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I have this expression:

$$ ψ(χ) = A\sin^3(\frac{πχ}{α}) $$

And somehow the book i read equalizes the previous equation to this one:

$$ ψ(χ) = \frac{A}{4}[3\sin(\frac{πχ}{α}) - \sin(\frac{3πχ}{α})] $$

What trigonometric identity was used to make this possible?

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The identity is $$\sin^3x=\frac{3\sin x-\sin(3x)}{4}$$ Proof: \begin{align*} 4\sin^3x-3\sin x &= \sin x(4\sin^2x-3) \\ &=\sin x(1-4\cos^2x) \quad \text{Pythagorean identity} \\ &=\sin x(2\cos^2x-\cos(2x)-4\cos^2x) \quad \text{Double angle cosine} \\ &=-\sin x(2\cos^2x+\cos(2x)) \\ &=-[(2\sin x\cos x)\cos x+\sin x\cos(2x)] \\ &=-[\sin (2x)\cos x+\sin x\cos(2x)]\quad \text{Double angle sine} \\ &=-\sin (2x+x)\quad \text{identity sum sine} \\ &=-\sin(3x) \end{align*}

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By DeMoivre's identity,

$\cos(3x)+i\sin(3x)=(\cos(x)+i\sin(x))^3=$

$\cos^3(x)+3i\cos^2(x)\sin(x)-3\cos(x)\sin^2(x)-i\sin^3(x)$.

Grouping the imaginary terms together, we see that $\sin(3x)=3\cos^2(x)\sin(x)-\sin^3(x)$.

We can rewrite $3\cos^2(x)\sin(x)$ as $3\sin(x)(1-\sin^2(x))$, which gives us $\sin(3x)=3\sin(x)-4\sin^3(x)$.

Finally, we can write $\sin^3(x)=\frac{1}{4}(3\sin(x)-\sin(3x))$.

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    $\begingroup$ +1 for demoivre $\endgroup$
    – user251865
    Jun 11 '19 at 0:13

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