cos 6^{circ} sin 24^{circ} cos 72^{circ} = | vedclass

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

Question

(C) We have the expression $E = \cos 6^{\circ} \sin 24^{\circ} \cos 72^{\circ}$.Since $\cos 72^{\circ} = \sin 18^{\circ}$,we can write $E = \cos 6^{\c

Vedclass · Accepted Answer

$\frac{1}{8}$

Vedclass · Answer

(C) We have the expression $E = \cos 6^{\circ} \sin 24^{\circ} \cos 72^{\circ}$.Since $\cos 72^{\circ} = \sin 18^{\circ}$,we can write $E = \cos 6^{\circ} \sin 24^{\circ} \sin 18^{\circ}$.Multiply and divide by $2$: $E = \frac{1}{2} (2 \sin 24^{\circ} \cos 6^{\circ}) \sin 18^{\circ}$.Using the identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:$E = \frac{1}{2} (\sin 30^{\circ} + \sin 18^{\circ}) \sin 18^{\circ}$.$E = \frac{1}{2} (\frac{1}{2} \sin 18^{\circ} + \sin^2 18^{\circ})$.Given $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$,then $\sin^2 18^{\circ} = \frac{5+1-2\sqrt{5}}{16} = \frac{6-2\sqrt{5}}{16} = \frac{3-\sqrt{5}}{8}$.Substituting these values:$E = \frac{1}{2} (\frac{1}{2} \cdot \frac{\sqrt{5}-1}{4} + \frac{3-\sqrt{5}}{8}) = \frac{1}{2} (\frac{\sqrt{5}-1}{8} + \frac{3-\sqrt{5}}{8}) = \frac{1}{2} (\frac{2}{8}) = \frac{1}{8}$.

$\cos 6^{\circ} \sin 24^{\circ} \cos 72^{\circ} = $

Similar Questions

If $y^2+z^2=3yz$,$z^2+x^2=8zx$,and $x^2+y^2=4xy$,then the value of $\frac{y^2}{xz}+\frac{xz}{y^2}$ is

$\cos ^2 76^{\circ}+\sin ^2 46^{\circ}+\sin 76^{\circ} \cos 46^{\circ} = $

$\operatorname{sech}^{-1}(\sin \theta)$ is equal to

If $(1+\tan 1^{\circ})(1+\tan 2^{\circ}) \ldots (1+\tan 45^{\circ})=2^n$,then $n=$

The value of $\sum_{k=1}^3 \cos ^2\left((2 k-1) \frac{\pi}{12}\right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module