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# 3.6: Half-Angle Identities

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Apply the half angle identities to expressions, equations and other identities.
• Use the half angle identities to find the exact value of trigonometric functions for certain angles.

Just as there are double angle identities, there are also half angle identities. For example: sin12a\begin{align*}\sin \frac{1}{2} a\end{align*} can be found in terms of the angle a′′\begin{align*}a''\end{align*}. Recall that 12a\begin{align*}\frac{1}{2} a\end{align*} and a2\begin{align*}\frac{a}{2}\end{align*} are the same thing and will be used interchangeably throughout this section.

## Deriving the Half Angle Formulas

In the previous lesson, one of the formulas that was derived for the cosine of a double angle is: cos2θ=12sin2θ\begin{align*}\cos 2 \theta = 1 - 2 \sin^2 \theta\end{align*}. Set θ=α2\begin{align*}\theta = \frac{\alpha}{2}\end{align*}, so the equation above becomes cos2α2=12sin2α2\begin{align*}\cos 2 \frac{\alpha}{2} = 1 - 2 \sin^2 \frac{\alpha}{2}\end{align*}.

Solving this for sinα2\begin{align*}\sin \frac{\alpha}{2}\end{align*}, we get:

cos2α2cosα2sin2α2sin2α2sinα2=12sin2α2=12sin2α2=1cosα=1cosα2=±1cosα2

sinα2=1cosα2\begin{align*}\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}}\end{align*} if α2\begin{align*}\frac{\alpha}{2}\end{align*} is located in either the first or second quadrant.

sinα2=1cosα2\begin{align*}\sin \frac{\alpha}{2} = - \sqrt{\frac{1 - \cos \alpha}{2}}\end{align*} if α2\begin{align*}\frac{\alpha}{2}\end{align*} is located in the third or fourth quadrant.

Example 1: Determine the exact value of sin15\begin{align*}\sin 15^\circ\end{align*}.

Solution: Using the half angle identity, α=30\begin{align*}\alpha = 30^\circ\end{align*}, and 15\begin{align*}15^\circ\end{align*} is located in the first quadrant. Therefore, sinα2=1cosα2\begin{align*}\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}}\end{align*}.

sin15=1cos302=1322=2322=234

Plugging this into a calculator, 2340.2588\begin{align*}\sqrt{\frac{2 - \sqrt{3}}{4}} \approx 0.2588\end{align*}. Using the sine function on your calculator will validate that this answer is correct.

Example 2: Use the half angle identity to find exact value of sin112.5\begin{align*}\sin 112.5^\circ\end{align*}

Solution: since sin2252=sin112.5\begin{align*}\sin \frac{225^\circ}{2} = \sin 112.5^\circ\end{align*}, use the half angle formula for sine, where α=225\begin{align*}\alpha = 225^\circ\end{align*}. In this example, the angle 112.5\begin{align*}112.5^\circ\end{align*} is a second quadrant angle, and the sin of a second quadrant angle is positive.

sin112.5=sin2252=±1cos2252=+1(22)2=22+222=2+24

One of the other formulas that was derived for the cosine of a double angle is:

cos2θ=2cos2θ1\begin{align*}\cos 2 \theta = 2 \cos^2 \theta - 1\end{align*}. Set θ=α2\begin{align*}\theta = \frac{\alpha}{2}\end{align*}, so the equation becomes cos2α2=1+2cos2α2\begin{align*}\cos 2 \frac{\alpha}{2} = - 1 + 2 \cos^2 \frac{\alpha}{2}\end{align*}. Solving this for cosα2\begin{align*}\cos \frac{\alpha}{2}\end{align*}, we get:

cos2α2cosα2cos2α2cos2α2cosα2=2cos2α21=2cos2α21=1+cosα=1+cosα2=±1+cosα2

cosα2=1+cosα2\begin{align*}\cos \frac{\alpha}{2} = \sqrt{\frac{1 + \cos \alpha}{2}}\end{align*} if α2\begin{align*}\frac{\alpha}{2}\end{align*} is located in either the first or fourth quadrant.

cosα2=1+cosα2\begin{align*}\cos \frac{\alpha}{2} = - \sqrt{\frac{1 + \cos \alpha}{2}}\end{align*} if α2\begin{align*}\frac{\alpha}{2}\end{align*} is located in either the second or fourth quadrant.

Example 3: Given that the cosθ=34\begin{align*}\cos \theta = \frac{3}{4}\end{align*}, and that θ\begin{align*}\theta\end{align*} is a fourth quadrant angle, find cos12 θ\begin{align*}\cos \frac{1}{2}\ \theta\end{align*}

Solution: Because θ\begin{align*}\theta\end{align*} is in the fourth quadrant, the half angle would be in the second quadrant, making the cosine of the half angle negative.

cosθ2=1+cosθ2=1+342=742=78=722=144

Example 4: Use the half angle formula for the cosine function to prove that the following expression is an identity: 2cos2x2cosx=1\begin{align*}2 \cos^2 \frac{x}{2} - \cos x = 1\end{align*}

Solution: Use the formula cosα2=1+cosα2\begin{align*}\cos \frac{\alpha}{2} = \sqrt{\frac{1 + \cos \alpha}{2}}\end{align*} and substitute it on the left-hand side of the expression.

2(1+cosθ2)2cosθ2(1+cosθ2)cosθ1+cosθcosθ1=1=1=1=1

The half angle identity for the tangent function begins with the reciprocal identity for tangent.

tanα=sinαcosαtanα2=sinα2cosα2

The half angle formulas for sine and cosine are then substituted into the identity.

tanα2=1cosα21+cosα2=1cosα1+cosα

At this point, you can multiply by either 1cosα1cosα\begin{align*}\frac{\sqrt{1 - \cos \alpha}}{\sqrt{1 - \cos \alpha}}\end{align*} or 1+cosα1+cosα\begin{align*}\frac{\sqrt{1 + \cos \alpha}}{\sqrt{1 + \cos \alpha}}\end{align*}. We will show both, because they produce different answers.

=1cosα1+cosα1cosα1cosα=1cosα1cos2α=1cosαsin2α=1cosαsinαor=1cosα1+cosα1+cosα1+cosα=1cos2α1+cosα=sin2α1+cosα=sinα1+cosα

So, the two half angle identities for tangent are tanα2=1cosαsinα\begin{align*}\tan \frac{\alpha}{2} = \frac{1 - \cos \alpha}{\sin \alpha}\end{align*} and tanα2=sinα1+cosα\begin{align*}\tan \frac{\alpha}{2} = \frac{\sin \alpha}{1 + \cos \alpha}\end{align*}.

Example 5: Use the half-angle identity for tangent to determine an exact value for tan7π12\begin{align*}\tan \frac{7 \pi}{12}\end{align*}.

Solution:

tanα2tan7π12tan7π12tan7π12=1cosαsinα=1cos7π6sin7π6=1+3212=23

Example 6: Prove the following identity: tanx=1cos2xsin2x\begin{align*}\tan x = \frac{1 - \cos 2x}{\sin 2x}\end{align*}

Solution: Substitute the double angle formulas for cos2x\begin{align*}\cos 2x\end{align*} and sin2x\begin{align*}\sin 2x\end{align*}.

tanx=1cos2xsin2x=1(12sin2x)2sinxcosx=11+2sin2x2sinxcosx=2sin2x2sinxcosx=sinxcosx=tanx

## Solving Trigonometric Equations Using Half Angle Formulas

Example 7: Solve the trigonometric equation sin2θ=2sin2θ2\begin{align*}\sin^2 \theta = 2 \sin^2 \frac{\theta}{2}\end{align*} over the interval [0,2π)\begin{align*}[0, 2\pi)\end{align*}.

Solution:

sin2θsin2θ1cos2θcosθcos2θcosθ(1cosθ)=2sin2θ2=2(1cosθ2)=1cosθ=0=0Half angle identityPythagorean identity

Then cosθ=0\begin{align*}\cos \theta = 0\end{align*} or 1cosθ=0\begin{align*}1 - \cos \theta = 0\end{align*}, which is cosθ=1\begin{align*}\cos \theta = 1\end{align*}.

θ=0,π2,3π2,or 2π\begin{align*}\theta = 0, \frac{\pi}{2}, \frac{3\pi}{2}, \text{or } 2\pi\end{align*}.

## Points to Consider

• Can you derive a third or fourth angle formula?
• How do 12sinx\begin{align*}\frac{1}{2} \sin x\end{align*} and sin12x\begin{align*}\sin \frac{1}{2}x\end{align*} differ? Is there a formula for 12sinx\begin{align*}\frac{1}{2} \sin x\end{align*}?

## Review Questions

1. Find the exact value of:
1. cos112.5\begin{align*}\cos 112.5^\circ\end{align*}
2. sin105\begin{align*}\sin 105^\circ\end{align*}
3. tan7π8\begin{align*}\tan \frac{7 \pi}{8}\end{align*}
4. tanπ8\begin{align*}\tan \frac{\pi}{8}\end{align*}
5. sin67.5\begin{align*}\sin 67.5^\circ\end{align*}
6. tan165\begin{align*}\tan 165^\circ\end{align*}
2. If sinθ=725\begin{align*}\sin \theta = \frac{7}{25}\end{align*} and θ\begin{align*}\theta\end{align*} is in Quad II, find sinθ2,cosθ2,tanθ2\begin{align*}\sin \frac{\theta}{2}, \cos \frac{\theta}{2}, \tan \frac{\theta}{2}\end{align*}
3. Prove the identity: tanb2=secbsecbcscb+cscb\begin{align*}\tan \frac{b}{2} = \frac{\sec b}{\sec b \csc b + \csc b}\end{align*}
4. Verify the identity: cotc2=sinc1cosc\begin{align*}\cot \frac{c}{2} = \frac{\sin c}{1 - \cos c}\end{align*}
5. Prove that sinxtanx2+2cosx=2cos2x2\begin{align*}\sin x \tan \frac{x}{2} + 2 \cos x = 2 \cos^2 \frac{x}{2}\end{align*}
6. If sinu=813\begin{align*}\sin u = - \frac{8}{13}\end{align*}, find cosu2\begin{align*}\cos \frac{u}{2}\end{align*}
7. Solve 2cos2x2=1\begin{align*}2 \cos^2 \frac{x}{2} = 1\end{align*} for 0x<2π\begin{align*}0 \le x < 2 \pi\end{align*}
8. Solve tana2=4\begin{align*}\tan \frac{a}{2} = 4\end{align*} for 0a<360\begin{align*}0^\circ \le a < 360^\circ\end{align*}
9. Solve the trigonometric equation cosx2=1+cosx\begin{align*}\cos \frac{x}{2} = 1 + \cos x\end{align*} such that 0x<2π\begin{align*}0 \le x < 2 \pi\end{align*}.
10. Prove sinx1+cosx=1cosxsinx\begin{align*}\frac{\sin x}{1 + \cos x} = \frac{1 - \cos x}{\sin x}\end{align*}.

Feb 23, 2012

## Last Modified:

Aug 06, 2015
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CK.MAT.ENG.SE.2.Trigonometry.3.6