May 2026

Find the Value of sin(3π/10) Question: \[ \sin\frac{3\pi}{10} \] Solution First convert the angle into degrees: \[ \frac{3\pi}{10} = \frac{3\times180^\circ}{10} = 54^\circ \] Therefore, \[ \sin\frac{3\pi}{10} = \sin54^\circ \] Using the standard exact value, \[ \sin54^\circ = \cos36^\circ \] and \[ \cos36^\circ = \frac{1+\sqrt5}{4} \] Hence, \[ \sin\frac{3\pi}{10} = \frac{1+\sqrt5}{4} \] Answer \[ \boxed{\frac{1+\sqrt5}{4}} \] […]

Find the Value of cos(π/5) cos(2π/5) cos(4π/5) cos(8π/5) Question: \[ \cos\frac{\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}\cos\frac{8\pi}{5} \] Solution Using periodicity of cosine, \[ \cos\frac{8\pi}{5} = \cos\left(2\pi-\frac{2\pi}{5}\right) = \cos\frac{2\pi}{5} \] Also, \[ \cos\frac{4\pi}{5} = \cos\left(\pi-\frac{\pi}{5}\right) = -\cos\frac{\pi}{5} \] Therefore, \[ P= \cos\frac{\pi}{5}\cos\frac{2\pi}{5} \left(-\cos\frac{\pi}{5}\right) \cos\frac{2\pi}{5} \] \[ =- \cos^2\frac{\pi}{5} \cos^2\frac{2\pi}{5} \] Now use the standard identity \[ \cos\frac{\pi}{5}\cos\frac{2\pi}{5} = \frac{1}{4} \] Hence,

If tan x = 1/7, tan y = 1/3 and cos 2x = sin ky, Then Find k Question: \[ \tan x=\frac{1}{7}, \qquad \tan y=\frac{1}{3} \] and \[ \cos 2x=\sin ky \] Find the value of \(k\). Solution Using the identity \[ \cos 2x=\frac{1-\tan^2x}{1+\tan^2x} \] Substituting \(\tan x=\frac{1}{7}\), \[ \cos 2x = \frac{1-\frac{1}{49}} {1+\frac{1}{49}} =

If tan θ = a/b, Then Find a sin 2θ + b cos 2θ Question: \[ \text{If } \tan\theta=\frac{a}{b}, \text{ then find } a\sin2\theta+b\cos2\theta. \] Solution Given \[ \tan\theta=\frac{a}{b} \] Using double-angle identities, \[ \sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta} \] \[ =\frac{2(a/b)}{1+a^2/b^2} =\frac{2ab}{a^2+b^2} \] Also, \[ \cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta} \] \[ =\frac{1-a^2/b^2}{1+a^2/b^2} =\frac{b^2-a^2}{a^2+b^2} \] Therefore, \[ a\sin2\theta+b\cos2\theta \] \[ = a\left(\frac{2ab}{a^2+b^2}\right)

If tan θ = t, Then Find tan 2θ + sec 2θ Question: \[ \text{If } \tan\theta=t, \text{ then find } \tan 2\theta+\sec 2\theta. \] Solution Using the identities \[ \tan 2\theta=\frac{2t}{1-t^2} \] and \[ \sec 2\theta=\frac{1+t^2}{1-t^2} \] Therefore, \[ \tan 2\theta+\sec 2\theta = \frac{2t}{1-t^2} + \frac{1+t^2}{1-t^2} \] \[ = \frac{1+2t+t^2}{1-t^2} \] \[ = \frac{(1+t)^2}{(1-t)(1+t)}

Find the Value of (cot x – tan x)/cot 2x Question: \[ \frac{\cot x-\tan x}{\cot 2x} \] Solution First simplify the numerator: \[ \cot x-\tan x = \frac{\cos x}{\sin x} -\frac{\sin x}{\cos x} \] \[ = \frac{\cos^2 x-\sin^2 x} {\sin x\cos x} \] \[ = \frac{\cos 2x} {\sin x\cos x} \] Using \[ \sin 2x=2\sin

Find the Value of (1 + cos π/8)(1 + cos 3π/8)(1 + cos 5π/8)(1 + cos 7π/8) Question: \[ (1+\cos \frac{\pi}{8}) (1+\cos \frac{3\pi}{8}) (1+\cos \frac{5\pi}{8}) (1+\cos \frac{7\pi}{8}) \] Solution Use the identity \[ 1+\cos\theta = 2\cos^2\frac{\theta}{2} \] Therefore, \[ P=(1+\cos \frac{\pi}{8}) (1+\cos \frac{3\pi}{8}) (1+\cos \frac{5\pi}{8}) (1+\cos \frac{7\pi}{8}) \] \[ =16 \cos^2\frac{\pi}{16} \cos^2\frac{3\pi}{16} \cos^2\frac{5\pi}{16} \cos^2\frac{7\pi}{16} \]

If cos⁶x + sin⁶x + k sin²2x = 1, Then Find k Question: \[ \cos^6 x+\sin^6 x+k\sin^2 2x=1 \] Find the value of \(k\). Solution Use the identity \[ a^3+b^3=(a+b)^3-3ab(a+b) \] Let \[ a=\cos^2 x,\qquad b=\sin^2 x \] Then \[ \cos^6 x+\sin^6 x =(\cos^2 x+\sin^2 x)^3 -3\sin^2 x\cos^2 x(\sin^2 x+\cos^2 x) \] \[ =1-3\sin^2 x\cos^2

Find the Least Value of 2 sin²θ + 3 cos²θ Question: \[ 2\sin^2\theta+3\cos^2\theta \] Find its least value. Solution Using the identity \[ \sin^2\theta+\cos^2\theta=1 \] Write \[ 2\sin^2\theta+3\cos^2\theta \] \[ =2(\sin^2\theta+\cos^2\theta)+\cos^2\theta \] \[ =2+\cos^2\theta \] Since \[ 0\le \cos^2\theta \le 1 \] the minimum value of \(\cos^2\theta\) is \(0\). Therefore, \[ 2+\cos^2\theta \ge 2 \]

Find the Value of cos²48° – sin²12° Question: \[ \cos^2 48^\circ-\sin^2 12^\circ \] Solution Use the identity \[ \sin^2\theta=\cos^2(90^\circ-\theta) \] Therefore, \[ \sin^2 12^\circ=\cos^2 78^\circ \] Hence, \[ \cos^2 48^\circ-\sin^2 12^\circ = \cos^2 48^\circ-\cos^2 78^\circ \] Using \[ \cos^2 A=\frac{1+\cos 2A}{2} \] \[ =\frac{1+\cos96^\circ}{2} -\frac{1+\cos156^\circ}{2} \] \[ =\frac{\cos96^\circ-\cos156^\circ}{2} \] Using the identity \[ \cos C-\cos