3.6: Half Angle Identities

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.

Power Reduction and Half Angle Identities

Another use of the cosine double angle identities is to use them in reverse to rewrite a squared sine or cosine in terms of the double angle. Starting with one form of the cosine double angle identity:

\[\cos (2\alpha )=2\cos ^{2} (\alpha )-1\nonumber\]Isolate the cosine squared term
\[\cos (2\alpha )+1=2\cos ^{2} (\alpha )\nonumber\] Add 1
\[\cos ^{2} (\alpha )=\dfrac{\cos (2\alpha )+1}{2}\nonumber\]Divide by 2
\[\cos ^{2} (\alpha )=\dfrac{\cos (2\alpha )+1}{2}\nonumber\] This is called a power reduction identity

Exercise \(\PageIndex{1}\)

Use another form of the cosine double angle identity to prove the identity \(\sin ^{2} (\alpha )=\dfrac{1-\cos (2\alpha )}{2}\).

Answer

\[\begin{array}{l} {\dfrac{1-\cos (2\alpha )}{2} } \\ {\dfrac{1-\left(\cos ^{2} (\alpha )-\sin ^{2} (\alpha )\right)}{2} } \\ {\dfrac{1-\cos ^{2} (\alpha )+\sin ^{2} (\alpha )}{2} } \\ {\dfrac{\sin ^{2} (\alpha )+\sin ^{2} (\alpha )}{2} } \\ {\dfrac{2\sin ^{2} (\alpha )}{2} =\sin ^{2} (\alpha )} \end{array}\nonumber\]

The cosine double angle identities can also be used in reverse for evaluating angles that are half of a common angle. Building from our formula \(\cos ^{2} (\alpha )=\dfrac{\cos (2\alpha )+1}{2}\), if we let \(\theta =2\alpha\), then \(\alpha =\dfrac{\theta }{2}\) this identity becomes \(\cos ^{2} \left(\dfrac{\theta }{2} \right)=\dfrac{\cos (\theta )+1}{2}\). Taking the square root, we obtain

\[\cos \left(\dfrac{\theta }{2} \right)=\pm \sqrt{\dfrac{\cos (\theta )+1}{2} }\nonumber\]where the sign is determined by the quadrant.

This is called a half-angle identity.

Exercise \(\PageIndex{2}\)

Use your results from the last Try it Now to prove the identity \(\sin \left(\dfrac{\theta }{2} \right)=\pm \sqrt{\dfrac{1-\cos (\theta )}{2} }\).

Answer

\[\begin{array}{l} {\sin ^{2} (\alpha )=\dfrac{1-\cos (2\alpha )}{2} } \\ {\sin (\alpha )=\pm \sqrt{\dfrac{1-\cos (2\alpha )}{2} } } \\ {\alpha =\dfrac{\theta }{2} } \\ {\sin \left(\dfrac{\theta }{2} \right)=\pm \sqrt{\dfrac{1-\cos \left(2\left(\dfrac{\theta }{2} \right)\right)}{2} } } \\ {\sin \left(\dfrac{\theta }{2} \right)=\pm \sqrt{\dfrac{1-\cos (\theta )}{2} } } \end{array}\nonumber\]

IDENTITIES

Half-Angle Identities

\[\cos \left(\dfrac{\theta }{2} \right)=\pm \sqrt{\dfrac{\cos (\theta )+1}{2} }\]

\[\sin \left(\dfrac{\theta }{2} \right)=\pm \sqrt{\dfrac{1-\cos (\theta )}{2} }\]

Power Reduction Identities

\[\cos ^{2} (\alpha )=\dfrac{\cos (2\alpha )+1}{2}\]

\[\sin ^{2} (\alpha )=\dfrac{1-\cos (2\alpha )}{2}\]

Since these identities are easy to derive from the double-angle identities, the power reduction and half-angle identities are not ones you should need to memorize separately.

Example \(\PageIndex{1}\)

Rewrite \(\cos ^{4} (x)\) without any powers.

Solution

\[\cos ^{4} (x)=\left(\cos ^{2} (x)\right)^{2}\nonumber\]Using the power reduction formula
\[=\left(\dfrac{\cos (2x)+1}{2} \right)^{2}\nonumber\]Square the numerator and denominator
\[=\dfrac{\left(\cos (2x)+1\right)^{2} }{4}\nonumber\]Expand the numerator
\[=\dfrac{\cos ^{2} (2x)+2\cos (2x)+1}{4}\nonumber\]Split apart the fraction
\[=\dfrac{\cos ^{2} (2x)}{4} +\dfrac{2\cos (2x)}{4} +\dfrac{1}{4}\nonumber\]Apply the formula above to \(\cos ^{2} (2x)\)
\[\cos ^{2} (2x)=\dfrac{\cos (2\cdot 2x)+1}{2}\nonumber\]
\[=\dfrac{\left(\dfrac{\cos (4x)+1}{2} \right)}{4} +\dfrac{2\cos (2x)}{4} +\dfrac{1}{4}\nonumber\]Simplify
\[=\dfrac{\cos (4x)}{8} +\dfrac{1}{8} +\dfrac{1}{2} \cos (2x)+\dfrac{1}{4}\nonumber\]Combine the constants
\[=\dfrac{\cos (4x)}{8} +\dfrac{1}{2} \cos (2x)+\dfrac{3}{8}\nonumber\]

Example \(\PageIndex{2}\)

Find an exact value for \(\cos \left(15{}^\circ \right)\).

Solution

Since 15 degrees is half of 30 degrees, we can use our result from above:

\[\cos (15{}^\circ )=\cos \left(\dfrac{30{}^\circ }{2} \right)=\pm \sqrt{\dfrac{\cos (30{}^\circ )+1}{2} }\nonumber\]

We can evaluate the cosine. Since 15 degrees is in the first quadrant, we need the positive result.

\[\sqrt{\dfrac{\cos (30{}^\circ )+1}{2} } =\sqrt{\dfrac{\dfrac{\sqrt{3} }{2} +1}{2} }\nonumber\]
\[=\sqrt{\dfrac{\sqrt{3} }{4} +\dfrac{1}{2} }\nonumber\]

Exercise \(\PageIndex{3}\)

If \(\csc \left(x\right)=7\) and \(90{}^\circ <x<180{}^\circ\), then find exact values for (without solving for \(x\)):

a. \(\sin \left(\dfrac{x}{2} \right)\)
b. \(\cos \left(\dfrac{x}{2} \right)\)
c. \(\tan \left(\dfrac{x}{2} \right)\)

Answer

a. \(\sqrt{\dfrac{1}{2}+\dfrac{2 + \sqrt{7}}{7}}\)
b. \(\sqrt{\dfrac{1}{2}-\dfrac{2 + \sqrt{7}}{7}}\)
c. \(\dfrac{1}{7 - 4\sqrt{3}}\)

Important Topics of This Section

• Power reduction identity
• Half angle identity
• Using identities
• Simplify equations
• Prove identities
• Solve equations