Evaluate: cos^2 48^circ - sin^2 12^circ | vedclass

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

Question

(B) We use the trigonometric identity: $\cos^2 A - \sin^2 B = \cos(A + B) \cos(A - B)$.Substituting $A = 48^\circ$ and $B = 12^\circ$:$\cos^2 48^\circ

Vedclass · Accepted Answer

$\frac{\sqrt{5} + 1}{8}$

Vedclass · Answer

(B) We use the trigonometric identity: $\cos^2 A - \sin^2 B = \cos(A + B) \cos(A - B)$.Substituting $A = 48^\circ$ and $B = 12^\circ$:$\cos^2 48^\circ - \sin^2 12^\circ = \cos(48^\circ + 12^\circ) \cos(48^\circ - 12^\circ)$$= \cos 60^\circ \cos 36^\circ$We know that $\cos 60^\circ = \frac{1}{2}$ and $\cos 36^\circ = \frac{\sqrt{5} + 1}{4}$.Therefore,the expression becomes:$= \frac{1}{2} 	imes \left( \frac{\sqrt{5} + 1}{4} ight) = \frac{\sqrt{5} + 1}{8}$.

Evaluate: $\cos^2 48^\circ - \sin^2 12^\circ$

Similar Questions

The value of $(\cos \alpha+\cos \beta)^2+(\sin \alpha+\sin \beta)^2$ is

The value of $\cos ^{2} 75^{\circ}+\cos ^{2} 45^{\circ}+\cos ^{2} 15^{\circ}-\cos ^{2} 30^{\circ}-\cos ^{2} 60^{\circ}$ is

If $\sin A = n \sin B,$ then $\frac{n - 1}{n + 1} \tan \frac{A + B}{2} = $

$\tan\, 20^{\circ} + \tan\, 40^{\circ} + \sqrt{3}\, \tan\, 20^{\circ} \tan\, 40^{\circ}$ is equal to

If $\sin A = -\frac{24}{25}$,$\cos B = \frac{15}{17}$,$A$ does not belong to the $4^{\text{th}}$ quadrant,and $B$ does not belong to the $1^{\text{st}}$ quadrant,then $(A+B)$ lies in which quadrant?

Vedclass Test Series

Exam Paper Generator

Online Exam Module