Expressing Sums as Products
Some problems require the reverse of the process we just used. The sum-to-product formulas allow us to express sums of sine or cosine as products. These formulas can be derived from the product-to-sum identities. For example, with a few substitutions, we can derive the sum-to-product identity for sine . Let [latex]\frac{u+v}{2}=\alpha[/latex] and [latex]\frac{u-v}{2}=\beta[/latex].
Then,
[latex]\begin{align}\alpha +\beta &=\frac{u+v}{2}+\frac{u-v}{2} \\ &=\frac{2u}{2} \\ &=u \\ \text{ } \\ \alpha -\beta &=\frac{u+v}{2}-\frac{u-v}{2} \\ &=\frac{2v}{2} \\ &=v \end{align}[/latex]
Thus, replacing [latex]\alpha[/latex] and [latex]\beta[/latex] in the product-to-sum formula with the substitute expressions, we have
[latex]\begin{align}&\sin \alpha \cos \beta =\frac{1}{2}\left[\sin \left(\alpha +\beta \right)+\sin \left(\alpha -\beta \right)\right] \\ &\sin \left(\frac{u+v}{2}\right)\cos \left(\frac{u-v}{2}\right)=\frac{1}{2}\left[\sin u+\sin v\right]&& \text{Substitute for}\left(\alpha +\beta \right)\text{ and }\left(\alpha -\beta \right) \\ &2\sin \left(\frac{u+v}{2}\right)\cos \left(\frac{u-v}{2}\right)=\sin u+\sin v \end{align}[/latex]
The other sum-to-product identities are derived similarly.
sum-to-product formulas
The sum-to-product formulas are as follows:
[latex]\sin \alpha +\sin \beta =2\sin \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)[/latex]
[latex]\sin \alpha -\sin \beta =2\sin \left(\frac{\alpha -\beta }{2}\right)\cos \left(\frac{\alpha +\beta }{2}\right)[/latex]
[latex]\cos \alpha -\cos \beta =-2\sin \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)[/latex]
[latex]\cos \alpha +\cos \beta =2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)[/latex]
Write the following difference of sines expression as a product: [latex]\sin \left(4\theta \right)-\sin \left(2\theta \right)[/latex].
Show Solution
We begin by writing the formula for the difference of sines.
[latex]\sin \alpha -\sin \beta =2\sin \left(\frac{\alpha -\beta }{2}\right)\cos \left(\frac{\alpha +\beta }{2}\right)[/latex]
Substitute the values into the formula, and simplify.
[latex]\begin{align}\sin \left(4\theta \right)-\sin \left(2\theta \right)&=2\sin \left(\frac{4\theta -2\theta }{2}\right)\cos \left(\frac{4\theta +2\theta }{2}\right) \\ &=2\sin \left(\frac{2\theta }{2}\right)\cos \left(\frac{6\theta }{2}\right) \\ &=2\sin \theta \cos \left(3\theta \right) \end{align}[/latex]
Use the sum-to-product formula to write the sum as a product: [latex]\sin \left(3\theta \right)+\sin \left(\theta \right)[/latex].
Show Solution
[latex]2\sin \left(2\theta \right)\cos \left(\theta \right)[/latex]
Evaluate [latex]\cos \left({15}^{\circ }\right)-\cos \left({75}^{\circ }\right)[/latex].
Show Solution
We begin by writing the formula for the difference of cosines.
[latex]\cos \alpha -\cos \beta =-2\sin \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)[/latex]
Then we substitute the given angles and simplify.
[latex]\begin{align}\cos \left({15}^{\circ }\right)-\cos \left({75}^{\circ }\right)&=-2\sin \left(\frac{{15}^{\circ }+{75}^{\circ }}{2}\right)\sin \left(\frac{{15}^{\circ }-{75}^{\circ }}{2}\right) \\ &=-2\sin \left({45}^{\circ }\right)\sin \left(-{30}^{\circ }\right) \\ &=-2\left(\frac{\sqrt{2}}{2}\right)\left(-\frac{1}{2}\right) \\ &=\frac{\sqrt{2}}{2}\end{align}[/latex]
Prove the identity:
[latex]\frac{\cos \left(4t\right)-\cos \left(2t\right)}{\sin \left(4t\right)+\sin \left(2t\right)}=-\tan t[/latex]
Show Solution
We will start with the left side, the more complicated side of the equation, and rewrite the expression until it matches the right side.
[latex]\begin{align}\frac{\cos \left(4t\right)-\cos \left(2t\right)}{\sin \left(4t\right)+\sin \left(2t\right)}&=\frac{-2\sin \left(\frac{4t+2t}{2}\right)\sin \left(\frac{4t - 2t}{2}\right)}{2\sin \left(\frac{4t+2t}{2}\right)\cos \left(\frac{4t - 2t}{2}\right)} \\ &=\frac{-2\sin \left(3t\right)\sin t}{2\sin \left(3t\right)\cos t} \\ &=\frac{-2\sin \left(3t\right)\sin t}{2\sin \left(3t\right)\cos t} \\ &=-\frac{\sin t}{\cos t} \\ &=-\tan t \end{align}[/latex]
Analysis of the Solution
Recall that verifying trigonometric identities has its own set of rules. The procedures for solving an equation are not the same as the procedures for verifying an identity. When we prove an identity, we pick one side to work on and make substitutions until that side is transformed into the other side.