Answer to Question #176572 in Trigonometry for Yusuf Idris

Question

cos³θ + sin³θ = (1/4)(cos³θ +3cosθ - sin³θ + 3sinθ)

Answer 1

cos³θ + sin³θ = (1/4)(cos³θ +3cosθ - sin³θ + 3sinθ)

(1/4)(3cosθ + cos3θ + 3sinθ - sin3θ) = (1/4)(cos³θ +3cosθ - sin³θ + 3sinθ)

3cosθ + cos3θ + 3sinθ - sin3θ = cos³θ +3cosθ - sin³θ + 3sinθ

cos3θ - sin3θ = cos³θ - sin³θ

cos3θ - sin3θ =(1/4)(3cosθ + cos3θ - 3sinθ - sin3θ)

4cos3θ - 4sin3θ = 3cosθ + cos3θ - 3sinθ - sin3θ

-3cosθ + 3cos3θ + 3sinθ - 5sin3θ = 0

-2sinθ(1 + 5cos2θ + 3sin2θ) = 0

sinθ = 0 or 1 + 5cos2θ + 3sin2θ = 0

θ = $\pi$n₁ for n₁ $\isin$ Z

substitute y = tanθ.

1 + 5/(y² + 1) + 6y/(y² + 1) - 5y²/(y² + 1) = 0

(2(2y² - 3y - 3))/(y² + 1) = 0

2y² - 3y - 3 = 0

tanθ = 3/4 - $\sqrt{33}$/4 or tanθ = 3/4 + $\sqrt{33}$/4

θ = tan^-1(3/4 - $\sqrt{33}$/4) + $\pi$n₂ for n₂ $\isin$ Z

θ = tan^-1(3/4 + $\sqrt{33}$/4) + $\pi$n₃ for n₃ $\isin$ Z

Answer:

θ = $\pi$n₁ for n₁ $\isin$ Z

θ = tan^-1(3/4 - $\sqrt{33}$/4) + $\pi$n₂ for n₂ $\isin$ Z

θ = tan^-1(3/4 + $\sqrt{33}$/4) + $\pi$n₃ for n₃ $\isin$ Z