The value of cos(18^{circ}-A) cos(18^{circ}+A | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Using the identity $\cos(x-y)\cos(x+y) = \cos^2 x - \sin^2 y$,we have: $\cos(18^{\circ}-A)\cos(18^{\circ}+A) = \cos^2 18^{\circ} - \sin^2 A$ $\cos

Vedclass · Accepted Answer

$\cos 36^{\circ}$

Vedclass · Answer

(B) Using the identity $\cos(x-y)\cos(x+y) = \cos^2 x - \sin^2 y$,we have: $\cos(18^{\circ}-A)\cos(18^{\circ}+A) = \cos^2 18^{\circ} - \sin^2 A$ $\cos(72^{\circ}-A)\cos(72^{\circ}+A) = \cos^2 72^{\circ} - \sin^2 A$ Subtracting these: $(\cos^2 18^{\circ} - \sin^2 A) - (\cos^2 72^{\circ} - \sin^2 A) = \cos^2 18^{\circ} - \cos^2 72^{\circ}$ Since $\cos 72^{\circ} = \sin 18^{\circ}$,this becomes: $\cos^2 18^{\circ} - \sin^2 18^{\circ} = \cos(2 	imes 18^{\circ}) = \cos 36^{\circ}$ Alternatively,using $\cos(90^{\circ}-	heta) = \sin 	heta$: $\cos(72^{\circ}-A) = \sin(18^{\circ}+A)$ and $\cos(72^{\circ}+A) = \sin(18^{\circ}-A)$ Expression $= \cos(18^{\circ}-A)\cos(18^{\circ}+A) - \sin(18^{\circ}+A)\sin(18^{\circ}-A)$ $= \cos((18^{\circ}-A) + (18^{\circ}+A)) = \cos 36^{\circ}$

The value of $\cos(18^{\circ}-A) \cos(18^{\circ}+A) - \cos(72^{\circ}-A) \cos(72^{\circ}+A)$ is equal to

Similar Questions

Prove that: $\sin x + \sin 3x + \sin 5x + \sin 7x = 4 \cos x \cos 2x \sin 4x$

Prove that $\sin (n+1) x \sin (n+2) x + \cos (n+1) x \cos (n+2) x = \cos x$.

If $\frac{\sin(x + y)}{\sin(x - y)} = \frac{a + b}{a - b},$ then $\frac{\tan x}{\tan y}$ is equal to

If $\sin \alpha = 1/\sqrt{5}$ and $\sin \beta = 3/5$,then $\beta - \alpha$ lies in the interval

If $A = \frac{\pi}{24}$,then $\frac{\cos A + \cos 3A + \cos 5A + \cos 7A}{\sin A + \sin 3A + \sin 5A + \sin 7A} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module