Class 11 Mathematics Trigonometrical Ratios, Functions and Identities Mix Examples-Trigonometrical Ratios, Functions and Identities

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Prove that: $(\sin 3x + \sin x) \sin x + (\cos 3x - \cos x) \cos x = 0$

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Solution

(N/A) $L.H.S.$ $= (\sin 3x + \sin x) \sin x + (\cos 3x - \cos x) \cos x$
$= \sin 3x \sin x + \sin^2 x + \cos 3x \cos x - \cos^2 x$
$= (\cos 3x \cos x + \sin 3x \sin x) - (\cos^2 x - \sin^2 x)$
$= \cos(3x - x) - \cos 2x$
$= \cos 2x - \cos 2x$
$= 0$
$= R.H.S.$

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Trigonometrical Ratios, Functions and Identities

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Mix Examples-Trigonometrical Ratios, Functions and Identities

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Prove that: $(\sin 3x + \sin x) \sin x + (\cos 3x - \cos x) \cos x = 0$

Similar Questions

Observe that,at any instant,the minute and hour hands of a clock make two angles between them whose sum is $360^{\circ}$. At $6:15$,the difference between these two angles is $....^{\circ}$.

If $\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5},$ then$(A) \tan ^2 x=\frac{2}{3}$ $(B) \frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}$$(C) \tan ^2 x=\frac{1}{3}$ $(D) \frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{2}{125}$

If $\sin x \cosh y = \cos \theta$ and $\cos x \sinh y = \sin \theta$,then $\sinh^2 y =$

$\tan 2 \alpha \cdot \tan \left(30^{\circ}-\alpha\right)+\tan 2 \alpha \cdot \tan \left(60^{\circ}-\alpha\right)+\tan \left(60^{\circ}-\alpha\right) \cdot \tan \left(30^{\circ}-\alpha\right)$ is equal to

Assertion $(A)$: If $\sqrt{4 \sin^4 \theta + \sin^2 2\theta} + 4 \cos^2\left(\frac{\pi}{4} - \frac{\theta}{2}\right) = 2$,then $\theta$ lies in the $3^{\text{rd}}$ quadrant or $4^{\text{th}}$ quadrant. Reason $(R)$: $\sqrt{\sin^2 \theta} = \sin \theta$

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If $\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5},$ then
$(A) \tan ^2 x=\frac{2}{3}$ $(B) \frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}$
$(C) \tan ^2 x=\frac{1}{3}$ $(D) \frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{2}{125}$

Assertion $(A)$: If $\sqrt{4 \sin^4 \theta + \sin^2 2\theta} + 4 \cos^2\left(\frac{\pi}{4} - \frac{\theta}{2}\right) = 2$,then $\theta$ lies in the $3^{\text{rd}}$ quadrant or $4^{\text{th}}$ quadrant.
Reason $(R)$: $\sqrt{\sin^2 \theta} = \sin \theta$