I can not get this question right for the life of me. I'm so flustered, please help.

Question

Use an Addition or Subtraction Formula to simplify the equation.

sin(3theta) cos(theta) - cos(theta) sin(theta) = 0

Answer 1

sin(3θ)cos(θ) - cos(θ)sin(θ) = 0

Obviously sin(3θ) = sin(2θ + θ) and, since the sine angle addition identity says:

sin(x + y) = sin(x)cos(y) + cos(x)sin(y), we can say

sin(3θ) = sin(2θ + θ) = sin(2θ)cos(θ) + cos(2θ)sin(θ)

We can repeat for sin(2θ) because sin(2θ) = sin(θ + θ) so:

sin(2θ) = sin(θ + θ) = sin(θ)cos(θ) + cos(θ)sin(θ) = 2cos(θ)sin(θ)

And, since the cosine angle addition identity is:

cos(x + y) = cos(x)cos(y) – sin(x)sin(y) we can say

cos(2θ) = cos(θ + θ) = cos(θ)cos(θ) – sin(θ)sin(θ) = cos²(θ) - sin²(θ)

Putting these all together by substituting into the original equation:

sin(3θ)cos(θ) - cos(θ)sin(θ) = 0

[sin(2θ)cos(θ) + cos(2θ)sin(θ)]cos(θ) - cos(θ)sin(θ) = 0

[2cos(θ)sin(θ)cos(θ) + (cos²(θ) - sin²(θ))(sin(θ)]cos(θ) - cos(θ)sin(θ) = 0

[2cos²(θ)sin(θ) + cos²(θ)sin(θ) - sin³(θ)]cos(θ) - cos(θ)sin(θ) = 0

2cos³(θ)sin(θ) + cos³(θ)sin(θ) - sin³(θ)cos(θ) - cos(θ)sin(θ) = 0

3cos³(θ)sin(θ) - sin³(θ)cos(θ) - cos(θ)sin(θ) = 0

[cos(θ)sin(θ)][3cos²(θ) - sin²(θ) - 1] = 0

Using the Pythagorean Identity sin²(θ) + cos²(θ) = 1, we can say sin²(θ) = 1 - cos²(θ) so:

[cos(θ)sin(θ)][3cos²(θ) - (1 - cos²(θ)) - 1] = 0

[cos(θ)sin(θ)][3cos²(θ) - 1 + cos²(θ) - 1] = 0

[cos(θ)sin(θ)][4cos²(θ) - 2] = 0

2cos(θ)sin(θ)[2cos²(θ) - 1] = 0

The question doesn't say to solve the equation, but when it is in this form, it is easily solvable:

We can use the zero product rule to set each piece equal to zero:

A. 2cos(θ) = 0 or cos(θ) = 0 which occurs at θ = π/2 and 3π/2

B. sin(θ) = 0 which occurs at θ = 0 and θ = π

C. 2cos²(θ) - 1 = 0

2cos²(θ) = 1

cos²(θ) = 1/2

cos(θ) = ±√(1/2) = ±√2/2 which occurs at θ = π/4, 3π/4, 5π/4, and 7π/4

So, summarizing, we have 8 answers on [0, 2π):

θ = 0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, and 7π/4