Question

Question #26bf4

Answer 1

Prove `sin(3theta) + sin(theta) = 2sin(2theta) cos(theta)`

Substitute `sin(3theta) = sin(2theta+theta)` and `sin(theta) = sin(2theta-theta)`

`sin(2theta+theta) + sin(2theta-theta) = 2sin(2theta) cos(theta)`

Substitute the identities `sin(A+B) = sin(A)cos(B)+cos(A)sin(B)` and `sin(A-B) = sin(A)cos(B)-cos(A)sin(B)` where `A = 2theta` and `B = theta`:

`sin(2theta)cos(theta)+cos(2theta)sin(theta) + sin(2theta)cos(theta)-cos(2theta)sin(theta) = 2sin(2theta) cos(theta)`

Combine like terms:

`2sin(2theta)cos(theta) = 2sin(2theta)cos(theta)` Q.E.D.