Solve this: TRIGONOMETRIC FUNCTION s 67
Since, none of the y and (x + y) IS multiple of n, we find that sin x sin y and
sin (x + y) are non-zeru Now,
cos y) _ eos xcos y— sin x sin y
cot y) —
sm (x + y) sin x eos y + eos x sin y
Dividing numerator and denominator hy sin sin y, we have
cot x cot y —I
cot (x + y) —
coly + cot x
cot x cot y +1
13.
cot (x —
if none of angles x, y and is a multiple of a
cot y — cot x
If we replace y by —y in identity 12, we get the result
cos2x coer — sin2 x
We know that
cos (x + y) = cos x
Replacing y hy we get
I — 2 sin! x
cos y — sm x sin y
I —tan2
+ tan x
Again.
We have
cos 2x = cosh — sm x
(I — = 2 cos2x— I
= cos x —
cos 2s— coe x —sn x
—I —Sin? x— Sin; —2 sin2x.
cos x—sm x
cos coe x —sm x
coe x
Dividing numerator and denominator by cos? x. we get
— tan 2 x
cos 2x =
x * nit* — . where n is an integer
14 tan'x'
2tan x
sin 2x = 2 sinx eos x =
nit + —
We have
sin (x + y) sin x cos y + cos x sin y
Replacing by x, we get sin 2x= 2 Sin x cos x.
where n is an integer
Again
2sin x eos x
sin 2x =
cos x-+sln x
