Educational

The Value of cos 52° + cos 68° + cos 172° \( \cos52^\circ+\cos68^\circ+\cos172^\circ \) Options: (a) \(0\) (b) \(1\) (c) \(2\) (d) \( \frac32 \) Solution: Using, \[ \cos(180^\circ-\theta)=-\cos\theta \] \[ \cos172^\circ=-\cos8^\circ \] Therefore, \[ \cos52^\circ+\cos68^\circ+\cos172^\circ \] \[ =\cos52^\circ+\cos68^\circ-\cos8^\circ \] Using identity, \[ \cos A+\cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \] \[ = 2\cos60^\circ\cos8^\circ-\cos8^\circ \] \[ = […]

If sin 2θ + sin 2ϕ = 1/2 and cos 2θ + cos 2ϕ = 3/2, then find cos²(θ − ϕ) If \( \sin2\theta+\sin2\phi=\frac12 \) and \( \cos2\theta+\cos2\phi=\frac32 \), then find \( \cos^2(\theta-\phi) \) Options: (a) \( \frac38 \) (b) \( \frac58 \) (c) \( \frac34 \) (d) \( \frac54 \) Solution: Using, \[ \sin

sin 163° cos 347° + sin 73° sin 167° \( \sin163^\circ\cos347^\circ+\sin73^\circ\sin167^\circ \) Options: (a) \(0\) (b) \(\frac12\) (c) \(1\) (d) none of these Solution: \[ =\sin163^\circ\cos347^\circ+\sin73^\circ\sin167^\circ \] \[ =\sin17^\circ\cos13^\circ+\sin73^\circ\sin13^\circ \] \[ =\sin17^\circ\cos13^\circ+\cos17^\circ\sin13^\circ \] Using identity, \[ \sin A\cos B+\cos A\sin B=\sin(A+B) \] \[ =\sin(17^\circ+13^\circ) \] \[ =\sin30^\circ \] \[ =\frac12 \] \[ \boxed{\frac12} \] Correct

cos 40° + cos 80° + cos 160° + cos 240° \( \cos40^\circ+\cos80^\circ+\cos160^\circ+\cos240^\circ \) Options: (a) \(0\) (b) \(1\) (c) \(\frac12\) (d) \(-\frac12\) Solution: We know that, \[ \cos(180^\circ-\theta)=-\cos\theta \] and \[ \cos(180^\circ+\theta)=-\cos\theta \] Therefore, \[ \cos160^\circ=-\cos20^\circ \] and \[ \cos240^\circ=-\cos60^\circ=-\frac12 \] Hence, \[ \cos40^\circ+\cos80^\circ+\cos160^\circ+\cos240^\circ \] \[ =\cos40^\circ+\cos80^\circ-\cos20^\circ-\frac12 \] Using identity, \[ \cos20^\circ=\cos40^\circ+\cos80^\circ \] Substituting,

Prove that tan x tan(π/3 − x) tan(π/3 + x) = tan 3x Prove that: \( \tan x \tan\left(\frac{\pi}{3}-x\right)\tan\left(\frac{\pi}{3}+x\right)=\tan3x \) Solution: \[ \tan x \tan\left(\frac{\pi}{3}-x\right)\tan\left(\frac{\pi}{3}+x\right) \] Using, \[ \tan\left(\frac{\pi}{3}-x\right)=\frac{\sqrt3-\tan x}{1+\sqrt3\tan x} \] and \[ \tan\left(\frac{\pi}{3}+x\right)=\frac{\sqrt3+\tan x}{1-\sqrt3\tan x} \] \[ =\tan x\cdot \frac{\sqrt3-\tan x}{1+\sqrt3\tan x} \cdot \frac{\sqrt3+\tan x}{1-\sqrt3\tan x} \] \[ =\tan x\cdot \frac{3-\tan^2x}{1-3\tan^2x} \] \[

If α + β = π/2, Show that the Maximum Value of cos α cos β is 1/2 If \( \alpha+\beta=\frac{\pi}{2} \), show that the maximum value of \( \cos\alpha\cos\beta \) is \( \frac12 \) Solution: Given, \[ \alpha+\beta=\frac{\pi}{2} \] Using identity, \[ 2\cos\alpha\cos\beta=\cos(\alpha+\beta)+\cos(\alpha-\beta) \] \[ 2\cos\alpha\cos\beta = \cos\frac{\pi}{2}+\cos(\alpha-\beta) \] \[ 2\cos\alpha\cos\beta = 0+\cos(\alpha-\beta) \]

Show that sin(B − C) cos(A − D) + sin(C − A) cos(B − D) + sin(A − B) cos(C − D) = 0 Show that: \( \sin(B-C)\cos(A-D)+\sin(C-A)\cos(B-D)+\sin(A-B)\cos(C-D)=0 \) Solution: \[ \sin(B-C)\cos(A-D)+\sin(C-A)\cos(B-D) \] \[ +\sin(A-B)\cos(C-D) \] Using identity, \[ \sin x \cos y=\frac12[\sin(x+y)+\sin(x-y)] \] \[ =\frac12[\sin(B-C+A-D)+\sin(B-C-A+D)] \] \[ +\frac12[\sin(C-A+B-D)+\sin(C-A-B+D)] \] \[ +\frac12[\sin(A-B+C-D)+\sin(A-B-C+D)] \] \[ =\frac12[\sin(A+B-C-D)+\sin(B+D-A-C)

Show that sin A sin(B − C) + sin B sin(C − A) + sin C sin(A − B) = 0 Show that: \( \sin A \sin(B-C)+\sin B \sin(C-A)+\sin C \sin(A-B)=0 \) Solution: \[ \sin A \sin(B-C)+\sin B \sin(C-A)+\sin C \sin(A-B) \] Using identity, \[ \sin(x-y)=\sin x \cos y-\cos x \sin y \] \[ =\sin

Prove that sin 10° sin 50° sin 60° sin 70° = √3/16 Prove that: \( \sin10^\circ \sin50^\circ \sin60^\circ \sin70^\circ = \frac{\sqrt3}{16} \) Solution: \[ \sin10^\circ \sin50^\circ \sin60^\circ \sin70^\circ \] Using identity, \[ \sin3\theta = 4\sin\theta \sin(60^\circ+\theta)\sin(60^\circ-\theta) \] Putting \( \theta=10^\circ \), \[ \sin30^\circ = 4\sin10^\circ \sin50^\circ \sin70^\circ \] \[ \frac12 = 4\sin10^\circ \sin50^\circ \sin70^\circ \]

Prove that sin 20° sin 40° sin 60° sin 80° = 3/16 Prove that: \( \sin20^\circ \sin40^\circ \sin60^\circ \sin80^\circ = \frac{3}{16} \) Solution: \[ \sin20^\circ \sin40^\circ \sin60^\circ \sin80^\circ \] Using identity, \[ \sin3\theta = 4\sin\theta \sin(60^\circ+\theta)\sin(60^\circ-\theta) \] Putting \( \theta=20^\circ \), \[ \sin60^\circ = 4\sin20^\circ \sin40^\circ \sin80^\circ \] \[ \frac{\sqrt3}{2} = 4\sin20^\circ \sin40^\circ \sin80^\circ \]