Find the value of $\sin 15^{\circ}$
Find the value of $\sin 15^{\circ}$
We have
$\sin {15^\circ } = \sin \left( {{{45}^\circ } - {{30}^\circ }} \right)$
$ = \sin {45^\circ }\cos {30^\circ } - \cos {45^\circ }\sin {30^\circ }$
$ = \frac{1}{{\sqrt 2 }} \times \frac{{\sqrt 3 }}{2} - \frac{1}{{\sqrt 2 }} \times \frac{1}{2} = \frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$
Similar Questions
If $\cos (\alpha - \beta ) = 1$ and $\cos (\alpha + \beta ) = \frac{1}{e}$, $ - \pi < \alpha ,\beta < \pi $, then total number of ordered pair of $(\alpha ,\beta )$ is
If $\cos (\alpha - \beta ) = 1$ and $\cos (\alpha + \beta ) = \frac{1}{e}$, $ - \pi < \alpha ,\beta < \pi $, then total number of ordered pair of $(\alpha ,\beta )$ is
- [IIT 2005]
Let the function $:(0, \pi) \rightarrow R$ be defined by
$f (\theta)=(\sin \theta+\cos \theta)^2+(\sin \theta-\cos \theta)^4$
Suppose the function $f$ has a local minimum at $\theta$ precisely when $\theta \in\left\{\lambda_1 \pi, \ldots, \lambda_{ T } \pi\right\}$, where $0<\lambda_1<\cdots<\lambda_r<1$. Then the value of $\lambda_1+\cdots+\lambda_r$ is. . . . .
Let the function $:(0, \pi) \rightarrow R$ be defined by
$f (\theta)=(\sin \theta+\cos \theta)^2+(\sin \theta-\cos \theta)^4$
Suppose the function $f$ has a local minimum at $\theta$ precisely when $\theta \in\left\{\lambda_1 \pi, \ldots, \lambda_{ T } \pi\right\}$, where $0<\lambda_1<\cdots<\lambda_r<1$. Then the value of $\lambda_1+\cdots+\lambda_r$ is. . . . .
- [IIT 2020]
Prove that $\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^{2} x$
Prove that $\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^{2} x$
$\frac{{2\sin \theta \,\tan \theta (1 - \tan \theta ) + 2\sin \theta {{\sec }^2}\theta }}{{{{(1 + \tan \theta )}^2}}} = $
$\frac{{2\sin \theta \,\tan \theta (1 - \tan \theta ) + 2\sin \theta {{\sec }^2}\theta }}{{{{(1 + \tan \theta )}^2}}} = $
Prove that:
$\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}=-\frac{1}{2}$
Prove that:
$\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}=-\frac{1}{2}$