# Double-Angle and Half-Angle Identities

### Double-Angle Identities

The Double-Angle Identities (these are really just special cases of Bhaskaracharya's formulas , when $u$ = $v$ )

$\begin{array}{l}\mathrm{sin}\left(2u\right)=2\mathrm{sin}\left(u\right)\mathrm{cos}\left(u\right)\\ \mathrm{cos}\left(2u\right)={\mathrm{cos}}^{2}\left(u\right)-{\mathrm{sin}}^{2}\left(u\right)\\ \mathrm{cos}\left(2u\right)=2{\mathrm{cos}}^{2}\left(u\right)-1\\ \mathrm{cos}\left(2u\right)=1-2{\mathrm{sin}}^{2}\left(u\right)\\ \mathrm{tan}\left(2u\right)=\frac{2\mathrm{tan}\left(u\right)}{1-{\mathrm{tan}}^{2}\left(u\right)}\end{array}$

Example 1:

Rewrite in a simpler form using a trigonometric identity:

$2\mathrm{sin}\left(5p\right)\mathrm{cos}\left(5p\right)$

Use the Double-Angle Formula for sine, where

$u=5p$

Apply the formula.

$\begin{array}{l}2\mathrm{sin}\left(5p\right)\mathrm{cos}\left(5p\right)=\mathrm{sin}\left(2\cdot 5p\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\mathrm{sin}\left(10p\right)\end{array}$

### Power-Reducing Identities

These can be derived from the identities above, by solving for ${\mathrm{sin}}^{2}\left(u\right)$ , ${\mathrm{cos}}^{2}\left(u\right)$ , or ${\mathrm{tan}}^{2}\left(u\right)$ .

$\begin{array}{l}{\mathrm{sin}}^{2}\left(u\right)=\frac{1-\mathrm{cos}\left(2u\right)}{2}\\ {\mathrm{cos}}^{2}\left(u\right)=\frac{1+\mathrm{cos}\left(2u\right)}{2}\\ {\mathrm{tan}}^{2}\left(u\right)=\frac{1-\mathrm{cos}\left(2u\right)}{1+\mathrm{cos}\left(2u\right)}\end{array}$

### Half-Angle Identities

These are the same as the identities above, but with the square root of both sides taken, and $\theta$ substituted for $2u$ .

$\begin{array}{cc}\mathrm{sin}\left(\frac{\theta }{2}\right)=±\sqrt{\frac{1-\mathrm{cos}\left(\theta \right)}{2}}& \mathrm{tan}\left(\frac{\theta }{2}\right)=±\sqrt{\frac{1-\mathrm{cos}\left(\theta \right)}{1+\mathrm{cos}\left(\theta \right)}}\\ \mathrm{cos}\left(\frac{\theta }{2}\right)=±\sqrt{\frac{1+\mathrm{cos}\left(\theta \right)}{2}}& \mathrm{tan}\left(\frac{\theta }{2}\right)=\frac{1-\mathrm{cos}\left(\theta \right)}{\mathrm{sin}\left(\theta \right)}\\ & \mathrm{tan}\left(\frac{\theta }{2}\right)=\frac{\mathrm{sin}\left(\theta \right)}{1+\mathrm{cos}\left(\theta \right)}\end{array}$

Example 2:

Determine the exact value of $\mathrm{cos}\left(15°\right)$ .

Since the angle $15°$ is in the first quadrant, where the cosine is positive, the value is

$=\frac{\sqrt{2+\sqrt{3}}}{2}$