Module 4: Double Angle-Half Angle Formulas

4.1 Computation of the Formulas

In the previous sections we computed the following double angle formulas. It is very useful to have these memorized.

$LaTeX: \begin{equation*}\boxed{ \begin{array}{lcl} \cos(2x)&=&\cos^2(x)-\sin^2(x)\\\\ \cos(2x)&=&2\cos^2(x)-1\\\\ \cos(2x)&=&1-2\sin^2(x)\\\\ \sin(2x)&=&2\sin(x)\cos(x) \end{array}} \end{equation*}$ $\begin{equation*}\boxed{ \begin{array}{lcl} \cos(2x)&=&\cos^2(x)-\sin^2(x)\\\\ \cos(2x)&=&2\cos^2(x)-1\\\\ \cos(2x)&=&1-2\sin^2(x)\\\\ \sin(2x)&=&2\sin(x)\cos(x) \end{array}} \end{equation*}$

Let’s rewrite the cosine formula.

$LaTeX: \begin{align} 2\cos^2 (x) -1&=\cos(2x) \\ \cos^2 (x) &={\frac{\cos(2x)+1}{2} \label{half}} \end{align}$ $\begin{align} 2\cos^2 (x) -1&=\cos(2x) \\ \cos^2 (x) &={\frac{\cos(2x)+1}{2} \label{half}} \end{align}$

Remark: How may equation (2) be useful in integrating? Suppose we are asked to compute

$LaTeX: \int \cos^2 (x)\; dx.$ $\int \cos^2 (x)\; dx.$

Employing equation (2), we can replace cosine squared by an expression which has the function cosine in the first power, something we know how to integrate.

$LaTeX: \begin{array}{lcl}{ \int \cos^2 (x)\; dx}&=& {\int \frac{\cos(2x)+1}{2}}\; dx\\ &=& {\frac{\sin(2x)}{4}+\frac{x}{2}+C.} \end{array}$ $\begin{array}{lcl}{ \int \cos^2 (x)\; dx}&=& {\int \frac{\cos(2x)+1}{2}}\; dx\\ &=& {\frac{\sin(2x)}{4}+\frac{x}{2}+C.} \end{array}$

4.2 Examples

Compute $LaTeX: \int \sin^4(x)\; dx$ $\int \sin^4(x)\; dx$ .

$LaTeX: \begin{array}{lcl}{ \int \sin^4 (x)\; dx}&=& { \int \left(\sin^2 (x)\right)^2\; dx}\\\\ &=&{\int \left(\frac{1-\cos(2x)}{2}\right)^2}\; dx\\\\ &=& \frac{1}{4}{\int \left(1-2\cos(2x)+\cos^2(2x)\right)\; dx}\\\\ &=&{\frac{x}{4}-\frac{\sin(2x)}{4}+\frac{1}{4}\int \left(\frac{\cos(4x)+1}{2}\right)}\; dx\\\\ &=&{\frac{x}{4}-\frac{\sin(2x)}{4}+\frac{1}{4}\left(\frac{\sin(4x)}{8}+\frac{x}{2}\right)}+C \end{array}$ $\begin{array}{lcl}{ \int \sin^4 (x)\; dx}&=& { \int \left(\sin^2 (x)\right)^2\; dx}\\\\ &=&{\int \left(\frac{1-\cos(2x)}{2}\right)^2}\; dx\\\\ &=& \frac{1}{4}{\int \left(1-2\cos(2x)+\cos^2(2x)\right)\; dx}\\\\ &=&{\frac{x}{4}-\frac{\sin(2x)}{4}+\frac{1}{4}\int \left(\frac{\cos(4x)+1}{2}\right)}\; dx\\\\ &=&{\frac{x}{4}-\frac{\sin(2x)}{4}+\frac{1}{4}\left(\frac{\sin(4x)}{8}+\frac{x}{2}\right)}+C \end{array}$