Solve the equation by first using a Sum-to-Product Formula.

Question

sin θ + sin 3θ = 0?

Answer 1

The equation sin(θ) + sin(3θ) = 0

Apply the formula sin(A) +sin(B) = 2 [sin(A+B)/2] [cos(A-B)/2]

A = θ and B = 3θ

sin(θ) + sin(3θ) = 2 sin[(θ+3θ)/2] cos[(θ-3θ)/2]

= 2 sin(4θ)/2 cos(-2θ)/2

= 2 sin(2θ) cos(-θ)

= 2 sin(2θ) cos(θ)

Apply the double angle formula sin(2x) = 2sin(x) cos(x)

= 2 [2 sin(θ) cos(θ)] cos(θ)

= 4sin(θ) cos²(θ)

Now the equation is 4sin(θ) cos²(θ) = 0

4 sin(θ)  =  0 and cos²(θ) = 0

sin(θ) = 0 and cos²(θ) = 0

Solve sin(θ) = 0

The general solution of sin(θ) = 0 then θ = nπ. where n  is an integer.

For n = 0, θ = 0

For n = 1, θ = π

Solve cos²(θ) = 0

[cos(θ)]² = 0

cos(θ) = 0

The general solution of cos(θ) = 0 then θ = (2n+1)π/2. where n  is an integer.

For n = 0, θ = (0+1)π/2 = π/2

For n = 1, θ = (2+1)π/2 = 3π/2.