Question

prove that sin(60-A) sin(60+A)= 1/4(3-4sin^2A​

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