Conversion Relations of Trigonometric Ratios

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.

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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:

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

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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:

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

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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:

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

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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.

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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:

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

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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:

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

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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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