Question

sin 18° cos 39° + sin 6° cos 15° = sin 24° cos 33°​

Answer 1

Answer:

We have to prove that,

sin 18° cos 39° + sin 6° cos 15° = sin 24° cos 33°​

L.H.S.

= sin 18° cos 39° + sin 6° cos 15°

$= \frac{1}{2} \{\sin (18+39) + \sin (18-39)\}+\frac{1}{2}\{\sin (6+15) + \sin (6-15)\}$                                              

( ∵ 2sinA cosB = sin(A + B) + sin(A - B) )

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

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

( ∵ sin (-x) = -sin x )    

$=\frac{1}{2} \{\sin (57) - \sin (9)\}$

Now, Let X + Y = 57 and X - Y = 9

By adding,

2X = 66 ⇒ X = 33

By subtracting,

2Y = 48 ⇒ Y = 24

$\implies \frac{1}{2} \{\sin (57) - \sin (9)\}=\frac{1}{2} \{\sin (33+24) - \sin (33-24)\}$

$=\frac{1}{2} \{\sin (24+33) + \sin (24-33)\}$

Again by the formula,

2sinA cosB = sin(A + B) + sin(A - B)

$\frac{1}{2} \{\sin (24+33) + \sin (24-33)\}=\sin 24.\cos 33$

= R.H.S.

Hence, proved....