Ravi Kant Kumar

Prove that cos(π/65) cos(2π/65) cos(4π/65) cos(8π/65) cos(16π/65) cos(32π/65) = 1/64 Prove that: \[ \cos\frac{\pi}{65} \cos\frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} = \frac{1}{64} \] Question Prove that \[ \cos\frac{\pi}{65} \cos\frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} = \frac{1}{64} \] Solution Using the identity \[ 2\sin\theta\cos\theta=\sin2\theta \] Start with \[ \cos\frac{\pi}{65} \cos\frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} \] Multiply and divide

Prove that cos(π/5) cos(2π/5) cos(4π/5) cos(8π/5) = -1/16 Prove that: cos(π/5) cos(2π/5) cos(4π/5) cos(8π/5) = -1/16 Question Prove that \[ \cos\frac{\pi}{5} \cos\frac{2\pi}{5} \cos\frac{4\pi}{5} \cos\frac{8\pi}{5} = -\frac{1}{16} \] Solution Using the identity \[ 2\sin\theta\cos\theta=\sin2\theta \] Start with \[ \cos\frac{\pi}{5} \cos\frac{2\pi}{5} \cos\frac{4\pi}{5} \cos\frac{8\pi}{5} \] Multiply and divide by \[ \sin\frac{\pi}{5} \] \[ = \frac{ \sin\frac{\pi}{5} \cos\frac{\pi}{5} \cos\frac{2\pi}{5}

Prove that cos(2π/15) cos(4π/15) cos(8π/15) cos(16π/15) = 1/16 Prove that: cos(2π/15) cos(4π/15) cos(8π/15) cos(16π/15) = 1/16 Question Prove that \[ \cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos\frac{8\pi}{15} \cos\frac{16\pi}{15} = \frac{1}{16} \] Solution Using the identity \[ 2\sin\theta\cos\theta=\sin2\theta \] Start with \[ \cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos\frac{8\pi}{15} \cos\frac{16\pi}{15} \] Multiply and divide by \[ \sin\frac{2\pi}{15} \] \[ = \frac{ \sin\frac{2\pi}{15} \cos\frac{2\pi}{15} \cos\frac{4\pi}{15}

Prove that cos 7° cos 14° cos 28° cos 56° = sin 68° / 16 cos 83° Prove that: cos 7° cos 14° cos 28° cos 56° = sin 68° / 16 cos 83° Question Prove that \[ \cos 7^\circ \cos 14^\circ \cos 28^\circ \cos 56^\circ = \frac{\sin 68^\circ}{16\cos 83^\circ} \] Solution We use the

If tan A = 1/7 and tan B = 1/3, Show that cos 2A = sin 4B If tan A = 1/7 and tan B = 1/3, Show that cos 2A = sin 4B Question If \[ \tan A=\frac{1}{7} \quad \text{and} \quad \tan B=\frac{1}{3}, \] show that \[ \cos 2A=\sin 4B. \] Solution Using the

If tan x = b/a, Find the Value of √((a+b)/(a-b)) + √((a-b)/(a+b)) If tan x = b/a, Find the Value of √((a+b)/(a-b)) + √((a-b)/(a+b)) Question If \[ \tan x = \frac{b}{a}, \] then find the value of \[ \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}. \] Solution Given, \[ \tan x = \frac{b}{a} \] Using the identity, \[ \frac{1+\tan x}{1-\tan x}=\tan\left(\frac{\pi}{4}+x\right)

If sin x = 4/5 and 0 < x < π/2, Find the Value of sin 4x If sin x = 4/5 and 0 < x < π/2, Find the Value of sin 4x Question If \[ \sin x = \frac{4}{5} \] and \[ 0 < x < \frac{\pi}{2}, \] then find the value of

If cos x = 4/5 and x is acute, find tan 2x If \[ \cos x=\frac45 \] and \(x\) is acute, find \[ \tan2x \] Solution: Given, \[ \cos x=\frac45 \] Using \[ \sin^2x+\cos^2x=1 \] we get \[ \sin^2x = 1-\left(\frac45\right)^2 \] \[ = 1-\frac{16}{25} \] \[ = \frac9{25} \] \[ \sin x=\pm\frac35 \] Since

If sin x = 1/4 and x lies in the IInd quadrant, find cos(x/2), sin(x/2) and tan(x/2) If \[ 0\le x\le \pi \] and \(x\) lies in the IInd quadrant such that \[ \sin x=\frac14 \] find the values of \[ \cos\frac{x}{2},\quad \sin\frac{x}{2},\quad \tan\frac{x}{2} \] Solution: Given, \[ \sin x=\frac14 \] Using \[ \sin^2x+\cos^2x=1 \]

If sin x = √5/3 and x lies in the IInd quadrant, find cos(x/2), sin(x/2) and tan(x/2) If \[ \sin x=\frac{\sqrt5}{3} \] and \(x\) lies in the IInd quadrant, find the values of \[ \cos\frac{x}{2},\quad \sin\frac{x}{2},\quad \tan\frac{x}{2} \] Solution: Given, \[ \sin x=\frac{\sqrt5}{3} \] Using \[ \sin^2x+\cos^2x=1 \] we get \[ \cos^2x = 1-\left(\frac{\sqrt5}{3}\right)^2 \]