The value of $\sin \left(\frac{5 \pi}{24}\right) \cdot \cos \left(\frac{\pi}{24}\right)$ is

Prove that: $\sin 3x + \sin 2x - \sin x = 4 \sin x \cos \frac{x}{2} \cos \frac{3x}{2}$

In an isosceles right-angled triangle,a straight line is drawn from the midpoint of one of the equal sides to the opposite vertex. Then a pair of possible values of the cotangents of the two angles so formed at that vertex are

$1 - 2{\sin ^2}\left( {\frac{\pi }{4} + \theta } \right) = $

If $\cot \alpha = \frac{1}{2}$ and $\sec \beta = -\frac{5}{3}$,where $\alpha \in \left(\pi, \frac{3\pi}{2}\right)$ and $\beta \in \left(\frac{\pi}{2}, \pi\right)$,then the value of $\tan(\alpha + \beta)$ is:

If $\sin A = \frac{4}{5}$ and $\cos B = -\frac{12}{13}$,where $A$ and $B$ lie in the first and third quadrant respectively,then $\cos(A + B) = $