Trig identities and double angles

From $\sin^2\theta+\cos^2\theta=1$, $1-\sin^2\theta=\cos^2\theta$. If $\sin\theta=\frac{3}{5}$ and $\theta$ is acute, then $\cos\theta>0$, so $\cos\theta=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$.

The double-angle formulas are the addition formulas with $\beta=\alpha$.

Solving $\cos 2\theta=1-2\sin^2\theta$ for $\sin^2$ gives the half-angle formula $\sin^2\frac{\theta}{2}=\frac{1-\cos\theta}{2}$, used in integration to lower the power of $\sin^2 x$.

Examples

1. 01Write using $\cos$ only:$\displaystyle 1 - \sin^2\theta$AnswerHide

   Answer$\displaystyle \cos^2\theta$

   Steps

   1. $\displaystyle \sin^2\theta + \cos^2\theta$$\displaystyle = 1$$\displaystyle \cos^2\theta$$\displaystyle = 1 - \sin^2\theta$
2. 02$\theta$ is acute. What is $\cos\theta$?$\displaystyle \sin^2\theta+\cos^2\theta=1,\; \sin\theta = \frac{8}{17}$AnswerHide

   Answer$\displaystyle \frac{15}{17}$

   Steps

   1. $\displaystyle \cos^2\theta$$\displaystyle = 1 - \left(\frac{8}{17}\right)^2$$\displaystyle 1 - \left(\frac{8}{17}\right)^2$$\displaystyle = \frac{225}{289}$
   2. θ is acute, so take the positive root
   3. $\displaystyle \cos\theta$$\displaystyle = \frac{15}{17}$
3. 03Use this. If $\sin\theta = \frac{1}{4}$, what is $\cos 2\theta$?$\displaystyle \cos 2\theta = 1 - 2\sin^2\theta$AnswerHide

   Answer$\displaystyle \frac{7}{8}$

   Steps

   1. $\displaystyle \left(\frac{1}{4}\right)^2$$\displaystyle = \frac{1}{16}$
   2. Put this into 1 - 2sin²θ
   3. $\displaystyle 1 - 2 \cdot \frac{1}{16}$$\displaystyle = \frac{7}{8}$
4. 04Write using $\sin\theta$ and $\cos\theta$:$\displaystyle \sin 2\theta$AnswerHide

   Answer$\displaystyle 2\sin\theta\cos\theta$

   Steps

   1. $\displaystyle \sin 2\theta$$\displaystyle = 2\sin\theta\cos\theta$
   2. The 2 does not come out front. Check at θ = π/6
   3. $\displaystyle \theta = \frac{\pi}{6}$$\displaystyle \Rightarrow \sin 2\theta = \frac{\sqrt{3}}{2}$$\displaystyle \theta = \frac{\pi}{6}$$\displaystyle \Rightarrow 2\sin\theta = 1$
5. 05What is this equal to?$\displaystyle 2\sin\theta\cos\theta$AnswerHide

   Answer$\displaystyle \sin 2\theta$

   Steps

   1. $\displaystyle 2\sin\frac{\pi}{6}\cos\frac{\pi}{6}$$\displaystyle = \frac{\sqrt{3}}{2}$$\displaystyle \sin \frac{\pi}{3}$$\displaystyle = \frac{\sqrt{3}}{2}$
   2. Same value, and 2θ = π/3; so this is sin 2θ
   3. $\displaystyle \sin 2\theta$$\displaystyle = 2\sin\theta\cos\theta$
   4. Meanwhile cos 2θ is a different value
   5. $\displaystyle \cos \frac{\pi}{3}$$\displaystyle = \frac{1}{2}$
6. 06What goes in □?$\displaystyle \sin^2\theta + \cos^2\theta = \square$AnswerHide

   Answer$\displaystyle 1$

   Steps

   1. $\displaystyle \sin \frac{\pi}{6}$$\displaystyle = \frac{1}{2}$$\displaystyle \cos \frac{\pi}{6}$$\displaystyle = \frac{\sqrt{3}}{2}$
   2. Square them and add
   3. $\displaystyle \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$$\displaystyle = 1$

FAQ

Q1Is $\cos\theta$ equal to $\frac{4}{5}$ or $-\frac{4}{5}$?AnswerHide

A

It depends on the range of $\theta$. If $\theta$ is acute ($0<\theta<\frac{\pi}{2}$), $\cos\theta>0$, so it is $\frac{4}{5}$. If $\theta$ is obtuse, it is $-\frac{4}{5}$.

Q2Why are there three formulas for $\cos 2\theta$?AnswerHide

A

Replacing $\cos^2\theta$ or $\sin^2\theta$ in $\cos^2\theta-\sin^2\theta$ using $\sin^2\theta+\cos^2\theta=1$ gives $1-2\sin^2\theta$ or $2\cos^2\theta-1$. They are all equal; pick the one that fits the value you are given.