Help with a trigonometry problem.

Question

I have encountered some problems to solve the left side to the right.

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$$ \cos^ 2x \sin x = \frac{\sin 3x + \sin x}{4}$$

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I trying to solve a differential equation on the form $$ y'' + y = \cos^ 2x * \sin x$$ and need to rewrite it to $$ y'' + y = \frac{\sin 3x + \sin x}{4}$$

Have tried with different combinations but not really got it to $$\frac{\sin 3x + \sin x}{4}$$

I have also tried to start backwards, but it seems not to be the method I would have chosen if I started from the left.

\begin{align} \frac{\sin 3x+\sin x}{4} & =\frac{\sin x+\sin 3x}{4} \\ & =\frac{\sin x+\sin \left( 2x+x \right)}{4} \\ & =\frac{\sin x+\left( \sin 2x\cos x+\sin x\cos 2x \right)}{4} \\ & =\frac{\sin x+\left( \left( 2\sin x\cos x \right)\cos x+\sin x\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right) \right)}{4} \\ & =\frac{\sin x+\left( 2\sin x{{\cos }^{2}}x+\sin x\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right) \right)}{4} \\ & =\frac{\sin x+\left( 2\sin x\left( 1-{{\sin }^{2}}x \right)+\sin x\left( 1-2{{\sin }^{2}}x \right) \right)}{4} \\ & =\frac{\sin x+\left( (2\sin x-2{{\sin }^{3}}x)+(\sin x-2{{\sin }^{3}}x) \right)}{4} \\ & =\frac{\sin x+\left( 3\sin x-4{{\sin }^{3}}x \right)}{4} \\ & =\frac{4\sin x-4{{\sin }^{3}}x}{4} \\ & =\frac{4\sin x(1-{{\sin }^{2}}x)}{4} \\ & =\sin x{{\cos }^{2}}x \\ \end{align}

Would be grateful if you could give out some kind of guidance.

Answer 1

Note that by product to sum formula

$$2\cos \theta \sin \varphi = {{\sin(\theta + \varphi) - \sin(\theta - \varphi)} }\implies \cos 2x \cdot \sin x = \frac{\sin 3x - \sin x}{2} $$

Answer 2

Consider \begin{align} \cos^2x\sin x&=(1-\sin^2x)\sin x\\ &=\sin x-\sin^3 x\\ &=\sin x-\frac14(3\sin x-\sin 3x) &(*)\\ &=\frac{\sin3x-\sin x}{4} \end{align}

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For $(*)$, \begin{align} \sin 3x&=\sin(x+2x)\\ &=\sin x\cos 2x+\sin 2x\cos x \\ &= \sin x (1-2\sin^2x)+2\sin x\cos^2x\\ &= \sin x-2\sin^3x+2\sin x(1-\sin^2x)\\ &=3\sin x-4\sin^3x \end{align}

Hence, $$\sin^3x=\frac14(3\sin x-\sin3x)$$

Basically, you will have to make some sort of detour around $\sin 3x$ and $\sin^3x$ no matter how you approach this problem...