show that
sinx - sin 3x /sin²x - cos²x = 2 sin x
Answers
Solution :-
On taking LHS
(sin x - sin 3x) / ( sin²x - cos²x) ------------(1)
The numerator in (1) = sin x - sin 3x
This is in the form of sin C - sin D
where , C = x and D = 3x
We know that
sin C-sin D = 2 cos (C+D)/2 sin (C-D)/2
Therefore,
sin x - sin 3x
= 2 cos (x+3x)/2 sin (x-3x)/2
= 2 cos (4x/2) sin (-2x)/2
= 2 cos 2x sin (-x)
We know that
sin (-A) = - sin A
Therefore,
sin x - sin 3x = - 2 cos 2x sin x ----------(2)
The denominator in (1) = sin² x - cos²x
=> - ( cos² x - sin² x)
We know that
cos 2x = cos² x - sin² x
Therefore, sin² x - cos²x = - cos 2x -------(3)
Now (1) becomes from (2)&(3)
[ - 2 cos 2x sin x] / (- cos 2x)
On cancelling - cos 2x in both numerator and denominator then
= 2 sin x
= RHS
=> LHS = RHS
Hence, Proved.
Answer :-
(sin x - sin 3x) / ( sin²x - cos²x) = 2 sin x
Used formulae:-
→ sin C-sin D=2 cos (C+D)/2 sin (C-D)/2
→ cos 2x = cos² x - sin² x