# 3.4.6: Trigonometric Equations Using Half Angle Formulas

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Simplifying all six trigonometric functions with half a given angle.

As you've seen many times, the ability to find the values of trig functions for a variety of angles is a critical component to a course in Trigonometry. If you were given an angle as the argument of a trig function that was half of an angle you were familiar with, could you solve the trig function?

For example, if you were asked to find

\(\sin 22.5^{\circ}\)

would you be able to do it? Keep reading, and in this section you'll learn how to do this.

### using Half Angle Formulas on Trigonometric Equations

It is easy to remember the values of trigonometric functions for certain common values of \(\theta \). However, sometimes there will be fractional values of known trig functions, such as wanting to know the sine of half of the angle that you are familiar with. In situations like that, a **half angle identity** can prove valuable to help compute the value of the trig function.

In addition, half angle identities can be used to simplify problems to solve for certain angles that satisfy an expression. To do this, first remember the half angle identities for sine and cosine:

\(\sin \alpha 2=\sqrt{\dfrac{1−\cos \alpha }{2}}\) if \(\dfrac{\alpha}{2}\) is located in either the first or second quadrant.

\(\sin \alpha 2=−\sqrt{\dfrac{1−\cos \alpha}{ 2}}\) if \(\dfrac{\alpha}{2}\) is located in the third or fourth quadrant.

\(\cos \alpha 2=\sqrt{\dfrac{1+\cos \alpha}{ 2}}\) if \(\dfrac{\alpha}{2}\) is located in either the first or fourth quadrant.

\(\cos \alpha 2=−\sqrt{\dfrac{1+\cos \alpha }{2}}\) if \(\dfrac{\alpha}{2}\) is located in either the second or fourth quadrant.

When attempting to solve equations using a half angle identity, look for a place to substitute using one of the above identities. This can help simplify the equation to be solved.

Let's look at some problems that use the half angle formula.

1. Solve the trigonometric equation \(\sin ^2\theta =2\sin ^2 \dfrac{\theta }{2}\) over the interval \([0,2\pi )\).

\(\begin{aligned} \sin ^2\theta&=2\sin ^2 \dfrac{\theta }{2} \\ \sin ^2\theta&=2\left(\dfrac{1−\cos \theta }{2}\right) && \text{ Half angle identity}\\ 1−\cos ^2\theta &=1−\cos \theta &&\text{Pythagorean identity}\\ \cos \theta −\cos ^2\theta&=0 \\ \cos \theta (1−\cos \theta )&=0 \end{aligned}\)

Then \(\cos \theta =0\) or \(1−\cos \theta =0\), which is \(\cos \theta\).

\(\theta =0,\; \dfrac{\pi }{2},\; \dfrac{3\pi }{2},\; \text{ or } 2\pi \).

2. Solve \(2\cos ^2\dfrac{x}{2}=1\) for \(0\leq x<2\pi\)

To solve \(2\cos ^2\dfrac{x}{2}=1\), first we need to isolate cosine, then use the half angle formula.

\(\begin{aligned} 2\cos ^2\dfrac{x}{2}&=1 \\ \cos ^2 \dfrac{x}{2}&=\dfrac{1}{2} \\ \dfrac{1+\cos x}{2}&=\dfrac{1}{2} \\ 1+\cos x&=1 \\ \cos x&=0 \end{aligned}\)

\(\cos x=0\) when \(x=\dfrac{\pi }{2},\; \dfrac{3\pi }{2}\)

3. Solve \(\tan \dfrac{a}{2}=4\) for \(0^{\circ} \leq a<360^{\circ}\)

To solve \(\tan \dfrac{a}{2}=4\), first isolate tangent, then use the half angle formula.

\(\begin{aligned}

\tan \dfrac{a}{2} &=4 \\

\sqrt{\dfrac{1-\cos a}{1+\cos a}} &=4 \\

\dfrac{1-\cos a}{1+\cos a} &=16 \\

16+16 \cos a &=1-\cos a \\

17 \cos a &=-15 \\

\cos a &=-\dfrac{15}{17}

\end{aligned}\)

using your graphing calculator, \(\cos a=−\dfrac{15}{17}\) when \(a=152^{\circ} ,\; 208^{\circ}\)

Earlier, you were asked to solve sin \(22.5^{\circ}\).

**Solution**

Knowing the half angle formulas, you can compute \(\sin 22.5^{\circ} \) easily:

\(\begin{aligned}

\sin 22.5^{\circ} &=\sin \left(\dfrac{45^{\circ}}{2}\right) \\

&= \sqrt{\dfrac{1-\cos 45^{\circ}}{2}} \\

&=\sqrt{\dfrac{1-\dfrac{\sqrt{2}}{2}}{2}} \\

&=\sqrt{\dfrac{\dfrac{2-\sqrt{2}}{2}}{2}} \\

&=\sqrt{\dfrac{2-\sqrt{2}}{4}} \\

&=\dfrac{\sqrt{2-\sqrt{2}}}{2}

\end{aligned}\)

Find the exact value of \(\cos 112.5^{\circ}\)

**Solution**

\(\begin{aligned}

\sin 105^{\circ} &= \sin \dfrac{210^{\circ}}{2} \\

&=\sqrt{\dfrac{1-\cos 210^{\circ}}{2}} \\

&=\sqrt{\dfrac{1-\dfrac{\sqrt{3}}{2}}{2}} \\

&=\sqrt{\dfrac{\dfrac{2-\sqrt{3}}{2}}{2}} \\

&=\sqrt{\dfrac{2-\sqrt{3}}{4}} \\

&=\dfrac{\sqrt{2-\sqrt{3}}}{2}

\end{aligned}\)

Find the exact value of \(\sin 105^{\circ}\)

**Solution**

\(\begin{aligned}

\sin 105^{\circ} &= \sin \dfrac{210^{\circ}}{2} \\

&=\sqrt{\dfrac{1-\cos 210^{\circ}}{2}} \\

&=\sqrt{\dfrac{1-\dfrac{\sqrt{3}}{2}}{2}} \\

&=\sqrt{\dfrac{\dfrac{2-\sqrt{3}}{2}}{2}} \\

&=\sqrt{\dfrac{2-\sqrt{3}}{4}} \\

&=\dfrac{\sqrt{2-\sqrt{3}}}{2}

\end{aligned}\)

Find the exact value of \(\tan \dfrac{7\pi }{8}\).

**Solution**

\(\begin{aligned} \tan \dfrac{7\pi }{8} &=\tan \dfrac{1}{2}\cdot \dfrac{7\pi }{4} \\ &=\dfrac{1−\cos \dfrac{7\pi }{4} }{\sin \dfrac{7\pi }{4}} \\ &=\dfrac{1−\dfrac{\sqrt{2}}{2}}{−\dfrac{\sqrt{2}}{2}} \\ &=\dfrac{\dfrac{2−\sqrt{2}}{2}}{−\dfrac{\sqrt{2}}{2}} \\&=−\dfrac{2−\sqrt{2}}{\sqrt{2}}=\dfrac{−2\sqrt{2}+2}{2} \\ &=−\sqrt{2}+1 \end{aligned}\)

### Review

Use half angle identities to find the exact value of each expression.

- \(\tan 15^{\circ}\)
- \(\tan 22.5^{\circ}\)
- \(\cot 75^{\circ}\)
- \(\tan 67.5^{\circ}\)
- \(\tan 157.5^{\circ}\)
- \(\tan 112.5^{\circ}\)
- \(\cos 105^{\circ}\)
- \(\sin 112.5^{\circ}\)
- \(\sec 15^{\circ}\)
- \(\csc 22.5^{\circ}\)
- \(\csc 75^{\circ}\)
- \(\sec 67.5^{\circ}\)
- \(\cot 157.5^{\circ}\)

Use half angle identities to help solve each of the following equations on the interval \([0,2\pi )\).

- \(3\cos ^2 \left(\dfrac{x}{2}\right)=3\)
- \(4\sin ^2x=8\sin ^2\left(\dfrac{x}{2}\right)\)

### Review (Answers)

To see the Review answers, open this PDF file and look for section 3.12.

## Vocabulary

Term | Definition |
---|---|

Half Angle Identity |
A half angle identity relates a trigonometric function of one half of an argument to a set of trigonometric functions, each containing the original argument. |