<meta http-equiv="refresh" content="1; url=/nojavascript/">

# 14.5: Double and Half Angle Formulas

Difficulty Level: At Grade Created by: CK-12

Objective

You’ll learn how to use the double and half angle formulas.

Review Queue

Use your calculator to find the value of the trig functions below. Round your answers to 4 decimal places.

1. $\sin 22.5^\circ$

2. $\tan 157.5$

Find the exact values of the trig expressions below.

3. $\cos \left(\frac{\pi}{4}+\frac{\pi}{4}\right)$

4. $\sin \left(\frac{2 \pi}{3} - \frac{\pi}{4}\right)$

## Finding Exact Trig Values using Double and Half Angle Formulas

Objective

Here you'll use the half and double angle formulas to find exact values of angles other than the critical angles.

Guidance

In the previous concept, we added two different angles together to find the exact values of trig functions. In this concept, we will learn how to find the exact values of the trig functions for angles that are half or double of other angles. Here we will introduce the Double-Angle $(2a)$ and Half-Angle $\left(\frac{a}{2}\right)$ Formulas.

Double-Angle and Half-Angle Formulas

$\cos 2a&=\cos^2 a- \sin^2a && \sin 2a=2 \sin a \cos a \\&=2 \cos^2 a-1 && \tan 2a=\frac{2 \tan a}{1- \tan^2 a} \\&=1- \sin^2a \\\sin \frac{a}{2}&= \pm \sqrt{\frac{1- \cos a}{2}} && \tan \frac{a}{2}=\frac{1- \cos a}{\sin a} \\\cos \frac{a}{2}&= \pm \sqrt{\frac{1+ \cos a}{2}} && \qquad \ \ =\frac{\sin a}{1+ \cos a}$

The signs of $\sin \frac{a}{2}$ and $\cos \frac{a}{2}$ depend on which quadrant $\frac{a}{2}$ lies in. For $\cos 2a$ and $\tan \frac{a}{2}$ any formula can be used to solve for the exact value.

Example A

Find the exact value of $\cos \frac{\pi}{8}$.

Solution: $\frac{\pi}{8}$ is half of $\frac{\pi}{4}$ and in the first quadrant.

$\cos \left(\frac{1}{2} \cdot \frac{\pi}{4}\right)&=\sqrt{\frac{1+ \cos \frac{\pi}{4}}{2}} \\&=\sqrt{\frac{1+ \frac{\sqrt{2}}{2}}{2}} \\&=\sqrt{\frac{1}{2} \cdot \frac{2+\sqrt{2}}{2}} \\&=\frac{\sqrt{2+ \sqrt{2}}}{2}$

Example B

Find the exact value of $\sin 2a$ if $\cos a=- \frac{4}{5}$ and $\frac{3 \pi}{2} \le a < 2 \pi$.

Solution: To use the sine double-angle formula, we also need to find $\sin a$, which would be $\frac{3}{5}$ because $a$ is in the $4^{th}$ quadrant.

$\sin 2a&=2 \sin a \cos a \\&=2 \cdot \frac{3}{5} \cdot - \frac{4}{5} \\&=- \frac{24}{25}$

Example C

Find the exact value of $\tan 2a$ for $a$ from Example B.

Solution: Use $\tan a=\frac{\sin a}{\cos a}=\frac{\frac{3}{5}}{- \frac{4}{5}}=- \frac{3}{4}$ to solve for $\tan 2a$.

$\tan 2a=\frac{2 \cdot - \frac{3}{4}}{1- \left(- \frac{3}{4}\right)^2}=\frac{- \frac{3}{2}}{\frac{7}{16}}= - \frac{3}{2} \cdot \frac{16}{7}=- \frac{24}{7}$

Guided Practice

1. Find the exact value of $\cos \left(-\frac{5 \pi}{8}\right)$.

2. $\cos a=\frac{4}{7}$ and $0 \le a < \frac{\pi}{2}$. Find:

a) $\sin 2a$

b) $\tan \frac{a}{2}$

1. $- \frac{5 \pi}{8}$ is in the $3^{rd}$ quadrant.

$- \frac{5 \pi}{8}=\frac{1}{2} \left(- \frac{5 \pi}{4}\right) \rightarrow \cos \frac{1}{2} \left(- \frac{5 \pi}{4}\right)=- \sqrt{\frac{1+ \cos \left(- \frac{5 \pi}{4}\right)}{2}}=-\sqrt{\frac{1- \frac{\sqrt{2}}{2}}{2}}=\sqrt{\frac{1}{2} \cdot \frac{2-\sqrt{2}}{2}}=\frac{\sqrt{2- \sqrt{2}}}{2}$

2. First, find $\sin a$. $4^2+y^2=7^2\rightarrow y=\sqrt{33}$, so $\sin a=\frac{\sqrt{33}}{7}$

a) $\sin 2a=2 \cdot \frac{\sqrt{33}}{7} \cdot \frac{4}{7}=\frac{8 \sqrt{33}}{49}$

b) You can use either $\tan \frac{a}{2}$ formula.

$\tan \frac{a}{2}=\frac{1- \frac{4}{7}}{\frac{\sqrt{33}}{7}}=\frac{3}{7} \cdot \frac{7}{\sqrt{33}}=\frac{3}{\sqrt{33}}=\frac{\sqrt{33}}{11}$

Vocabulary

Double-Angle and Half-Angle Formulas
$\cos 2a&=\cos^2 a- \sin^2a && \sin 2a=2 \sin a \cos a \\&=2 \cos^2 a-1 && \tan 2a=\frac{2 \tan a}{1- \tan^2 a} \\&=1- \sin^2a \\\sin \frac{a}{2}&= \pm \sqrt{\frac{1- \cos a}{2}} && \tan \frac{a}{2}=\frac{1- \cos a}{\sin a} \\\cos \frac{a}{2}&= \pm \sqrt{\frac{1+ \cos a}{2}} && \qquad \ \ =\frac{\sin a}{1+ \cos a}$

Problem Set

Find the exact value of the following angles.

1. $\sin 105^\circ$
2. $\tan \frac{\pi}{8}$
3. $\cos \frac{5 \pi}{12}$
4. $\cos 165^\circ$
5. $\sin \frac{3 \pi}{8}$
6. $\tan \left(- \frac{\pi}{12}\right)$
7. $\sin \frac{11 \pi}{8}$
8. $\cos \frac{19 \pi}{12}$

The $\cos a=- \frac{5}{13}$ and $\frac{3 \pi}{2} \le a < 2 \pi$. Find:

1. $\sin 2a$
2. $\cos \frac{a}{2}$
3. $\tan \frac{a}{2}$
4. $\cos 2a$

The $\sin a=\frac{8}{11}$ and $\frac{\pi}{2} \le a < \pi$. Find:

1. $\tan 2a$
2. $\sin \frac{a}{2}$
3. $\cos \frac{a}{2}$
4. $\sin 2a$

## Simplifying Trig Expressions using Double and Half Angle Formulas

Objective

Here you'll use the half and double angle formulas to simplify more complicated expressions.

Guidance

We can also use the double-angle and half-angle formulas to simplify trigonometric expressions.

Example A

Simplify $\frac{\cos 2x}{\sin x \cos x}$.

Solution: Use $\cos 2a=\cos^2a-\sin^2a$ and then factor.

$\frac{\cos 2x}{\sin x \cos x}&=\frac{\cos^2x- \sin^2x}{\sin x+ \cos x} \\&=\frac{\left(\cos x- \sin x\right) \cancel{\left(\cos x + \sin x\right)}}{\cancel{\sin x+ \cos x}} \\&=\cos x- \sin x$

Example B

Find the formula for $\sin 3x$.

Solution: You will need to use the sum formula and the double-angle formula. $\sin 3x=\sin(2x+x)$

$\sin 3x&=\sin (2x+x) \\&=\sin 2x \cos x + \cos 2x \sin x \\&=2 \sin x \cos x \cos x+ \sin x(2 \cos^2x-1) \\&=2 \sin x \cos^2x+2 \sin x \cos^2 x- \sin x \\&=4 \sin x \cos^2x- \sin x \\&=\sin x(4 \cos^2x-1)$

We will explore other possibilities for the $\sin 3x$ because of the different formulas for $\cos 2a$ in the Problem Set.

Example C

Verify the identity $\cos x+2 \sin^2 \frac{x}{2}=1$.

Solution: Simplify the left-hand side use the half-angle formula.

$& \cos x+2 \sin^2 \frac{x}{2} \\& \cos x+2 \left(\sqrt{\frac{1- \cos x}{2}}\right)^2 \\& \cos x+2 \cdot \frac{1- \cos x}{2} \\& \cos x+1- \cos x \\& 1$

Guided Practice

1. Simplify $\frac{\sin 2x}{\sin x}$.

2. Verify $\cos x+2 \cos^2 \frac{x}{2}=1+ 2 \cos x$.

1. $\frac{\sin 2x}{\sin x}=\frac{2 \sin x \cos x}{\sin x}=2 \cos x$

2. $\cos x+2 \cos^2 \frac{x}{2}&=1+2 \cos x \\\cos x+2 \sqrt{\frac{1+ \cos x}{2}}^2&= \\\cos x+1 + \cos x&= \\1+2 \cos x&=$

Problem Set

Simplify the following expressions.

1. $\sqrt{2+2 \cos x} \left(\cos \frac{x}{2}\right)$
2. $\frac{\cos 2x}{\cos^2x}$
3. $\tan 2x(1+ \tan x)$
4. $\cos 2x- 3 \sin^2x$
5. $\frac{1+\cos 2x}{\cot x}$
6. $(1+ \cos x)^2 \tan \frac{x}{2}$

Verify the following identities.

1. $\cot \frac{x}{2}=\frac{\sin x}{1- \cos x}$
2. $\frac{\sin x}{1+ \cos x}=\frac{1- \cos x}{\sin x}$
3. $\frac{\sin 2x}{1+ \cos 2x}= \tan x$
4. $(\sin x+ \cos x)^2=1+ \sin 2x$
5. $\sin x \tan \frac{x}{2}+2 \cos x=2 \cos^2 \frac{x}{2}$
6. $\cot x+ \tan x=2 \csc 2x$
7. $\cos 3x=4 \cos^3x-3 \cos x$
8. $\cos 3x=\cos^3x-3 \sin^2x \cos x$
9. $\sin 2x-\tan x=\tan x \cos 2x$
10. $\cos^4x-\sin^4x=\cos 2x$

## Solving Trig Equations using Double and Half Angle Formulas

Objective

Here you'll solve trig equations using the half and double angle formulas.

Guidance

Lastly, we can use the half and double angle formulas to solve trigonometric equations.

Example A

Solve $\tan 2x+\tan x=0$ when $0\le x <2\pi$.

Solution: Change $\tan 2x$ and simplify.

$\tan 2x + \tan x &=0 \\\frac{2\tan x}{1-\tan ^2 x}+\tan x &=0 \\2\tan x +\tan x(1-\tan ^2x) &=0 \quad \rightarrow \text{Multiply everything by} \ 1-\tan^2x \text{ to eliminate denominator.}\\2\tan x +\tan x -\tan ^3 x &=0 \\3\tan x - \tan ^3 x &=0 \\\tan x(3-\tan^2 x) &=0$

Set each factor equal to zero and solve.

$&\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad 3-\tan ^2x =0 \\&\ \ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ -\tan ^2 x =-3 \\&\tan x=0 \qquad \qquad \qquad and \qquad \qquad \ \tan^2 x=3 \\& \qquad x=0 \ and \ \pi \ \qquad \qquad \qquad \quad \qquad \tan x =\pm \sqrt{3} \\& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ x =\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$

Example B

Solve $2\cos \frac{x}{2}+1=0$ when $0\le x<2\pi$.

Solution: In this case, you do not have to use the half-angle formula. Solve for $\frac{x}{2}$.

$2\cos \frac{x}{2}+1 &=0 \\2\cos \frac{x}{2} &=-1 \\\cos \frac{x}{2} &= -\frac{1}{2}$

Now, let’s find $\cos a = -\frac{1}{2}$ and then solve for $x$ by dividing by 2.

$\frac{x}{2} &=\frac{2\pi}{3},\frac{4\pi}{3} \\&=\frac{4\pi}{3}, \frac{8\pi}{3}$

Now, the second solution is not in our range, so the only solution is $x=\frac{4\pi}{3}$.

Example C

Solve $4\sin x \cos x = \sqrt{3}$ for $0\le x < 2\pi$.

Solution: Pull a 2 out of the left-hand side and use the $\sin 2x$ formula.

$4\sin x \cos x &=\sqrt{3} \\2\cdot 2\sin x \cos x &=\sqrt{3} \\2 \cdot \sin 2x &=\sqrt{3} \\\sin 2x &=\frac{\sqrt{3}}{2} \\2x &= \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3} \\x &= \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$

Guided Practice

Solve the following equations for $0\le x <2\pi$.

1. $\sin \frac{x}{2}=-1$

2. $\cos 2x-\cos x=0$

1. $\sin \frac{x}{2} &=-1 \\\frac{x}{2} &=\frac{3\pi}{2} \\x &=3\pi$

From this we can see that there are no solutions within our interval.

2. $\cos 2x - \cos x &=0 \\2\cos ^2x-\cos x -1 &=0 \\(2\cos x -1)(\cos x +1) &=0$

Set each factor equal to zero and solve.

$2\cos x-1 &=0 \\2\cos x &=1 \qquad \qquad \qquad \qquad\cos x +1=0\\\cos x &=\frac{1}{2} \qquad \qquad and \qquad \qquad \cos x =-1\\x &=\frac{\pi}{3}, \frac{5\pi}{3} \qquad \qquad \qquad \qquad \quad \ x=\pi \\$

Problem Set

Solve the following equations for $0\le x < 2\pi$.

1. $\cos x -\cos \frac{1}{2}x=0$
2. $\sin 2x \cos x=\sin x$
3. $\cos 3x = \cos ^3x=3\sin ^2x\cos x$
4. $\tan 2x - \tan x =0$
5. $\cos 2x -\cos x =0$
6. $2\cos ^2\frac{x}{2}=1$
7. $\tan \frac{x}{2}=4$
8. $\cos \frac{x}{2}=1+\cos x$
9. $\sin 2x +\sin x=0$
10. $\cos ^2x=\cos 2x =0$
11. $\frac{\cos 2x}{\cos ^2x}=1$
12. $\cos 2x-1=\sin^2x$
13. $\cos 2x =\cos x$
14. $\sin 2x-\cos 2x =1$
15. $\sin^2x-2=\cos 2x$
16. $\cot x+\tan x=2\csc 2x$

## Date Created:

Apr 23, 2013

Dec 16, 2014
You can only attach files to None which belong to you
If you would like to associate files with this None, please make a copy first.