Use Reduction Formulas to Simplify an Expression
The double-angle formulas can be used to derive the reduction formulas , which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the double-angle formulas.
We can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let’s begin with [latex]\cos \left(2\theta \right)=1 - 2{\sin }^{2}\theta[/latex]. Solve for [latex]{\sin }^{2}\theta :[/latex]
[latex]\begin{gathered}\cos \left(2\theta \right)=1 - 2{\sin }^{2}\theta \\ 2{\sin }^{2}\theta =1-\cos \left(2\theta \right) \\ {\sin }^{2}\theta =\frac{1-\cos \left(2\theta \right)}{2} \end{gathered}[/latex]
Next, we use the formula [latex]\cos \left(2\theta \right)=2{\cos }^{2}\theta -1[/latex]. Solve for [latex]{\cos }^{2}\theta :[/latex]
[latex]\begin{gathered}\cos \left(2\theta \right)=2{\cos }^{2}\theta -1 \\ 1+\cos \left(2\theta \right)=2{\cos }^{2}\theta \\ \frac{1+\cos \left(2\theta \right)}{2}={\cos }^{2}\theta \end{gathered}[/latex]
The last reduction formula is derived by writing tangent in terms of sine and cosine:
[latex]\begin{align}{\tan }^{2}\theta &=\frac{{\sin }^{2}\theta }{{\cos }^{2}\theta } \\ &=\frac{\frac{1-\cos \left(2\theta \right)}{2}}{\frac{1+\cos \left(2\theta \right)}{2}}&& \text{Substitute the reduction formulas.} \\ &=\left(\frac{1-\cos \left(2\theta \right)}{2}\right)\left(\frac{2}{1+\cos \left(2\theta \right)}\right) \\ &=\frac{1-\cos \left(2\theta \right)}{1+\cos \left(2\theta \right)} \end{align}[/latex]
reduction formulas
[latex]\begin{align}&{\sin }^{2}\theta =\frac{1-\cos \left(2\theta \right)}{2} \\ &{\cos }^{2}\theta =\frac{1+\cos \left(2\theta \right)}{2} \\ &{\tan }^{2}\theta =\frac{1-\cos \left(2\theta \right)}{1+\cos \left(2\theta \right)} \end{align}[/latex]
Write an equivalent expression for [latex]{\cos }^{4}x[/latex] that does not involve any powers of sine or cosine greater than 1.
Show Solution
We will apply the reduction formula for cosine twice.
[latex]\begin{align}{\cos }^{4}x&={\left({\cos }^{2}x\right)}^{2} \\ &={\left(\frac{1+\cos \left(2x\right)}{2}\right)}^{2}&& {\text{Substitute reduction formula for cos}}^{2}x. \\ &=\frac{1}{4}\left(1+2\cos \left(2x\right)+{\cos }^{2}\left(2x\right)\right) \\ &=\frac{1}{4}+\frac{1}{2}\cos \left(2x\right)+\frac{1}{4}\left(\frac{1+\cos \left(2\left(2x\right)\right)}{2}\right) && {\text{ Substitute reduction formula for cos}}^{2}x. \\ &=\frac{1}{4}+\frac{1}{2}\cos \left(2x\right)+\frac{1}{8}+\frac{1}{8}\cos \left(4x\right) \\ &=\frac{3}{8}+\frac{1}{2}\cos \left(2x\right)+\frac{1}{8}\cos \left(4x\right) \end{align}[/latex]
Analysis of the Solution
The solution is found by using the reduction formula twice, as noted, and the perfect square formula from algebra.
Use the power-reducing formulas to prove
[latex]{\sin }^{3}\left(2x\right)=\left[\frac{1}{2}\sin \left(2x\right)\right]\left[1-\cos \left(4x\right)\right][/latex]
Show Solution
We will work on simplifying the left side of the equation:
[latex]\begin{align}{\sin }^{3}\left(2x\right)&=\left[\sin \left(2x\right)\right]\left[{\sin }^{2}\left(2x\right)\right] \\ &=\sin \left(2x\right)\left[\frac{1-\cos \left(4x\right)}{2}\right]&& \text{Substitute the power-reduction formula}. \\ &=\sin \left(2x\right)\left(\frac{1}{2}\right)\left[1-\cos \left(4x\right)\right] \\ &=\frac{1}{2}\left[\sin \left(2x\right)\right]\left[1-\cos \left(4x\right)\right] \end{align}[/latex]
Analysis of the Solution
Note that in this example, we substituted
[latex]\frac{1-\cos \left(4x\right)}{2}[/latex]
for [latex]{\sin }^{2}\left(2x\right)[/latex]. The formula states
[latex]{\sin }^{2}\theta =\frac{1-\cos \left(2\theta \right)}{2}[/latex]
We let [latex]\theta =2x[/latex], so [latex]2\theta =4x[/latex].
Use the power-reducing formulas to prove that [latex]10{\cos }^{4}x=\frac{15}{4}+5\cos \left(2x\right)+\frac{5}{4}\cos \left(4x\right)[/latex].
Show Solution
[latex]\begin{align}10{\cos }^{4}x&=10{\left({\cos }^{2}x\right)}^{2} \\ &=10{\left[\frac{1+\cos \left(2x\right)}{2}\right]}^{2}&& {\text{Substitute reduction formula for cos}}^{2}x. \\ &=\frac{10}{4}\left[1+2\cos \left(2x\right)+{\cos }^{2}\left(2x\right)\right] \\ &=\frac{10}{4}+\frac{10}{2}\cos \left(2x\right)+\frac{10}{4}\left(\frac{1+\cos\left( 2\left(2x\right)\right)}{2}\right)&& {\text{Substitute reduction formula for cos}}^{2}x. \\ &=\frac{10}{4}+\frac{10}{2}\cos \left(2x\right)+\frac{10}{8}+\frac{10}{8}\cos \left(4x\right) \\ &=\frac{30}{8}+5\cos \left(2x\right)+\frac{10}{8}\cos \left(4x\right) \\ &=\frac{15}{4}+5\cos \left(2x\right)+\frac{5}{4}\cos \left(4x\right) \end{align}[/latex]