Trigonometric identities are useful for solving equations and rewriting expressions in a more useful form, for example for integration.
\begin{align} \tan \theta &= \dfrac{\sin \theta }{\cos \theta } \\ \sin^2\theta +\cos^2\theta &= 1 \\ \\ 2\sin \theta \cos \phi &= \sin (\theta +\phi ) + \sin (\theta -\phi ) \\ 2\cos \theta \cos \phi &= \cos (\theta -\phi ) + \cos (\theta +\phi ) \\ 2\sin \theta \sin \phi &= \cos (\theta -\phi ) - \cos (\theta +\phi ) \\ \\ \sin \theta + \sin \phi &= 2\sin \left(\dfrac{\theta +\phi }{2}\right) \cos \left(\dfrac{\theta -\phi }{2}\right) \\ \sin \theta - \sin \phi &= 2\sin \left(\dfrac{\theta -\phi }{2}\right) \cos \left(\dfrac{\theta +\phi }{2}\right) \\ \cos \theta + \cos \phi &= 2\cos \left(\dfrac{\theta +\phi }{2}\right) \cos \left(\dfrac{\theta -\phi }{2}\right) \\ \cos \theta - \cos \phi &= -2\sin \left(\dfrac{\theta +\phi }{2}\right) \sin \left(\dfrac{\theta - \phi }{2}\right) \end{align}
The addition formula refers to the addition of two terms inside the trigonometric function, as opposed the the above identities which add two separate trig functions together.
\begin{align} \sin(\theta \pm \phi) &= \sin \theta \cos \phi \pm \cos \theta \sin \phi \\ \cos(\theta \pm \phi) &= \cos \theta \cos \phi \mp \sin \theta \sin \phi \\ \tan(\theta \pm \phi) &= \dfrac{\tan \theta \pm \tan \phi}{1 \mp \tan \theta \tan \phi} \end{align}
The double angle formula is a special case of the addition formula where $\phi = \theta$.
For example, the addition formula for sine is \[\sin(\theta + \phi) = \sin \theta \cos \phi +\cos \theta \sin \phi\] Substituting $\phi = \theta$ gives \[\sin(\theta + \theta) = \sin \theta \cos \theta +\cos \theta \sin \theta\] Since $\theta + \theta = 2\theta$, this gives \[\sin 2\theta = \sin \theta \cos \theta +\sin \theta \cos \theta\] \[\sin 2\theta = 2\sin\theta\cos\theta\]
Similarly for cosine and tangent. This gives the double angle formula as:
\begin{align} \sin 2\theta &= 2\sin \theta\cos \theta \\ \cos 2\theta &= \begin{cases} \cos^2\theta - \sin^2\theta \\1-2\sin^2\theta\\ 2\cos^2\theta-1 \end{cases} \\ \tan 2\theta &= \dfrac{2\tan \theta}{1-\tan^2\theta} \end{align}
These are useful formulae which reduce the sum of two trigonometric functions to one trigonometric function, which is easier to deal with.
\[a\cos x + b\sin x = R\sin \left(x+\arctan\frac{a}{b} \right) = R\cos \left(x-\arctan\frac{b}{a} \right)\]
where $R = \displaystyle{\sqrt{a^2 + b^2} }$.
Given that $\sin^2\theta = \cos\theta \cdot (2-\cos\theta)$, prove that $\cos\theta = \dfrac{1}{2}$.
Expand the expression
\[\sin^2\theta = 2\cos\theta - \cos^2\theta\]
Use the identity $\cos^2\theta + \sin^2\theta = 1$ so only $\cos$ is used.
\begin{align} \sin^2\theta &= 1-\cos^2\theta \\ \Rightarrow 1-\cos^2\theta &= 2\cos\theta - \cos^2\theta \end{align}
Simplify and rearrange
\begin{align} 1-\cos^2\theta &= 2\cos\theta - \cos^2\theta\\ 1 &= 2\cos\theta\\ \cos\theta &= \frac{1}{2} \end{align}
As required.
Verify the identity $\cos(a+b)\cos(a-b) = \cos^2a - \sin^2b$
Use the addition formula and rewrite the left hand side.
\begin{align} \cos(a+b)\cos(a-b) &= (\cos a\cos b - \sin a\sin b)(\cos a\cos b+\sin a\sin b)\\ &=\cos^2a\cos^2b + \cos a\cos b\sin a\sin b - \cos a\cos b\sin a\sin b - \sin^2a\sin^2b\\ &=\cos^2a\cos^2b - \sin^2a\sin^2b \end{align}
Use $\cos^2x+\sin^2x=1$ to rewrite everything in terms of $\cos$.
\begin{align} \cos^2a\cos^2b - \sin^2a\sin^2b &= \cos^2a\cos^2b - (1-\cos^2a)(1-\cos^2b)\\ &= \cos^2a\cos^2b - (1 - \cos^2a - \cos^2b + \cos^2a\cos^2b)\\ &= \cos^2a + \cos^2b -1 \end{align}
Use the formula again to rewrite in the form specified.
\begin{align} \cos^2a + \cos^2b -1 &=\cos^2a -(1-\cos^2b)\\ &=\cos^2a-\sin^2b \end{align}
Derive the formula for $\cos 2x$ from the addition formula.
The addition formula is $cos(\theta + \phi) = \cos\theta\cos\phi - \sin\theta\sin\phi$.
Set $\theta = x$ and $\phi = x$
\begin{align} \cos 2x &= \cos (x+x)\\ &=\cos x\cos x - \sin x\sin x\\ &=\cos^2x - \sin^2x \end{align}
Use $\cos^2x+\sin^2x=1$
\begin{align} \cos 2x &= \cos^2x - (1 - \cos^2x)\\ &= 2\cos^2 x-1 \end{align}
Or
\begin{align} \cos 2x &= (1-\sin^2x) - \sin^2x\\ &= 1-2\sin^2x \end{align}
Prof. Robin Johnson expands $\cos 6x$ in terms of $\cos x $ using the addition and double angle forumlae.
This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.