Sin frac{π }{{14}}sin frac{{3π }}{{14}}sin frac{{5π }}{{14}}sin frac{{7π }}{{14}}sin frac{{9π }

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3.Trigonometrical Ratios, Functions and Identities

medium

$\sin \frac{\pi }{{14}}\sin \frac{{3\pi }}{{14}}\sin \frac{{5\pi }}{{14}}\sin \frac{{7\pi }}{{14}}\sin \frac{{9\pi }}{{14}}\sin \frac{{11\pi }}{{14}}\sin \frac{{13\pi }}{{14}} = . . . .$

$\frac{1}{8}$

$\frac{1}{{16}}$

$\frac{1}{{32}}$

$\frac{1}{{64}}$

(IIT-1991)

Solution

(d) $\sin \frac{\pi }{{14}}\sin \frac{{3\pi }}{{14}}\sin \frac{{5\pi }}{{14}}\sin \frac{{7\pi }}{{14}}\sin \frac{{9\pi }}{{14}}\sin \frac{{11\pi }}{{14}}\sin \frac{{13\pi }}{{14}}$

$ = \sin \frac{\pi }{{14}}\sin \frac{{3\pi }}{{14}}\sin \frac{{5\pi }}{{14}} \times 1$

$ \times \sin \left( {\pi – \frac{{5\pi }}{{14}}} \right)\sin \left( {\pi – \frac{{3\pi }}{{14}}} \right)\sin \left( {\pi – \frac{\pi }{{14}}} \right)$

$ = {\left[ {\sin \frac{\pi }{{14}}\sin \frac{{3\pi }}{{14}}\sin \frac{{5\pi }}{{14}}\sin \frac{{7\pi }}{{14}}} \right]^2} = \frac{1}{{64}}$.

Standard 11

Mathematics

$\frac{{\cos 12^\circ – \sin 12^\circ }}{{\cos 12^\circ + \sin 12^\circ }} + \frac{{\sin 147^\circ }}{{\cos 147^\circ }} = $

easy

$\frac{{\sin \theta + \sin 2\theta }}{{1 + \cos \theta + \cos 2\theta }} = $

easy

સાબિત કરો કે : $\cos 4 x=1-8 \sin ^{2} x \cos ^{2} x$

medium

જો $\cos \,(\theta – \alpha ) = a,\,\,\sin \,(\theta – \beta ) = b,\,\,$ તો ${\cos ^2}(\alpha – \beta ) + 2ab\,\sin \,(\alpha – \beta ) = . . . .$

hard

$ \cos ^{3}\left(\frac{\pi}{8}\right) \cdot \cos \left(\frac{3 \pi}{8}\right)+\sin ^{3}\left(\frac{\pi}{8}\right) \cdot \sin \left(\frac{3 \pi}{8}\right)$ ની કિમંત મેળવો.