# Trigonometric Identities

Trigonometric Identities are identities in mathematics that involve trigonometric functions such as $\sin(x)$, $\cos(x)$ and $\tan(x)$. Identities, as opposed to equations, are statements where the left hand side is equivalent to the right hand side. We use a $\equiv$ symbol, which means ‘equivalent’, instead of the usual ‘equals’ sign. We solve equations to find values of $x$, for instance. We don’t solve an identity, however, as no additional information has been given. But they can be used to solve equations – see Trigonometric Equations. It follows that Trigonometric Identities are simply equivalent trigonometric expressions.

## Basic Trigonometric Identities

Exam questions may require you to have memorised some or all of the following basic trigonometric identities and common values:

### Fundamental Identities

$\frac{\sin(\theta)}{\cos(\theta)}\equiv\tan(\theta)$

$\cos^2(\theta)+\sin^2(\theta)\equiv 1$.

$\begin{array}{l}\sin(-\theta)\equiv-\sin(\theta)\\

\cos(-\theta)\equiv\cos(\theta)\\

\sin(90^\circ-\theta)\equiv\cos(\theta)\\

\cos(90^\circ-\theta)\equiv\sin(\theta)\\

\sin(180^\circ-\theta)\equiv\sin(\theta)\\

\cos(180^\circ-\theta)\equiv-\cos(\theta)\\\end{array}$

### Trigonometric Ratios

$\begin{array}{c}\sin(30^\circ)=\frac{1}{2}\\\cos(30^\circ)=\frac{\sqrt{3}}{2}\\\tan(30^\circ)=\frac{\sqrt{3}}{3}\\\\\sin(45^\circ)=\frac{\sqrt{2}}{2}\\\cos(45^\circ)=\frac{\sqrt{2}}{2}\\\tan(45^\circ)=1\\\\\sin(60^\circ)=\frac{\sqrt{3}}{2}\\\cos(60^\circ)=\frac{1}{2}\\\tan(60^\circ)=\sqrt{3}\end{array}$

## Where do these Trigonometric Identities come from?

### Fundamental Identities

Firstly, the identity $\tan(\theta)\equiv\frac{\sin(\theta)}{\cos

(\theta)}$ can be seen using SOHCAHTOA. Recall that

$\sin(\theta)=\frac{OPP}{HYP}$ and $\cos(\theta)=\frac{ADH}{HYP}$.

It follows that $\tan(\theta)=\frac{OPP}{ADJ}=\frac{OPP}{HYP}\div \frac{ADJ}{HYP}=\frac{\sin(\theta)}{\cos(\theta)}$.

Secondly, recall that we can think of $\cos(\theta)$ and $\sin(\theta)$ as the $x$ and $y$ coordinates of a particle traversing a unit circle. It follows that the identity $\cos^2(\theta)+\sin^2(\theta)\equiv 1$ results from an application of Pythagoras. See Trigonometric Graphs.

Now consider $\sin(-\theta)\equiv-\sin(\theta)$, for example. We can see this type of identity by performing transformations to standard curves. See Transformations.

Firstly, consider the graph of $\sin(-\theta)$. We obtain it by replacing $\theta$ with $-\theta$. That is, a reflection in the $y$-axis. This is precisely the sin curve multiplied by -1, i.e. reflected across the $x$-axis. This is the above identity, as expected. Functions with this property are known as odd functions. We derive the other identities in a similar way. Beware – it is common for student to perform composite transformations incorrectly.

Click here to see other trigonometric graph transformations.

### Trigonometric Ratios

The Trigonometric Ratios seen above can be found without using a calculator. We can find them from two ‘special triangles’:

Firstly, the triangle on the left is an equilateral triangle. As you can see, all of the sides have length 2. This means that all of the angles are $60^\circ$. If this triangle is split down the middle, then each of the angles at the top is $30^\circ$. Note that the length of this vertical line is root 3 and it follows from Pythagoras. We find the sin, cos and tan values for $30^\circ$ and $60^\circ$ from this triangle using SOHCAHTOA. For example, $\cos(30^\circ)=\frac{ADJ}{HYP}=\frac{\sqrt{3}}{2}$.

Secondly, the triangle on the right is an isosceles triangle. As you can see, two of the sides have length 1. It follows from Pythagoras that the length of the hypoteneuse is root 2. Since one of the angles is a right angle, the remaining angles are both $45^\circ$ degrees. We find the sin, cos and tan values for $45^\circ$ from this triangle also using SOHCAHTOA. For example, $\sin(45^\circ)=\frac{OPP}{HYP}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$.

In more advanced trigonometry, other trigonometric values may also be found. For example, it is possible to find $\tan(15^\circ)$ using the compound angle identities below. We achieve this by choosing $\alpha=45^\circ$ and $\beta=30^\circ$, for instance. More on this soon.

## More Trigonometric Identities

In addition to the above, you may be require to master the following identities. These identities are more complicated than the ones seen above and so may feature later on in your course.

### Double Angle Identities

$\begin{array}{l}\sin(2\theta)\equiv2\sin(\theta)\cos(\theta)\\\cos(2\theta)\equiv\cos(\theta)^2-\sin^2(\theta)\\\cos(2\theta)\equiv2\cos^2(\theta)-1\\\cos(2\theta)\equiv1-2\sin^2(\theta)\end{array}$

### Compound Angle Indentities

$\begin{array}{l}\sin(\alpha\pm\beta)\equiv\sin(\alpha)\cos(\beta)\pm\cos(\alpha)\sin(\beta)\\\cos(\alpha\pm\beta)\equiv\sin(\alpha)\sin(\beta)\mp\cos(\alpha)\cos(\beta)\\\tan(\alpha\pm\beta)\equiv\frac{\tan(\alpha)\pm\tan(\beta)}{1\mp\tan(\alpha)\tan(\beta)}\end{array}$

For more trigonometric identities visit Wikipedia Trig. Identities.

## Examples on Trigonometric Identities

Simplify

$\frac{\sqrt{1-\cos^2(x)}}{\cos(x)}$

Firstly, using the identity

$\sin^2(x)+\cos^2(x)\equiv1$

we have

$1-\cos^2(x)\equiv\sin^2(x).$

It follows that

$\sqrt{1-\cos^2(x)}\equiv\sin(x)$

Hence,

$\frac{\sqrt{1-\cos^2(x)}}{\cos(x)}\equiv\frac{\sin(x)}{\cos(x)}=\tan(x)$

Show that $\frac{\sin^4(\theta)-\cos^4(\theta)}{\cos^2(\theta)}\equiv\tan^2(\theta)-1$.

The first step in the question is the trickiest to spot: $\sin^4(\theta)-\cos^4(\theta)$ is in fact a difference of squares. If you struggle to see this, consider $x^2-y^2\equiv(x+y)(x-y)$ and then $x^4-y^4\equiv(x^2+y^2)(x^2-y^2)$. It follows that $\sin^4(\theta)-\cos^4(\theta)$ is equivalent to $(\sin^2(\theta)+\cos^2(\theta))(\sin^2(\theta)-\cos^2(\theta))$

Hence, the left hand side of the above identity becomes:

$\frac{\sin^4(\theta)-\cos^4(\theta)}{\cos^2(\theta)}\equiv\frac{(\sin^2(\theta)+\cos^2(\theta))(\sin^2(\theta)-\cos^2(\theta))}{\cos^2(\theta)}$

Since $\sin^2(\theta)+\cos^2(\theta)\equiv1$, we are left with

$\frac{\sin^2(\theta)-\cos^2(\theta)}{\cos^2(\theta)}\equiv\tan^2(\theta)-1.$

This can be separated into

$\frac{\sin^2(\theta)}{\cos^2(\theta)}-\frac{\cos^2(\theta)}{\cos^2(\theta)}\equiv\tan^2(\theta)-1$, thus completing the proof of the identity. Note that $\tan(\theta)\equiv\frac{\sin(\theta)}{\cos(\theta)}$ and so $\tan^2(\theta)\equiv\frac{\sin(\theta)}{\cos(\theta)}\times\frac{\sin(\theta)}{\cos(\theta)}\equiv\frac{\sin^2(\theta)}{\cos^2(\theta)}$.

Given that $\theta$ is obtuse and $\cos(\theta)=-\frac{2}{5}$, find the value of $\tan(\theta)$.

Since $\tan(\theta)\equiv\frac{\sin(\theta)}{\cos(\theta)}$, we need to know the value of $\sin(\theta)$. Given that we know $\cos(\theta)=-\frac{2}{5}$, we can use the identity $\cos^2(\theta)+\sin^2(\theta)\equiv 1$ to find it:

$\left(-\frac{2}{5}\right)^2+\sin^2(\theta)= 1$

$\Longrightarrow \sin^2(\theta)= 1-\frac{4}{25}$

$\Longrightarrow \sin(\theta)= \pm\sqrt{\frac{21}{25}}=\pm\frac{\sqrt{21}}{5}$

We have been told that $\theta$ is obtuse – this is when $\sin(\theta)$ is positive – see Trigonometric Graphs. It follows that $\sin(\theta)=\frac{\sqrt{21}}{5}$ and so

$\tan(\theta)=\frac{\frac{\sqrt{21}}{5}}{-\frac{2}{5}}=-\frac{\sqrt{21}}{2}$

Given that $x=2\cos(\alpha)$ and $y=3\sin(\alpha)$, find the value of $9x^2+4y^2$.

$\begin{array}{l}9x^2+4y^2\\=9\left(2\cos(\alpha)\right)^2+4\left(3\sin(\alpha)\right)^2\\=9\times 4\cos^2(\alpha)+4\times9\sin^2(\alpha)\\=36\cos^2(\alpha)+36\sin^2(\alpha)\\=36\left(\cos^2(\alpha)+\sin^2(\alpha)\right)\\=36\end{array}$

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