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

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 can be found in terms of the angle a′′. Recall that 12a and a2 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. Set θ=α2, so the equation above becomes cos2α2=12sin2α2.

Solving this for sinα2, we get:


sinα2=1cosα2 if α2 is located in either the first or second quadrant.

sinα2=1cosα2 if α2 is located in the third or fourth quadrant.

Example 1: Determine the exact value of sin15.

Solution: Using the half angle identity, α=30, and 15 is located in the first quadrant. Therefore, sinα2=1cosα2.


Plugging this into a calculator, 2340.2588. 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

Solution: since sin2252=sin112.5, use the half angle formula for sine, where α=225. In this example, the angle 112.5 is a second quadrant angle, and the sin of a second quadrant angle is positive.


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

cos2θ=2cos2θ1. Set θ=α2, so the equation becomes cos2α2=1+2cos2α2. Solving this for cosα2, we get:


cosα2=1+cosα2 if α2 is located in either the first or fourth quadrant.

cosα2=1+cosα2 if α2 is located in either the second or fourth quadrant.

Example 3: Given that the cosθ=34, and that θ is a fourth quadrant angle, find cos12 θ

Solution: Because θ is in the fourth quadrant, the half angle would be in the second quadrant, making the cosine of the half angle negative.


Example 4: Use the half angle formula for the cosine function to prove that the following expression is an identity: 2cos2x2cosx=1

Solution: Use the formula cosα2=1+cosα2 and substitute it on the left-hand side of the expression.


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


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


At this point, you can multiply by either 1cosα1cosα or 1+cosα1+cosα. We will show both, because they produce different answers.


So, the two half angle identities for tangent are tanα2=1cosαsinα and tanα2=sinα1+cosα.

Example 5: Use the half-angle identity for tangent to determine an exact value for tan7π12.



Example 6: Prove the following identity: tanx=1cos2xsin2x

Solution: Substitute the double angle formulas for cos2x and sin2x.


Solving Trigonometric Equations Using Half Angle Formulas

Example 7: Solve the trigonometric equation sin2θ=2sin2θ2 over the interval [0,2π).


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

Then cosθ=0 or 1cosθ=0, which is cosθ=1.

θ=0,π2,3π2,or 2π.

Points to Consider

  • Can you derive a third or fourth angle formula?
  • How do 12sinx and sin12x differ? Is there a formula for 12sinx?

Review Questions

  1. Find the exact value of:
    1. cos112.5
    2. sin105
    3. tan7π8
    4. tanπ8
    5. sin67.5
    6. tan165
  2. If sinθ=725 and θ is in Quad II, find sinθ2,cosθ2,tanθ2
  3. Prove the identity: tanb2=secbsecbcscb+cscb
  4. Verify the identity: cotc2=sinc1cosc
  5. Prove that sinxtanx2+2cosx=2cos2x2
  6. If sinu=813, find cosu2
  7. Solve 2cos2x2=1 for 0x<2π
  8. Solve tana2=4 for 0a<360
  9. Solve the trigonometric equation cosx2=1+cosx such that 0x<2π.
  10. Prove sinx1+cosx=1cosxsinx.

Image Attributions

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.


Please wait...
Please wait...
Image Detail
Sizes: Medium | Original

Original text