Question

Whether the given equation $\sin {18}^{\circ }\cos {39}^{\circ }+\sin 6^{\circ }\cos {15}^{\circ }=\sin {24}^{\circ }\cos {33}^{\circ }$. is ?

• A. True
• B. False

Answer 1

The correct option is A True

Given equation:

$\sin {18}^o\cos {39}^o+\sin 6^o\cos {15}^o=\sin {24}^o\cos {33}^o$

LHS$=\sin {18}^o\cos {39}^o+\sin 6^o\cos {15}^o$

$=\frac{1}{2}\left[\sin \right({18}^o+{39}^o)+\sin ({18}^o-{39}^o\left)\right]+\frac{1}{2}\left[\sin \right(6^o+{15}^o)+\sin (6^o-{15}^o\left)\right]$

$=\frac{1}{2}\sin 57-\frac{1}{2}\sin 21+\frac{1}{2}\sin 21-\frac{1}{2}\sin 9$

$=\frac{1}{2}[\sin {57}^o-\sin 9^o]$

$=\frac{1}{2}\left[2\cos \left(\frac{57+9}{2}\right)\sin \left(\frac{57-9}{2}\right)\right]$

$=\cos \frac{66}{2}\sin \frac{48}{2}$

$=\cos {33}^o\sin {24}^o$

$=$RHS

$\therefore$ The given statements is TRUE.