(i) `sin (alpha + beta) = sin alpha cos beta + cos alpha sin beta)`
(ii) `sin (alpha-beta)= sin alpha cos beta- cos alpha sin beta`
Proof

If sin alpha = sin beta and cos alpha = cos beta then

sin alpha=sinbeta, cos alpha=cos beta then

If cos alpha + cos beta =0 = sin alpha + sin beta " then " cos (alpha - beta )=

2 sin^(2) beta + 4 cos( alpha+ beta) sin alpha sin beta + cos ( 2 alpha + 2 beta )=

The locus of the point (r cos alpha cos beta, r cos alpha sin beta, r sin alpha) is

Evaluate {:[( cos alpha cos beta , cos alpha sin beta , -sin alpha ),( -sin beta , cos beta, 0),( sin alpha cos beta, sin alpha sin beta, cos alpha ) ]:} =0

If sin alpha + sin beta = a , cos alpha + cos beta = b " then cos ( alpha + beta ) =

If alpha , beta are complementary angles , sin alpha = 3//5 , then sin alpha cos beta - cos alpha sin beta =

[(cos alpha-cos beta)+i(sin alpha-sinbeta)]^n+[(cos alpha-cos beta)-i(sin alpha-sin beta)]^n=

If x= sin alpha + sin betga , y = cos alpha + cos beta tan alpha + tan beta =