# Conversion Relations of Trigonometric Ratios

Conversion Relations of Trigonometric Ratios
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## Introduction:

Coterminal angles are angles in standard position (angles with the initial side on the positive  $$x$$-axis) that have a common terminal side.  For example, $$30^\circ$$, $$- 330^\circ$$ and $$390^\circ$$ are all coterminal.

Note: To find a positive and a negative angle coterminal with a given angle, you can add and subtract $$360^\circ$$ if the angle is measured in degrees or $$2\pi$$ if the angle is measured in radians.

Let's discuss some more conversion relations of trigonometric ratios.

## Conversion Relations:

What will be the value of $$\sin \left( { - \theta } \right)$$ in terms of $$\sin \theta$$? Consider the following figure:

We have:

\begin{align}&\sin \theta = \frac{{PQ}}{{OP}} = \frac{{PQ}}{1} = PQ\\&\sin \left( { - \theta } \right) = \frac{{P'Q}}{{OP'}} = \frac{{PQ}}{1} = PQ\end{align}

Thus,

$\boxed {\sin \left( { - \theta } \right) = \sin \theta }$

This also means that

$\boxed{\text{cosec}\,\left( { - \theta } \right) = \text{cosec}\,\theta}$

What about $$cos\left( { - \theta } \right)$$? From the figure above, we have

\begin{align}&\cos \theta = \frac{{OQ}}{{OP}} = \frac{{OQ}}{1} = OQ\\&cos\left( { - \theta } \right) = \frac{{OQ}}{{OP'}} = \frac{{OQ}}{1} = OQ\end{align}

Thus,

\boxed{\begin{align}&\cos\left( { - \theta } \right) = \cos \theta \\&\sec \left( { - \theta } \right) = \sec \theta \end{align}}

Finally, from the same figure, we have:

\begin{align}\tan \theta &= \frac{{PQ}}{{OQ}} \\\tan \left( { - \theta } \right) &= \frac{{P'Q}}{{OQ}} = \frac{{PQ}}{{OQ}}\end{align}

Thus,

\boxed{\begin{align}&\tan \left( { - \theta } \right) = - \tan \theta \\&\cot \left( { - \theta } \right) = - \cot \theta \end{align}}

Challenge 1: Find the values of (a)  $$\sin \left( { - \frac{\pi }{6}} \right)$$  (b)  $$\cos \left( { - \frac{\pi }{3}} \right)$$  (c)  $$\tan \left( { - \frac{\pi }{4}} \right)$$.

Tip: Use conversion relation of $$\sin \left( { - \theta } \right)$$, $$\cos \left( { - \theta } \right)$$, and $$\tan \left( { - \theta } \right)$$.

We have already seen the following complementary angle relations (we used the degree scale earlier instead of the radian scale):

\begin{align}&\sin \left( {\frac{\pi }{2} - \theta } \right) = \cos \theta ,\,\,\,\cos \left( {\frac{\pi }{2} - \theta } \right) = \sin \theta \\&\tan \left( {\frac{\pi }{2} - \theta } \right) = \cot \theta ,\,\,\,\cot \left( {\frac{\pi }{2} - \theta } \right) = \tan \theta \\&\sec \left( {\frac{\pi }{2} - \theta } \right) = {\mathop{\rm \text{cosec}}\nolimits}\, \theta ,\,\,\,\text{cosec}\,\left( {\frac{\pi }{2} - \theta } \right) = \sec \theta \end{align}

We justified these relations for $$\theta$$ between 0 and $$\frac{\pi }{2}$$ radians. However, they hold true for $$\theta$$ of any arbitrary magnitude. Can you see how?

These relations above are just few of the many possible such relations which trigonometric ratios satisfy. For example, consider the following relation:

$\sin \left( {\frac{\pi }{2} + \theta } \right) = \cos \theta$

Let us prove this relation for acute $$\theta$$. Consider the following figure:

Note: $$\Delta OQP$$ and $$\Delta RSO$$ are congruent, which means that $$OQ = RS$$ and $$OS = PQ$$. We have:

\begin{align}&\sin \left( {\frac{\pi }{2} + \theta } \right) = \frac{{RS}}{{RO}} = RS = OQ\\&\cos \theta = \frac{{OQ}}{{OP}} = OQ\\&\Rightarrow \boxed{\sin \left( {\frac{\pi }{2} + \theta } \right) = \cos \theta }\end{align}

Using the same figure, we can prove that,

\begin{align}&\cos \left( {\frac{\pi }{2} + \theta } \right) = - \sin \theta \\&\tan \left( {\frac{\pi }{2} + \theta } \right) = - \cot \theta \end{align}

Let us prove the first of these relations. We have:

\begin{align}&\cos \left( {\frac{\pi }{2} + \theta } \right) = \frac{{OS}}{{OR}} = \frac{{ - PQ}}{1} = - PQ\\& - \sin \theta = - \frac{{PQ}}{{OP}} = - PQ\\& \Rightarrow \,\,\,\cos \left( {\frac{\pi }{2} + \theta } \right) = - \sin \theta \end{align}

Next, consider the following set of relations:

\begin{align}&\sin \left( {\pi + \theta } \right) = - \sin \theta \\&\cos \left( {\pi + \theta } \right) = - \cos \theta \\&\tan \left( {\pi + \theta } \right) = \tan \theta \end{align}

Can you prove these? Consider the following figure:

The point corresponding to the angle $$\pi + \theta$$ is diametrically opposite to the point corresponding to the angle $$\theta$$. We have:

\begin{align}&\sin \left( {\pi + \theta } \right) = \frac{{RS}}{{OR}} = \frac{{ - PQ}}{1} = - PQ = - \sin \theta \\&\cos \left( {\pi + \theta } \right) = \frac{{OS}}{{OR}} = \frac{{ - OQ}}{1} = - OQ = - \cos \theta \\&\tan \left( {\pi + \theta } \right) = \frac{{RS}}{{OS}} = \frac{{ - PQ}}{{ - OQ}} = \frac{{PQ}}{{OQ}} = \tan \theta \end{align}

## Solved Examples:

Example 1: Prove geometrically that:

\begin{align}&\sin \left( {\pi - \theta } \right) = \sin \theta \\&\cos \left( {\pi - \theta } \right) = - \cos \theta \\&\tan \left( {\pi - \theta } \right) = - \tan \theta \end{align}

Solution:Consider the following figure:

We have:

\begin{align}& \sin \left( {\pi - \theta } \right) = \frac{{RS}}{{OR}} = \frac{{PQ}}{1} = PQ = \sin \theta \\ &\cos \left( {\pi - \theta } \right) = \frac{{OS}}{{OR}} = \frac{{ - OQ}}{1} = - OQ = - \cos \theta \\ &\tan \left( {\pi - \theta } \right) = \frac{{RS}}{{OS}} = \frac{{PQ}}{{ - OQ}} = - \frac{{PQ}}{{OQ}} = - \tan \theta \end{align}

Example 2: Prove geometrically that:

\begin{align}&\sin \left( {2\pi - \theta } \right) = - \sin \theta \\&\cos \left( {2\pi - \theta } \right) = \cos \theta \\&\tan \left( {2\pi - \theta } \right) = - \tan \theta \end{align}

Solution: Consider the following figure:

We have:

\begin{align}&\sin \left( {2\pi - \theta } \right) = \frac{{RQ}}{{OR}} = \frac{{ - PQ}}{1} = - PQ = - \sin \theta \\&\cos \left( {2\pi - \theta } \right) = \frac{{OQ}}{{OR}} = \frac{{OQ}}{1} = OQ = \cos \theta \\&\tan \left( {2\pi - \theta } \right) = \frac{{RQ}}{{OQ}} = \frac{{ - PQ}}{{OQ}} = - \frac{{PQ}}{{OQ}} = - \tan \theta \end{align}

Challenge 2: Prove geometrically that:

\begin{align}&\sin \left( {2\pi + \theta } \right) = \sin \theta \\&\cos \left( {2\pi + \theta } \right) = \cos \theta \\&\tan \left( {2\pi + \theta } \right) = \tan \theta \end{align}

Tip: Show visually that the angle $$2\pi + \theta$$ geometrically corresponds to the same configuration as the angle $$\theta$$, and hence the trigonometric ratios of these angles are the same.

Example 3: Express the value of $$\sin \left( {\frac{{11\pi }}{2} + \theta } \right)$$ in terms of a trigonometric ratio of  $$\theta$$.

Solution: We have:

\begin{align}&\sin \left( {\frac{{11\pi }}{2} + \theta } \right) = \sin \left( {6\pi - \frac{\pi }{2} + \theta } \right)\\&\qquad\qquad\qquad\; = \sin \left( { - \frac{\pi }{2} + \theta } \right)\\&\qquad\qquad\qquad\; = - \sin \left( {\frac{\pi }{2} - \theta } \right) = \cos \theta \end{align}

Example 4: Express the value of $$\tan \left( {19\pi - \theta } \right)$$ in terms of a trigonometric ratio of  $$\theta$$.

Solution: We proved earlier that $$\tan \left( {\pi + \theta } \right)$$ is the same as $$\tan \theta$$. Thus, adding any multiple of $$\pi$$ to the original argument will not change the original $$\tan$$ value. This means that:

$\tan \left( {19\pi - \theta } \right) = \tan \left( { - \theta } \right) = - \tan \theta$

Challenge 3: Express the value of $$\cos \left( {\frac{{9\pi }}{2} + \theta } \right)$$ in terms of a trigonometric ratio of  $$\theta$$.

Tip: Use a similar approach as in example-3.

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