$\tcBal{Pythagorean Identities}$ $$ \begin{align} \sin^2{\theta}+\cos^2{\theta} &=1 \\ \tan^2{\theta}+1 &= \sec^2{\theta} \label{eq:pythagorean id tan} \\ \cot^2{\theta}+1 &= \csc^2{\theta} \end{align} $$
$\tcBal{Double-Angle Formulas}$ $$ \begin{align} \sin{2\theta} &= 2\sin\theta \cos\theta \\ \cos{2\theta} &= \cos^2{\theta}-\sin^2{\theta} \\ \tan{2\theta} &= \frac{2\tan\theta}{1-\tan^2{\theta}} \end{align} $$
$\tcBal{Half-Angle Formulas}$ $$ \begin{align} \sin\frac{\theta}{2} &= \sqrt{\frac{1-\cos\theta}{2}} \\ \cos\frac{\theta}{2} &= \sqrt{\frac{1+\cos\theta}{2}} \\ \tan\frac{\theta}{2} &= \frac{1-\cos\theta}{\sin\theta} = \frac{\sin\theta}{1+\cos\theta} \end{align} $$
$\tcBal{Sum and Difference Identities}$ $$ \begin{align} \sin{(\alpha \pm \beta)} &= \sin\alpha \cos\beta \pm \cos\alpha \sin\beta \\ \cos{(\alpha \pm \beta)} &= \cos\alpha \cos\beta \mp \sin\alpha \sin\beta \\ \tan{(\alpha \pm \beta)} &= \frac{\tan\alpha \pm \tan\beta}{1 \mp \tan\alpha \tan\beta} \end{align} $$
$\tcBal{Product-Sum Identities}$ $$ \begin{align} \sin\alpha \sin\beta &= \frac{1}{2}\left[ \cos(\alpha-\beta)-\cos(\alpha+\beta)\right] \\ \cos\alpha \cos\beta &= \frac{1}{2}\left[ \cos(\alpha+\beta)+\cos(\alpha-\beta)\right] \\ \sin\alpha \cos\beta &= \frac{1}{2}\left[ \sin(\alpha+\beta)+\sin(\alpha-\beta)\right] \end{align} $$
$\tcBal{Sum-Product Identities}$ $$ \begin{align} \sin\alpha + \sin\beta &= 2\sin\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} \\ \sin\alpha - \sin\beta &= 2\sin\frac{\alpha-\beta}{2} \cos\frac{\alpha+\beta}{2} \\ \cos\alpha + \cos\beta &= 2\cos\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} \\ \cos\alpha - \cos\beta &= -2\sin\frac{\alpha+\beta}{2} \sin\frac{\alpha-\beta}{2} \end{align} $$
$\example{1}$ Simplify the function $y = \frac{\sin 2x}{\cot x}$.
$\solution$
\[\begin{align*} y &= \frac{\sin 2x}{\cot x} \\ &= \frac{2\sin x \cos x}{\br{\dfrac{\cos x}{\sin x}}} \\ &= \frac{2\sin x \cancel{\cos x} \sin x}{\cancel{\cos x}} \\ &= 2\sin^2 x \tagans \end{align*}\]
$\example{2}$ Express the function $\tan x + \cot x$ in terms of sine and cosine.
$\solution$
\[\begin{align*} \tan x + \cot x &= \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \\ &= \frac{(\sin x) (\sin x) + (\cos x) (\ cos x)}{\sin x \cos x} \\ &= \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} \\ &= \frac{1}{\sin x \cos x} \tagans \end{align*}\]
$\example{3}$ Express the angle of the function as a single term $\cos 7x \cos x$.
$\solution$
Recall that the identity: $\cos\alpha \cos\beta = \frac{1}{2}\brk{\cos(\alpha+\beta)+\cos(\alpha-\beta)}$
\[\begin{align*} \cos 7x \cos x &= \frac{1}{2} \brk{ \cos(7x+x)+\cos(7x-x) }\\ &= \frac{1}{2} \brk{ \cos(8x)+\cos(6x) } \tagans \end{align*}\]$\example{4}$ Prove the equation $\frac{\tan{x}\csc^2{x}}{1+\tan^2{x}}=\cot{x}$.
$\solution$
In proving identites, the first thing that we want to do is to choose one side of the equation that is more complicated. Then try to break it down in terms of sine and cosine to simplify the function.
\[\begin{align*} \cot{x} &= \frac{\tan{x}\csc^2{x}}{1+\tan^2{x}} \\ &= \frac{\dfrac{\cancel{\sin x}}{\cos x} \cdot \dfrac{1}{\sin^{\cancel{2}}x}}{\sec^2{x}} \\ &= \frac{\dfrac{1}{\cos x \sin x}}{\dfrac{1}{\cos^2 x}} \\ % &= \frac{\frac{1}{\cos{x}\cdot\sin{x}}}{\frac{\cos{x}+\sin{x}}{\cos{x}}} \\ &= \frac{1}{\cancel{\cos{x}}\cdot\sin{x}} \cdot \cos^{\cancel2}{x} \\ &= \frac{\cos x}{\sin x} \\ &= \cot x \tagans \end{align*}\]$\example{5}$ Prove the equation $1+2\sin\beta = (\sin\beta+\cos\beta)^2$.
$\solution$
In this example, the function is already expressed in terms of sine and cosine. So, what we can do now is we can first expand the binomial.
\[\begin{align*} 1+2\sin\beta &= (\sin\beta+\cos\beta)^2 \\ &= \sin^2\beta + 2\sin\beta\cos\beta + \cos^2\beta \\ &= (\sin^2\beta + \cos^2\beta) + 2\sin\beta\cos\beta \\ &= 1 + 2\sin\beta \tagans \end{align*}\]