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Problem: Find a formula for $\sin(3a)$ in terms of $\sin(a)$ and $\cos(a)$. Use this to calculate $\sin(π/3)$ and $\cos(π/3)$.

My attempt:

\begin{align} \sin(3a) &= \sin(2a + a) = \sin(2a)\cos(a) + \cos(2a)\sin(a) \\ &= \sin(a + a)\cos(a) + \cos(a + a)\sin(a) \\ &= [\sin(a)\cos(a) + \cos(a)\sin(a)]\cos(a) + [\cos(a)\cos(a) - \sin(a)\sin(a)]\sin(a). \end{align}

It can then be simplified to

$$2\sin(a)\cos^2(a) + \sin(a)\cos^2(a) - \sin^3(a) = 3\sin(a)\cos^2(a) - \sin^3(a).$$

My question is this: How am I supposed to use this formula to find $\sin(π/3)$ and $\cos(π/3)$?

Xander Henderson
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2 Answers2

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$$\sin 3a = \sin (2a+a) = \sin(2a)\cos(a)+ \cos(2a)\sin(a)= 2\sin(a)\cos^2(a)+\sin(a)(1-2\sin^2(a))=2\sin(a)(1-\sin^2(a))+\sin(a)(1-2\sin^2(a)) =3\sin(a)-4\sin^3(a) $$

So...

$$\sin(\pi) = 0 = 3\sin(\dfrac{\pi}{3})-4\sin^3(\dfrac{\pi}{3})\implies \dfrac{3}{4} =\sin^2(\dfrac{\pi}{3})\implies \sin(\dfrac{\pi}{3})=\dfrac{\sqrt{3}}{2}$$

Using $cos^2(a)+\sin^2(a)=1$

$$\cos^2(\dfrac{\pi}{3})+\dfrac{3}{4} = 1\implies \cos^2(\dfrac{\pi}{3}) =\dfrac{1}{4}\implies \cos(\dfrac{\pi}{3}) =\dfrac{1}{2}$$

3d0
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hint: Put $x = \sin \left(\frac{\pi}{3}\right)$, then the equation $0=\sin\left(3\cdot\dfrac{\pi}{3}\right)= 3\sin\left(\frac{\pi}{3}\right) - 4\sin^3(\frac{\pi}{3})= 3x - 4x^3$ gives: $4x^3-3x = 0 \to x(4x^2-3) = 0$. Can you solve for $x$. Note that $x \neq 0$.

DeepSea
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