cos(36^{circ}-A) cos(36^{circ}+A) + cos(54^{c | vedclass

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

Question

(A) Using the identity $\cos(x)\cos(y) + \sin(x)\sin(y) = \cos(x-y)$ and the co-function identity $\cos(90^{\circ}-	heta) = \sin(	heta)$:Given expre

Vedclass · Accepted Answer

$\cos(2A)$

Vedclass · Answer

(A) Using the identity $\cos(x)\cos(y) + \sin(x)\sin(y) = \cos(x-y)$ and the co-function identity $\cos(90^{\circ}-	heta) = \sin(	heta)$:Given expression: $\cos(36^{\circ}-A) \cos(36^{\circ}+A) + \cos(54^{\circ}+A) \cos(54^{\circ}-A)$Since $54^{\circ}+A = 90^{\circ}-(36^{\circ}-A)$ and $54^{\circ}-A = 90^{\circ}-(36^{\circ}+A)$,we have:$\cos(54^{\circ}+A) = \sin(36^{\circ}-A)$ and $\cos(54^{\circ}-A) = \sin(36^{\circ}+A)$Substituting these into the expression:$\cos(36^{\circ}-A) \cos(36^{\circ}+A) + \sin(36^{\circ}-A) \sin(36^{\circ}+A)$This is in the form $\cos(x)\cos(y) + \sin(x)\sin(y)$ where $x = 36^{\circ}-A$ and $y = 36^{\circ}+A$.$= \cos((36^{\circ}-A) - (36^{\circ}+A))$$= \cos(36^{\circ}-A-36^{\circ}-A)$$= \cos(-2A)$Since $\cos(-	heta) = \cos(	heta)$,the result is $\cos(2A)$.

$\cos(36^{\circ}-A) \cos(36^{\circ}+A) + \cos(54^{\circ}+A) \cos(54^{\circ}-A) = $

Similar Questions

Prove that $\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x$.

If $\cos(\alpha + \beta) = \frac{4}{5}$ and $\sin(\alpha - \beta) = \frac{5}{13}$,where $0 \le \alpha, \beta \le \frac{\pi}{4}$,then $\tan 2\alpha = $

$\cos \left(\frac{3 \pi}{4}+x\right)-\sin \left(\frac{\pi}{4}-x\right) = $

If $(1+\tan \alpha)(1+\tan 4 \alpha)=2$ and $\alpha \in \left(0, \frac{\pi}{16}\right)$,then $\alpha$ is equal to

If $A+B=\frac{\pi}{4}$,then $\frac{\cos B-\sin B}{\cos B+\sin B}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module