Question

Prove that sin (60 - A). sin (60 + A) =1/4[3-4sin^A​

Answer 1

Answer:

As we know that sin(A+B)sin(A−B)=sin

2

A−sin

2

B

LHS=sinAsin(60−A)sin(60+A)=sinA(sin

2

60−sin

2

A)=sinA(

4

3

−sin

2

A)=

4

3sinA−4sin

3

A

=

4

1

sin3A=RHS