The sum to product identities are identities for turning sums of trigonometric functions into products. The main identities are
cosa+cosbcosa−cosbsina+sinbsina−sinb====2cos(2a+b)cos(2a−b)−2sin(2a+b)sin(2a−b)2sin(2a+b)cos(2a−b)2cos(2a+b)sin(2a−b).
The following related identities convert products into sums:
sin(a+b)+sin(a−b)cos(a+b)+cos(a−b)cos(a−b)−cos(a+b)=2sinacosb=2cosacosb=2sinasinb.
Find all x in the range [0,2π) such that
cos(3x)+cos(x)=0.
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)=2cos(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 2π, and thus, the solutions to cos(2x)=0 are x=4π,43π,45π,47π. Therefore, the set of solutions are x=4π,2π,43π,45π,23π,47π. □
Write the sum sin(5x)+sin(6x) as a product of trigonometric functions.
By the sum to product formula sin(a−b)+sin(a+b)=2sinacosb, 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). □
\end{align}]
Prove the identity cos(3k)+cos(k)cos(3k)−cos(k)=−tan(2k)tan(k).
Using the second identity above, the numerator is
cos(3k)−cos(k)=−2sin(23k+k)sin(23k−k)=−2sin(2k)sin(k).
Using the first identity, the denominator can be changed to
cos(3k)+cos(k)=2cos(23k+k)cos(23k−k)=2cos(2k)cos(k).
Substituting for our numerator and denominator yields
2cos(2k)cos(k)−2sin(2k)sin(k).
Using the identity cos(a)sin(a)=tan(a), the above can be simplified to
2(−2)⋅(cos(2k)sin(2k))⋅(cos(k)sin(k))=−tan(2k)tan(k). □