Trigonometric Equations - Sum to Product

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

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

The following related identities convert products into sums:

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

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

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