Trigonometric Equations - Sum to Product

The sum to product identities are identities for turning sums of trigonometric functions into products. The main identities are

$\begin{aligned} \cos a+\cos b & =&2\cos{\left(\dfrac{a+b}{2}\right)}\cos{\left(\dfrac{a-b}{2}\right)}\\ \cos a-\cos b & =&-2\sin{\left(\dfrac{a+b}{2}\right)}\sin{\left(\dfrac{a-b}{2}\right)}\\ \sin a+\sin b & =&2\sin{\left(\dfrac{a+b}{2}\right)}\cos{\left(\dfrac{a-b}{2}\right)}\\ \sin a-\sin b & =&2\cos{\left(\dfrac{a+b}{2}\right)}\sin{\left(\dfrac{a-b}{2}\right)}. \end{aligned}$

The following related identities convert products into sums:

$\begin{aligned} \sin (a+b) + \sin (a-b) &= 2\sin a\cos b\\ \cos (a+b) + \cos (a-b) &= 2\cos a\cos b\\ \cos (a-b) - \cos (a+b) &= 2\sin a\sin b. \end{aligned}$

Find all $x$ in the range $[0, 2\pi)$ such that

$\cos(3x) + \cos(x) = 0.$

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By the sum to product identity for cosine, the equation can be rewritten as

$0 = \cos(3x) + \cos(x) = \cos (2x+x) + \cos (2x-x) = 2\cos(2x) \cos(x).$

Therefore, we would like to find the values of $x$ such that $\cos(2x) = 0$ or $\cos(x) = 0$. Now, $\cos (x) = 0$ for $x$ equal to odd multiples of $\frac{\pi}{2}$, and thus, the solutions to $\cos(2x)=0$ are $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$. Therefore, the set of solutions are $x = \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}$. $_\square$

Write the sum $\sin(5x) + \sin(6x)$ as a product of trigonometric functions.

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By the sum to product formula $\sin (a-b) + \sin (a+b) = 2\sin a\cos b,$ we have

[\begin{align} \sin(5x) + \sin(6x) &= \sin \left( \frac{11x}{2}-\frac{x}{2} \right) + \sin \left( \frac{11x}{2}+ \frac{x}{2}\right) \ &= 2 \sin \left( \frac{11x}{2} \right) \cos \left( \frac{x}{2} \right). $_\square$ \end{align}]

Prove the identity $\dfrac{\cos(3k)-\cos(k)}{\cos (3k)+ \cos (k)}=-\tan (2k)\tan (k)$.

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Using the second identity above, the numerator is

$\cos(3k)-\cos(k)=-2\sin{\left(\dfrac{3k+k}{2}\right)}\sin{\left(\dfrac{3k-k}{2}\right)}=-2\sin(2k)\sin (k).$

Using the first identity, the denominator can be changed to

$\cos (3k)+ \cos (k)=2\cos{\left(\dfrac{3k+k}{2}\right)}\cos{\left(\dfrac{3k-k}{2}\right)}=2\cos (2k)\cos(k).$

Substituting for our numerator and denominator yields

$\dfrac{-2\sin (2k) \sin (k)}{2\cos (2k)\cos (k)}.$

Using the identity $\cfrac{\sin (a)}{\cos (a)}=\tan (a),$ the above can be simplified to

$\dfrac{(-2)}{2} \cdot \left(\dfrac{\sin (2k)}{\cos(2k)}\right) \cdot \left( \dfrac{\sin (k)}{\cos (k)}\right) =-\tan (2k)\tan (k).\ _\square$