Question #176572

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


1
Expert's answer
2021-05-07T14:21:04-0400

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 = π\pin1 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 - 33\sqrt{33}/4 or tanθ = 3/4 + 33\sqrt{33}/4

θ = tan-1(3/4 - 33\sqrt{33}/4) + π\pin2 for n2 \isin Z

θ = tan-1(3/4 + 33\sqrt{33}/4) + π\pin3 for n3 \isin Z


Answer:

θ = π\pin1 for n1 \isin Z

θ = tan-1(3/4 - 33\sqrt{33}/4) + π\pin2 for n2 \isin Z

θ = tan-1(3/4 + 33\sqrt{33}/4) + π\pin3 for n3 \isin Z



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