May 2026

Value of cos(π/65) cos(2π/65) cos(4π/65) cos(8π/65) cos(16π/65) cos(32π/65) Value of cos(π/65) cos(2π/65) cos(4π/65) cos(8π/65) cos(16π/65) cos(32π/65) Question Find the value of \[ \cos\frac{\pi}{65} \cos\frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} \] (a) \(\frac{1}{8}\) (b) \(\frac{1}{16}\) (c) \(\frac{1}{32}\) (d) none of these Solution Use the standard identity: \[ \sin(2^n x) = 2^n \sin x \cos x \cos 2x […]

(sec 8A – 1)/(sec 4A – 1) is Equal to What? MCQ Solution \((\sec 8A – 1)/(\sec 4A – 1)\) is Equal to What? Question \[ \frac{\sec 8A-1}{\sec 4A-1} \] is equal to (a) \(\dfrac{\tan 2A}{\tan 8A}\) (b) \(\dfrac{\tan 8A}{\tan 2A}\) (c) \(\dfrac{\cot 8A}{\cot 2A}\) (d) none of these Solution Use the identity: \[ \sec\theta-1

8 sin(x/8) cos(x/2) cos(x/4) cos(x/8) is Equal to What? MCQ Solution 8 sin(x/8) cos(x/2) cos(x/4) cos(x/8) is Equal to What? Question \(8\sin\frac{x}{8}\cos\frac{x}{2}\cos\frac{x}{4}\cos\frac{x}{8}\) is equal to (a) \(8\cos x\) (b) \(\cos x\) (c) \(8\sin x\) (d) \(\sin x\) Solution Use the identity: \[ 2\sin A\cos A=\sin 2A \] \[ 8\sin\frac{x}{8}\cos\frac{x}{8} =4\sin\frac{x}{4} \] \[ 4\sin\frac{x}{4}\cos\frac{x}{4} =2\sin\frac{x}{2} \]

Prove that cos(π/15) cos(2π/15) cos(3π/15) cos(4π/15) cos(5π/15) cos(6π/15) cos(7π/15) = 1/128 Prove that: \[ \cos\frac{\pi}{15} \cos\frac{2\pi}{15} \cos\frac{3\pi}{15} \cos\frac{4\pi}{15} \cos\frac{5\pi}{15} \cos\frac{6\pi}{15} \cos\frac{7\pi}{15} = \frac1{128} \] Solution Let \[ P= \cos\frac{\pi}{15} \cos\frac{2\pi}{15} \cos\frac{3\pi}{15} \cos\frac{4\pi}{15} \cos\frac{5\pi}{15} \cos\frac{6\pi}{15} \cos\frac{7\pi}{15} \] Using \[ \sin2\theta=2\sin\theta\cos\theta \] we successively write: \[ \sin\frac{2\pi}{15} = 2\sin\frac{\pi}{15}\cos\frac{\pi}{15} \] \[ \sin\frac{4\pi}{15} = 2\sin\frac{2\pi}{15}\cos\frac{2\pi}{15} \] \[ \sin\frac{8\pi}{15}

Value of Trigonometric Functions at Multiples and Submultiples of an angle – Multiple Choice Questions (MCQs) Exercise Solutions   Mark the correct alternative in each of the following: 8 sin(x/8) cos(x/2) cos(x/4) cos(x/8) is equal to(a) 8 cos x(b) cos x(c) 8 sin x(d) sin x Watch Solution (sec 8A − 1)/(sec 4A − 1) is

Prove that sin(π/5) sin(2π/5) sin(3π/5) sin(4π/5) = 5/16 Prove that: \[ \sin\frac{\pi}{5} \sin\frac{2\pi}{5} \sin\frac{3\pi}{5} \sin\frac{4\pi}{5} = \frac{5}{16} \] Solution Using \[ \sin(\pi-\theta)=\sin\theta \] we get \[ \sin\frac{3\pi}{5} = \sin\frac{2\pi}{5} \] \[ \sin\frac{4\pi}{5} = \sin\frac{\pi}{5} \] Therefore, \[ \sin\frac{\pi}{5} \sin\frac{2\pi}{5} \sin\frac{3\pi}{5} \sin\frac{4\pi}{5} = \sin^2\frac{\pi}{5} \sin^2\frac{2\pi}{5} \] Now use the identity \[ 2\sin A\sin B = \cos(A-B)-\cos(A+B)

Prove that cos36° cos42° cos60° cos78° = 1/16 Prove that: \[ \cos36^\circ\cos42^\circ\cos60^\circ\cos78^\circ = \frac1{16} \] Solution Since \[ \cos60^\circ=\frac12 \] therefore, \[ \cos36^\circ\cos42^\circ\cos60^\circ\cos78^\circ = \frac12\cos36^\circ\cos42^\circ\cos78^\circ \] Also, \[ \cos78^\circ=\sin12^\circ \] Hence, \[ = \frac12\cos36^\circ\cos42^\circ\sin12^\circ \] Using the identity \[ 2\sin A\cos B = \sin(A+B)+\sin(A-B) \] with \[ A=12^\circ,\qquad B=42^\circ \] \[ 2\sin12^\circ\cos42^\circ = \sin54^\circ+\sin(-30^\circ) \]

Prove that sin6° sin42° sin66° sin78° = 1/16 Prove that: \[ \sin6^\circ\sin42^\circ\sin66^\circ\sin78^\circ = \frac1{16} \] Solution Using \[ \sin78^\circ=\cos12^\circ \] \[ \sin66^\circ=\cos24^\circ \] Therefore, \[ \sin6^\circ\sin42^\circ\sin66^\circ\sin78^\circ = \sin6^\circ\sin42^\circ\cos24^\circ\cos12^\circ \] Now use the identity \[ 2\sin A\cos B = \sin(A+B)+\sin(A-B) \] For \[ A=42^\circ,\qquad B=24^\circ \] \[ 2\sin42^\circ\cos24^\circ = \sin66^\circ+\sin18^\circ \] \[ = \cos24^\circ+\sin18^\circ \] Hence,

Prove that cos6° cos42° cos66° cos78° = 1/16 Prove that: \[ \cos6^\circ\cos42^\circ\cos66^\circ\cos78^\circ = \frac1{16} \] Solution Using \[ \cos78^\circ=\sin12^\circ \] \[ \cos66^\circ=\sin24^\circ \] therefore, \[ \cos6^\circ\cos42^\circ\cos66^\circ\cos78^\circ = \cos6^\circ\cos42^\circ\sin24^\circ\sin12^\circ \] Now use \[ 2\sin A\cos B = \sin(A+B)+\sin(A-B) \] For \[ A=24^\circ,\qquad B=42^\circ \] \[ 2\sin24^\circ\cos42^\circ = \sin66^\circ+\sin(-18^\circ) \] \[ = \sin66^\circ-\sin18^\circ \] \[ = \cos24^\circ-\sin18^\circ

Prove that cos(π/15) cos(2π/15) cos(4π/15) cos(7π/15) = 1/16 Prove that: \[ \cos\frac{\pi}{15} \cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos\frac{7\pi}{15} = \frac1{16} \] Solution Using the identity \[ 2\sin A\cos A=\sin2A \] we successively write: \[ \sin\frac{2\pi}{15} = 2\sin\frac{\pi}{15}\cos\frac{\pi}{15} \] \[ \sin\frac{4\pi}{15} = 2\sin\frac{2\pi}{15}\cos\frac{2\pi}{15} \] \[ \sin\frac{8\pi}{15} = 2\sin\frac{4\pi}{15}\cos\frac{4\pi}{15} \] \[ \sin\frac{16\pi}{15} = 2\sin\frac{8\pi}{15}\cos\frac{8\pi}{15} \] Multiply all these equations: \[