TIE - Double Angle Identities Lesson

Double Angle Identities

Let's derive the double angle identities:

Recall	What if u = v?
$LaTeX: \sin\left(u+v\right)=\sin u\cos v+\cos u\sin v$ $\sin\left(u+v\right)=\sin u\cos v+\cos u\sin v$	$LaTeX: \sin\left(u+u\right)=\sin u\cos u+\cos u\sin u=2\sin$ $\sin\left(u+u\right)=\sin u\cos u+\cos u\sin u=2\sin$
$LaTeX: \cos\left(u+v\right)=\cos u\cos v-\sin u\sin v$ $\cos\left(u+v\right)=\cos u\cos v-\sin u\sin v$	$LaTeX: \cos(u+u)=\cos u\cos u-\sin u\sin\:u\\ =\cos^2u-\sin^2u\\ =2\cos^2u-1\\ =1-2\sin^2u$ $\cos(u+u)=\cos u\cos u-\sin u\sin\:u\\ =\cos^2u-\sin^2u\\ =2\cos^2u-1\\ =1-2\sin^2u$
$LaTeX: \tan\left(u+v\right)=\frac{\tan u+\tan v}{1-\tan u\tan v}$ $\tan\left(u+v\right)=\frac{\tan u+\tan v}{1-\tan u\tan v}$	$LaTeX: \tan\left(u+u\right)=\frac{\tan u+\tan u}{1-\tan u\tan u\:}$ $\tan\left(u+u\right)=\frac{\tan u+\tan u}{1-\tan u\tan u\:}$

Double Angle Identities

equation image indicator $LaTeX: \sin\left(2\theta\right)=2\sin\theta\cos\theta;\\ \cos\left(2\theta\right)=\cos^2\theta-\sin^2\theta\\ =2\cos^2\theta-1 \\ =1-2\sin^2\theta;\\ \tan\left(2\theta\right)=\frac{2\tan\theta}{1-\tan^2\theta}$ $\sin\left(2\theta\right)=2\sin\theta\cos\theta;\\ \cos\left(2\theta\right)=\cos^2\theta-\sin^2\theta\\ =2\cos^2\theta-1 \\ =1-2\sin^2\theta;\\ \tan\left(2\theta\right)=\frac{2\tan\theta}{1-\tan^2\theta}$

Let's try this problem:

If equation image indicator $LaTeX: \sin\theta=-\frac{7}{25}$ $\sin\theta=-\frac{7}{25}$ and $LaTeX: \theta$ $\theta$ is located in the Quadrant III, find $LaTeX: \sin\left(2\theta\right),\:\cos\left(2\theta\right),\:\tan\left(2\theta\right)$ $\sin\left(2\theta\right),\:\cos\left(2\theta\right),\:\tan\left(2\theta\right)$ .

1. Since equation image indicator $LaTeX: \sin\left(2\theta\right)=\frac{y}{r}=-\frac{7}{25}$ $\sin\left(2\theta\right)=\frac{y}{r}=-\frac{7}{25}$ and our angle is in the third quadrant, we know that y = -7 and r = 25. So, let's solve for x.

equation image indicator $LaTeX: x^2+y^2=r^2\\ x^2+\left(-7\right)^2=\left(25\right)^2\\ :x^2+49=625\\ x^2=576\\ x=\pm24$ $x^2+y^2=r^2\\ x^2+\left(-7\right)^2=\left(25\right)^2\\ :x^2+49=625\\ x^2=576\\ x=\pm24$

2. Since our angle is in the third quadrant, we know that x = -24 and thus: equation image indicator $LaTeX: \cos\theta=-\frac{24}{5}\:and\:\tan\theta=\frac{7}{24}$ $\cos\theta=-\frac{24}{5}\:and\:\tan\theta=\frac{7}{24}$ .

3. So, now let's use these facts to find equation image indicator $LaTeX: \sin\left(2\theta\right),\:\cos\left(2\theta\right),\:\tan\left(2\theta\right)$ $\sin\left(2\theta\right),\:\cos\left(2\theta\right),\:\tan\left(2\theta\right)$ .

equation image indicator $LaTeX: \sin\left(2\theta\right)=2\sin\theta\cos\theta=2\left(-\frac{7}{25}\right)\left(-\frac{24}{25}\right)=\frac{336}{625}\\ \cos\left(2\theta\right)=\cos^2\theta-\sin^2\theta=\left(-\frac{24}{25}\right)^2-\left(-\frac{7}{25}\right)^2=\frac{527}{625} \\ \tan\left(2\theta\right)=\frac{2\tan\theta}{1-\tan^2\theta}\\ =\frac{2\left(\frac{7}{24}\right)}{1-\left(\frac{7}{24}\right)^2}=\frac{336}{527}$ $\sin\left(2\theta\right)=2\sin\theta\cos\theta=2\left(-\frac{7}{25}\right)\left(-\frac{24}{25}\right)=\frac{336}{625}\\ \cos\left(2\theta\right)=\cos^2\theta-\sin^2\theta=\left(-\frac{24}{25}\right)^2-\left(-\frac{7}{25}\right)^2=\frac{527}{625} \\ \tan\left(2\theta\right)=\frac{2\tan\theta}{1-\tan^2\theta}\\ =\frac{2\left(\frac{7}{24}\right)}{1-\left(\frac{7}{24}\right)^2}=\frac{336}{527}$

Note, you could find equation image indicator $LaTeX: \tan2\theta$ $\tan2\theta$ using the method below: $LaTeX: \tan\left(2\theta\right)=\frac{\sin2\theta}{\cos2\theta}=\frac{\frac{336}{625}}{\frac{527}{625}}=\frac{336}{527}$ $\tan\left(2\theta\right)=\frac{\sin2\theta}{\cos2\theta}=\frac{\frac{336}{625}}{\frac{527}{625}}=\frac{336}{527}$ .