 - (-2/3)}{2}} = \sqrt{\frac{5/3}{2}} = \sqrt{\frac{5}{6}} \).
\( \cos\frac{x}{2} = \sqrt{\frac{1 + \cos x}{2}} = \sqrt{\frac{1 + (-2/3)}{2}} = \sqrt{\frac{1/3}{2}} = \frac{1}{\sqrt{6}} \).
\( \tan\frac{x}{2} = \frac{\sin(x/2)}{\cos(x/2)} = \frac{\sqrt{5}/\sqrt{6}}{1/\sqrt{6}} = \sqrt{5} \).
In simple words: First find the value of cos x using standard trigonometric formulas, then use the half-angle identities to calculate the sine, cosine, and tangent values while minding the quadrant of the half angle.

Exam Tip: Be sure to establish the correct quadrant for \( \frac{x}{2} \) first, as it dictates the signs of all half-angle ratios.