Trig question 3U (1 Viewer)

Jizin · Nov 13, 2019

It can be show that: sin3Θ = 3sinΘ-4sin^3Θ (No need to prove)

Use this result to solve sin3Θ + sin2Θ = sinΘ, 0<x<2pi

fan96 · Nov 13, 2019

Rewriting (and using $\sin 2x = 2 \sin x \cos x$ ), we have

$( 3 \sin \theta - 4 \sin^3\theta) + 2 \sin \theta \cos \theta = \sin \theta$

or

$\sin \theta (3 - 4 \sin^2\theta + 2\cos \theta - 1) = 0$ .

Use the Pythagorean identity to rewrite $\sin^2$ in terms of $\cos^2$ :

$(2 \sin \theta) (2 \cos ^2\theta + \cos \theta - 1) = 0$ .

Now all you need to do is solve separately the equations

$\begin{cases} 2 \sin \theta = 0 \\ 2 \cos ^2\theta + \cos \theta - 1 = 0\end{cases}$

and combine all your solutions (discarding those that are not in the range $0 < \theta < 2\pi$ ).

Trig question 3U (1 Viewer)

Jizin

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fan96

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