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

$\cos ^2 48^{\circ}-\sin ^2 12^{\circ} = $ . . . . . . ,if $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$

$\frac{\cos 9^\circ + \sin 9^\circ}{\cos 9^\circ - \sin 9^\circ} = $

If $\cot \alpha = \frac{1}{2}$ and $\sec \beta = -\frac{5}{3}$,where $\alpha \in \left(\pi, \frac{3\pi}{2}\right)$ and $\beta \in \left(\frac{\pi}{2}, \pi\right)$,then the value of $\tan(\alpha + \beta)$ is:

If $\cos \theta = \frac{8}{17}$ and $\theta$ lies in the $1^{st}$ quadrant,then the value of $\cos (30^\circ + \theta) + \cos (45^\circ - \theta) + \cos (120^\circ - \theta)$ is

If $\cos (\theta+\phi)=\frac{3}{5}$ and $\sin (\theta-\phi)=\frac{5}{13}$, where $0 < \theta, \phi < \frac{\pi}{4}$, then $\cot (2 \theta)$ has the value:

Vedclass Test Series

Exam Paper Generator

Online Exam Module