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:

\[\begin{align} \sin x+\sin y & =2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) \\ \cos x+\cos y &=2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) \\ \end{align}\]

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

\[\begin{align} \sin x-\sin y & = 2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) \\ \cos x-\cos y &= -2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) = 2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{y-x}{2}\right)\\ \end{align}\]

and the tangent sum and difference to product identity:

\[ \tan (x) + \tan (y) = (\tan(x+y))(1-\tan(x)\tan(y))\] \[\tan(x)-\tan(y)=(\tan(x-y))(1+\tan(x)\tan(y))\]

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

\[\begin{align} 2 \sin \left({\frac {\alpha + \beta} 2}\right) \cos \left({\frac {\alpha - \beta} 2}\right) & = 2 \frac {\sin \left({\dfrac {\alpha + \beta} 2 + \dfrac {\alpha - \beta} 2}\right) + \sin \left({\dfrac {\alpha + \beta} 2 - \dfrac {\alpha - \beta} 2}\right)} 2\\ &= \sin \frac {2 \alpha} 2 + \sin \frac {2 \beta} 2\\ &= \sin \alpha + \sin \beta \end{align}\]

Thus, \[\sin \alpha+\sin \beta =2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)\]

The proofs for cosine and tangent are similar.

Simplify \[\dfrac{\cos3x-\cos x}{\cos3x+\cos x}\]

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We have

\[\begin{align} \dfrac{\cos3x-\cos x}{\cos3x+\cos x} &= \dfrac{-2\sin\left(\dfrac{3x+x}{2}\right)\sin\left(\dfrac{3x-x}{2}\right)}{2\cos\left(\dfrac{3x+x}{2}\right)\cos\left(\dfrac{3x-x}{2}\right)} \\ & = -\dfrac{2\sin2x\sin x}{2\cos2x\cos x} \\ & = -\tan2x\tan x \end{align}\]

In triangle \(\Delta ABC\), 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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Solution: We have \[\begin{align} \sin A+\sin B+\sin C &=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) +2\sin\frac{C}{2}\cos\frac{C}{2} \\ &= 2\cos\frac{C}{2}\cos\left(\frac{A-B}{2}\right) + 2\sin\frac{C}{2}\cos\frac{C}{2} \\ & = 2\cos\frac{C}{2}\left(\cos\left(\frac{A}{2}-\frac{B}{2}\right)+\sin\frac{C}{2}\right) \\ &= 2\cos\frac{C}{2}\left(\cos\left(\frac{A}{2}-\frac{B}{2}\right)+\cos\left(\frac{A}{2}+\frac{B}{2}\right)\right) \\ &= 2\cos\frac{C}{2}\left(2\cos\frac{A}{2}\cos\frac{B}{2}\right) \\ &= 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}. \end{align}\]

Find all solutions in the domain \([-2\pi,2\pi]\) to the equation \[\cos(4x) + cos(2x) =0.\]

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Solution:

We have

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

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

For \(\cos(x)=0,~x=2n\pi\pm\frac{\pi}{2}\rightarrow x=-\frac{3\pi}{2},~-\frac{\pi}{2},~\frac{\pi}{2},~\frac{3\pi}{2}.\)

Hence, the solutions are \(x = -\frac{3\pi}{2},~-\frac{5\pi}{6},~-\frac{\pi}{2},~-\frac{\pi}{6},~\frac{\pi}{6},~\frac{\pi}{2},~\frac{5\pi}{6},~\frac{3\pi}{2}\).

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

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Solution: We have

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