Sum to Product Trigonometric Identities

The sum-to-product trigonometric identities are similar to the product-to-sum trigonometric identities.

The basic sum-to-product identities for sine and cosine are as follows:

$$\begin{aligned} \sin x+\sin y & =2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) \\ \cos x+\cos y &=2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right). \end{aligned}$$

From these identities, we can also infer the difference-to-product identities:

$$\begin{aligned} \sin x-\sin y & = 2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) \\ \cos x-\cos y &= -2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) \\&= 2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{y-x}{2}\right) \end{aligned}$$

and the tangent sum and difference-to-product identities:

$$\begin{aligned} \tan (x) + \tan (y) &= \tan(x+y)\big(1-\tan(x)\tan(y)\big)\\ \tan(x)-\tan(y)&=\tan(x-y)\big(1+\tan(x)\tan(y)\big). \end{aligned}$$

The proof of the basic sum-to-product identity for sine proceeds as follows:

$$\begin{aligned} 2 \sin \left({\frac {\alpha + \beta} 2}\right) \cos \left({\frac {\alpha - \beta} 2}\right) & = 2 \frac {\sin \left({\frac {\alpha + \beta} 2 + \frac {\alpha - \beta} 2}\right) + \sin \left({\frac {\alpha + \beta} 2 - \frac {\alpha - \beta} 2}\right)} 2\\ &= \sin \left(\frac {2 \alpha} 2\right) + \sin \left(\frac {2 \beta} 2\right)\\ &= \sin \alpha + \sin \beta. \end{aligned}$$

The proofs for cosine and tangent are similar. $_\square$

Simplify

$$\dfrac{\cos3x-\cos x}{\cos3x+\cos x}.$$

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We have

$$\begin{aligned} \frac{\cos3x-\cos x}{\cos3x+\cos x} &= \frac{-2\sin\left(\frac{3x+x}{2}\right)\sin\left(\frac{3x-x}{2}\right)}{2\cos\left(\frac{3x+x}{2}\right)\cos\left(\frac{3x-x}{2}\right)} \\ & = -\frac{2\sin2x\sin x}{2\cos2x\cos x} \\\\ & = -\tan2x\tan x.\ _\square \end{aligned}$$

In triangle $\triangle ABC$, prove that

$$\sin A+\sin B+\sin C=4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}.$$

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We have

$$\begin{aligned} \sin A+\sin B+\sin C &=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) +2\sin\frac{C}{2}\cos\frac{C}{2} \\ &= 2\cos\frac{C}{2}\cos\left(\frac{A-B}{2}\right) + 2\sin\frac{C}{2}\cos\frac{C}{2} \\ & = 2\cos\frac{C}{2}\left(\cos\Big(\frac{A}{2}-\frac{B}{2}\Big)+\sin\frac{C}{2}\right) \\ &= 2\cos\frac{C}{2}\left(\cos\Big(\frac{A}{2}-\frac{B}{2}\Big)+\cos\Big(\frac{A}{2}+\frac{B}{2}\Big)\right) \\ &= 2\cos\frac{C}{2}\left(2\cos\frac{A}{2}\cos\frac{B}{2}\right) \\ &= 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}.\ _\square \end{aligned}$$

Find all solutions in the domain $[-2\pi,2\pi]$ to the equation

$$\cos(4x) + \cos(2x) =0.$$

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We have

$$\begin{aligned} \cos(4x)+\cos(2x) &=0\\ 2\cos\left(\frac{4x+2x}{2}\right)\cos\left(\frac{4x-2x}{2}\right) &=0 \\ \cos(3x)\cos(x) &=0. \\ \end{aligned}$$

For $\cos(3x)=0,~x=\frac{4n\pi\pm\pi}{6}\implies x=-\frac{5\pi}{6},~-\frac{\pi}{6},~\frac{\pi}{6},~\frac{5\pi}{6}.$
For $\cos(x)=0,~x=2n\pi\pm\frac{\pi}{2}\implies x=-\frac{3\pi}{2},~-\frac{\pi}{2},~\frac{\pi}{2},~\frac{3\pi}{2}.$

Hence, the solutions are $x = -\frac{3\pi}{2},~-\frac{5\pi}{6},~-\frac{\pi}{2},~-\frac{\pi}{6},~\frac{\pi}{6},~\frac{\pi}{2},~\frac{5\pi}{6},~\frac{3\pi}{2}$. $_\square$

Write $\sin(x) + \sin(2x) + \sin(3x) + \sin(4x)$ as a product of trigonometric functions.

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We have

\[\begin{align}
\sin(x)+ \sin(2x)+\sin(3x)+\sin(4x) &=\big(\sin(4x)+\sin(x)\big)+\big(\sin(3x)+\sin(2x)\big)\\ & =2\sin\left(\frac{4x+x}{2}\right)\cos\left(\frac{4x-x}{2}\right)+2\sin\left(\frac{3x+2x}{2}\right)\cos\left(\frac{3x-2x}{2}\right)\\ &=2\sin\left(\frac{5x}{2}\right)\cos\left(\frac{3x}{2}\right)+2\sin\left(\frac{5x}{2}\right)\cos\left(\frac{x}{2}\right)\\ &=2\sin\left(\frac{5x}{2}\right)\left(\cos\Big(\frac{3x}{2}\Big)+\cos\Big(\frac{x}{2}\Big)\right)\\ &=2\sin\left(\frac{5x}{2}\right)\left(2\cos\left(\frac{\frac{3x}{2}+\frac{x}{2}}{2}\right)\cos\left(\frac{\frac{3x}{2}-\frac{x}{2}}{2}\right)\right)\\ &=2\sin\left(\frac{5x}{2}\right)\left(2\cos(x)\cos\Big(\frac{x}{2}\Big)\right)\\ &=4\sin\left(\frac{5x}{2}\right)\cos(x)\cos\left(\frac{x}{2}\right).\ _\square \end{align}\]