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Trigonometric Identities and Formulae



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

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 Forumla

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}

$R \sin(x+\alpha)$ and $R \cos(x-\alpha)$

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 $R = \displaystyle{\sqrt{a^2 + b^2} }$.

Worked Examples

Example 1

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.

Example 2

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}

Example 3

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}


\begin{align} \cos 2x &= (1-\sin^2x) - \sin^2x\\ &= 1-2\sin^2x \end{align}

Video Example

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.

External Resources

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