cos ^2left(frac{pi}{6}+thetaright)-sin ^2left | vedclass

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

Question

(A) We use the trigonometric identity $\cos^2 A - \sin^2 B = \cos(A+B) \cos(A-B)$.Let $A = \frac{\pi}{6} + 	heta$ and $B = \frac{\pi}{6} - 	heta$.Th

Vedclass · Accepted Answer

$\frac{1}{2} \cos 2 	heta$

Vedclass · Answer

(A) We use the trigonometric identity $\cos^2 A - \sin^2 B = \cos(A+B) \cos(A-B)$.Let $A = \frac{\pi}{6} + 	heta$ and $B = \frac{\pi}{6} - 	heta$.Then $A+B = \left(\frac{\pi}{6} + 	hetaight) + \left(\frac{\pi}{6} - 	hetaight) = \frac{2\pi}{6} = \frac{\pi}{3}$.And $A-B = \left(\frac{\pi}{6} + 	hetaight) - \left(\frac{\pi}{6} - 	hetaight) = 2	heta$.Substituting these into the identity:$\cos^2\left(\frac{\pi}{6}+	hetaight)-\sin^2\left(\frac{\pi}{6}-	hetaight) = \cos\left(\frac{\pi}{3}ight) \cos(2	heta)$.Since $\cos\left(\frac{\pi}{3}ight) = \frac{1}{2}$,the expression equals $\frac{1}{2} \cos 2	heta$.

$\cos ^2\left(\frac{\pi}{6}+\theta\right)-\sin ^2\left(\frac{\pi}{6}-\theta\right)$ is equal to

Similar Questions

$\tan 5x \tan 3x \tan 2x = $

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

If $\frac{\pi}{2} < \alpha < \pi$ and $\pi < \beta < \frac{3\pi}{2}$,with $\sin \alpha = \frac{15}{17}$ and $\tan \beta = \frac{12}{5}$,then the value of $\sin(\beta - \alpha)$ is (in $/221$)

$\tan 75^\circ - \cot 75^\circ = $

Prove that $\cos \left( \frac{\pi}{4} + x \right) + \cos \left( \frac{\pi}{4} - x \right) = \sqrt{2} \cos x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module