Find the radian measures corresponding to the following degree measures:
$-47^{\circ} 30^{\prime}$
$-47^{\circ} 30^{\prime}-47 \frac{1}{2}$
$=\frac{-95}{2}$ degree
since $180^{\circ}=\pi$ radian
$\frac{-95}{2}$ degree $=\frac{\pi}{180} \times\left(\frac{-95}{2}\right)$ radian $=\left(\frac{-19}{36 \times 2}\right) \pi$ radian $=\frac{-19}{72} \pi$ radian
$\therefore-47^{\circ} 30^{\prime}=\frac{-19}{72} \pi$ radian
If $\cos \theta = \frac{1}{2}\left( {x + \frac{1}{x}} \right)$, then $\frac{1}{2}\left( {{x^2} + \frac{1}{{{x^2}}}} \right) = $
Prove that: $(\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}=4 \sin ^{2} \frac{x-y}{2}$
If $\tan \theta + \sin \theta = m$ and $\tan \theta - \sin \theta = n,$ then
The value of $\cos A - \sin A$ when $A = \frac{{5\pi }}{4},$ is
If $\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5},$ then
$(A)$ $\tan ^2 x=\frac{2}{3}$ $(B)$ $\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}$
$(C)$ $\tan ^2 x=\frac{1}{3}$ $(D)$ $\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{2}{125}$