Trignomometry Identity Question

Question

I need help showing that $(4\sin\theta)(\sin(\theta-\frac{\pi}{3}))(\sin(\theta-\frac{2\pi}{3}))=\sin(3\theta)$.

Cheers.

Answer 1

Expand both sides. Left side:

$$4\sin\theta\left(\sin\theta\cos\frac{\pi}{3}-\cos\theta\sin\frac{\pi}{3}\right)\left(\sin\theta\cos\frac{2\pi}{3}-\cos\theta\sin\frac{2\pi}{3}\right)$$ Use complementary angles: $$=4\sin\theta\left(\sin\theta\cos\frac{\pi}{3}-\cos\theta\sin\frac{\pi}{3}\right)\left(-\sin\theta\cos\frac{\pi}{3}-\cos\theta\sin\frac{\pi}{3}\right)$$ Difference of squares: $$=-4\sin\theta\left(\sin^2\theta\cos^2\frac{\pi}{3}-\cos^2\theta\sin^2\frac{\pi}{3}\right)$$ $$=-4\sin\theta\left(\sin^2\theta\cos^2\frac{\pi}{3}-\left(1-\sin^2\theta\right)\sin^2\frac{\pi}{3}\right)$$ $$=-4\sin\theta\left(\sin^2\theta-\sin^2\frac{\pi}{3}\right)$$ $$=-4\sin\theta\left(\sin^2\theta-\frac34\right)$$ $$=3\sin\theta-4\sin^3\theta$$

Right side: $$\sin3\theta=\sin\theta\cos2\theta+\cos\theta\sin2\theta=$$ $$=\sin\theta(\cos^2\theta-\sin^2\theta)+2\cos^2\theta\sin\theta$$ $$=\sin\theta(1-2\sin^2\theta)+2(1-\sin^2\theta)\sin\theta$$ $$=3\sin\theta-4\sin^3\theta$$

You can also use triple angle formula if you know it.

Answer 2

$\displaystyle\sin\left(x-\frac{2\pi}3\right)=\sin\left[\left(\frac\pi3+x\right)-\pi\right]=-\sin\left[\pi-\left(x+\frac\pi3\right)\right]$ as $\sin(-y)=-\sin y$

$\displaystyle\sin\left(x-\frac{2\pi}3\right)=-\sin\left(x+\frac\pi3\right)$

Now using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,

$\displaystyle\sin\left(x-\frac\pi3\right)\sin\left(x-\frac{2\pi}3\right)=- \sin\left(x-\frac\pi3\right)\sin\left(x+\frac\pi3\right)$ $\displaystyle=-\left(\sin^2x-\sin^2\frac\pi3\right)=\frac34-\sin^2x$

Now we know $\displaystyle\sin3x=3\sin x-4\sin^3x$

Answer 3

Let $\displaystyle\sin3x=\sin3A$

So we have $\displaystyle3x=n\pi+(-1)^n3A$ where $n$ is any integer

$\displaystyle x=\frac{n\pi}3+(-1)^nA$ where $n$ can assume any three in-congruent values $\pmod 3$

If choose $n=1,0,-1$ we get $\displaystyle x=\frac\pi3-A,A,-\frac\pi3-A$

Again as $\sin3x=3\sin x−4\sin^3x,$ we have

$\displaystyle3\sin x−4\sin^3x=\sin3A\iff4\sin^3x-3\sin x+\sin3A=0$

which is a Cubic Equation in $\sin x$

So using Vieta's formula, $\displaystyle\sin\left(\frac\pi3-A\right)\cdot\sin A\cdot\sin\left(-\frac\pi3-A\right)=-\frac{\sin3A}4$

Now using $\displaystyle\sin(-u)=-\sin u,$ $\displaystyle\sin\left(\frac\pi3-A\right)=-\sin\left(A-\frac\pi3\right)$

and $\displaystyle\sin\left(-\frac\pi3-A\right)=\sin\left[-\pi-\left(A-\frac{2\pi}3\right)\right]$

$\displaystyle=\sin\left[-2\pi+\pi-\left(A-\frac{2\pi}3\right)\right]$

$\displaystyle=\sin\left[\pi-\left(A-\frac{2\pi}3\right)\right] =\sin\left(A-\frac{2\pi}3\right)$

Two observations :

I think this is how the problem came into being

There is no reason why this can not be generalized to $$(-1)^{n-1}\sin mA=2^{n-1}\prod_{n=0}^{m-1}\sin\left(\frac{n\pi}m+(-1)^nA\right)$$ where $m$ is a positive integer and $n$ can assume any three in-congruent values $\pmod m$

or $$\sin mA=2^{n-1}\prod_{n=0}^{m-1}\sin\left(\frac{n\pi}m+A\right)$$

Answer 4

By the Simpson's formulas:

$4\sin(x-\pi/3)\sin(x-2\pi/3)=2\cos(\pi/3)-2\cos(2x-\pi)=1+2\cos(2x)$

$\sin(x)+2\sin(x)\cos(2x)=\sin(x)+\sin(3x)-\sin(x)$