Question: "a problem with condition sin⁡(2θ)=tan⁡(θ)−cos⁡(2θ)"

Emmanuella Browne

Answered question

2022-03-02

a problem with condition

\sin (2 θ) = \tan (θ) - \cos (2 θ)

lilwayne10j6o

Beginner2022-03-03Added 8 answers

Hint:
Using The Weierstrass Substitution,

t = \tan θ

2 t + 1 - t^{2} = t (1 + t^{2}) \Leftrightarrow 0 = t^{3} + t^{2} - t - 1 = t^{2} (t + 1) - (t + 1) = {(t + 1)}^{2} (t - 1)

\Rightarrow t = \pm 1

Chance Mill

Beginner2022-03-04Added 8 answers

We have

\sin (2 θ) = \tan (θ) - \cos (2 θ)

2 \sin θ \cos θ = \frac{\sin θ}{\cos θ} - 1 + 2 \sin^{2} θ

2 \sin θ \cos^{2} θ = \sin θ - \cos θ + 2 \sin^{2} θ \cos θ

Hence

\sin θ - \cos θ = 2 \sin θ \cos θ (\cos θ - \sin θ) = - \sin 2 θ (\sin θ - \cos θ)

giving

(1 + 2 \sin 2 θ) (\sin θ - \cos θ) = 0

So either

\sin θ - \cos θ = 0

\sin 2 θ = - \frac{1}{2} \Rightarrow θ = k π - \frac{π}{12}

for

k \in Z

thus

\sin θ - \cos θ = \pm \frac{\sqrt{6}}{2}

(consider the cases when k is even and odd)
Therefore

\sin θ - \cos θ = - \frac{\sqrt{6}}{2}, 0, \frac{\sqrt{6}}{2}

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