Trigonometric Power Reduction Identities

The trigonometric power reduction identities allow us to rewrite expressions involving trigonometric terms with trigonometric terms of smaller powers. This becomes important in several applications such as integrating powers of trigonometric expressions in calculus.

Power Reduction Identities

For any angle $\theta$,

$$\cos^2 \theta = \frac{1 + \cos(2\theta)}{2},\quad \sin^2 \theta = \frac{1 - \cos(2\theta)}{2}.$$

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Proof:

Subtracting the two identities gives

$$\cos(2\theta) = \cos^2 \theta - \sin^2 \theta.$$

Then we have

$$\begin{aligned} \dfrac{1 + \cos(2\theta)}{2}& = \dfrac{1+\cos^2 \theta - \sin^2 \theta}{2}\\ & = \dfrac{\big(\sin^2 \theta + \cos^2 \theta\big) + \cos^2 \theta - \sin^2 \theta}{2}\\ & = \dfrac{2 \cos^2 \theta}{2}\\\\ & = \cos^2 \theta.\ _\square \end{aligned}$$

Using the power reduction formulas, we can derive the following half-angle formulas:

Half-angle Formulas

For any angle $\theta$,

$$\begin{aligned} \cos \left( \frac{\theta}{2} \right) &= \pm \sqrt{ \frac{1 + \cos\theta}{2} }\\ \sin \left( \frac{\theta}{2} \right) &= \pm \sqrt{ \frac{1 - \cos\theta}{2} }\\ \tan \left( \frac{\theta}{2} \right) &= \pm \sqrt{ \frac{1 - \cos\theta}{1 + \cos\theta} }. \end{aligned}$$

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Proof:

From the power reduction formulas, we have $\cos^{2}\theta=\frac{1+\cos(2\theta)}{2}$. Replacing $\theta$ with $\frac{\theta}{2}$, we obtain

$$\begin{aligned} \cos^{2}\left(\frac{\theta}{2}\right) &=\frac{1+\cos\theta}{2}\\ \cos\left(\frac{\theta}{2}\right) &=\pm\sqrt{\frac{1+\cos\theta}{2}}.\ _\square \end{aligned}$$

Use the power reduction identities to express $\cos^2 \theta \sin^2 \theta$ using only cosines and sines to the first power.

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We have

$$\begin{aligned} \left(\cos\theta\sin\theta\right)^{2} &=\frac{1}{4}\left(2\cos\theta\sin\theta\right)^{2}\\ &=\frac{1}{4}\left(\sin^{2}(2\theta)\right)\\ &=\frac{1}{4}\left(\frac{1-\cos(4\theta)}{2}\right)\\ &=\frac{1}{8}\big(1-\cos(4\theta)\big).\ _\square \end{aligned}$$

Use the power reduction identities to express $\cos^4 \theta$ using only cosines and sines to the first power.

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We have

$$\begin{aligned} \cos^{4}\theta &=\left(\cos^{2}\theta\right)^{2} \\ &=\left(\frac{1+\cos(2\theta)}{2}\right)\\ &=\frac{1+2\cos(2\theta)+\cos^{2}(2\theta)}{4}\\ &=\frac{1+2\cos(2\theta)+\frac{1+\cos(4\theta)}{2}}{4}\\ &=\frac{2+4\cos(2\theta)+1+\cos(4\theta)}{8}\\ &=\frac{1}{8}\big(\cos(4\theta)+4\cos(2\theta)+3\big).\ _\square \end{aligned}$$

Use the half-angle formulas to evaluate $\cos \left( \frac{\pi}{12} \right)$ and $\sin \left( \frac{\pi}{12} \right).$

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In the formulas $\cos \left( \frac{\theta}{2} \right) \pm \sqrt{ \frac{1 + \cos\theta}{2} }$ and $\sin \left( \frac{\theta}{2} \right) \pm \sqrt{ \frac{1 - \cos\theta}{2} },$ we replace $\theta$ with $\frac{\pi}{6}:$

$$\begin{aligned} \cos\left(\frac{\pi}{12}\right) &=\pm\sqrt{\frac{1+\cos\left(\frac{\pi}{6}\right)}{2}} \\ & =\pm\frac{\sqrt{2+\sqrt{3}}}{2} \end{aligned}$$

$$\begin{aligned} \sin\left(\frac{\pi}{12}\right) &=\pm\sqrt{\frac{1-\cos\left(\frac{\pi}{6}\right)}{2}} \\ &=\pm\frac{\sqrt{2-\sqrt{3}}}{2}. \end{aligned}$$

As $\frac{\pi}{12}$ lies in the first quadrant, we assign positive values to $\sin\left(\frac{\pi}{12}\right)$ and $\cos\left(\frac{\pi}{12}\right)$.

Hence $\cos\left(\frac{\pi}{12}\right)=\frac{\sqrt{2+\sqrt{3}}}{2}$ and $\sin\left(\frac{\pi}{12}\right)=\frac{\sqrt{2-\sqrt{3}}}{2}$. $_\square$

Verify the identities

$$\tan \left( \frac{\theta}{2} \right) = \frac{1-\cos \theta }{\sin\theta},\quad \tan \left( \frac{\theta}{2} \right) = \frac{\sin\theta}{1+\cos \theta }.$$

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We have

$$\begin{aligned} \tan \left( \frac{\theta}{2} \right) &= \pm \sqrt{ \frac{1 - \cos\theta}{1 + \cos\theta} }. \end{aligned}$$

Multiply both numerator and denominator by $\sqrt{1-\cos\theta}$

$$\begin{aligned} \tan \left( \frac{\theta}{2} \right) &= \pm \sqrt{ \frac{(1 - \cos \theta)^2}{1 - \cos^2\theta} }\\ &=\pm \sqrt{ \frac{(1 - \cos \theta)^2}{\sin^2\theta} }\\ &=\frac{1-\cos\theta}{\sin \theta}. \end{aligned}$$

Similarly, by multiplying numerator and denominator by $\sqrt{1+\cos \theta}$,

$$\begin{aligned} \tan \left( \frac{\theta}{2} \right) &= \pm \sqrt{ \frac{1 - \cos^2\theta}{(1 + \cos\theta)^2} }\\ &=\pm \sqrt{ \frac{\sin^2\theta}{(1 + \cos\theta)^2} }\\ &=\frac{\sin \theta}{1+\cos \theta}.\ _\square \end{aligned}$$