The value of 36(4 cos^2 9^{circ}-1)(4 cos^2 2 | vedclass

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

Question

(D) Using the identity $4 \cos^2 	heta - 1 = \frac{\sin 3	heta}{\sin 	heta}$,we can simplify each term in the product.The expression becomes:$36

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(D) Using the identity $4 \cos^2 	heta - 1 = \frac{\sin 3	heta}{\sin 	heta}$,we can simplify each term in the product.The expression becomes:$36 	imes \left( \frac{\sin 27^{\circ}}{\sin 9^{\circ}} ight) 	imes \left( \frac{\sin 81^{\circ}}{\sin 27^{\circ}} ight) 	imes \left( \frac{\sin 243^{\circ}}{\sin 81^{\circ}} ight) 	imes \left( \frac{\sin 729^{\circ}}{\sin 243^{\circ}} ight)$Canceling the common terms in the numerator and denominator,we get:$36 	imes \frac{\sin 729^{\circ}}{\sin 9^{\circ}}$Since $\sin 729^{\circ} = \sin(2 	imes 360^{\circ} + 9^{\circ}) = \sin 9^{\circ}$,the expression simplifies to:$36 	imes \frac{\sin 9^{\circ}}{\sin 9^{\circ}} = 36$

The value of $36(4 \cos^2 9^{\circ}-1)(4 \cos^2 27^{\circ}-1)(4 \cos^2 81^{\circ}-1)(4 \cos^2 243^{\circ}-1)$ is

Similar Questions

If $\sin \theta + \cos \theta = \frac{1}{2}$,then $16(\sin(2\theta) + \cos(4\theta) + \sin(6\theta))$ is equal to:

If $\sec(x) = \cosh(\theta)$,then find $\tanh^2\left(\frac{\theta}{2}\right)$.

If $5 \sinh x - \cosh x = 5$,then one of the values of $\tanh x$ is

For $0 \leq x \leq \pi$,if $81^{\sin ^2 x}+81^{\cos ^2 x}=30$,then $x=$

${\sin ^4}\frac{\pi }{8} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module