Supplementary Angle Identities (π – θ): Converting from Obtuse to Acute Angles

Angles that add up to 180^\circ (\pi) are called supplementary angles. For example, since \theta + (\pi - \theta) = \pi, we can say that \theta and \pi - \theta are supplementary to each other. The trigonometric functions have important identities related to these supplementary angles:

\sin{(\pi - \theta)} = \sin{\theta} \tag{1}

\cos{(\pi - \theta)} = -\cos{\theta} \tag{2}

\tan{(\pi - \theta)} = -\tan{\theta} \tag{3}

In this article, we will learn the basics of how to use these identities. If you are curious about why these identities hold true, please refer to the following article:

1 Features of the Identities

By using the supplementary angle identity, we can rewrite the angle x in a trigonometric function like y = \sin x so that an obtuse angle is expressed as an acute angle instead. For example, applying this identity results in:

y = \sin 150^\circ = \sin 30^\circ

Here, 150^\circ is an obtuse angle, while 30^\circ is an acute angle. This means that in the above equation, the value of y can be found using the acute angle 30^\circ instead of the obtuse angle 150^\circ (I will explain why this equation holds true later).

Expressing the value of a trigonometric function with an obtuse angle in terms of an acute angle offers the following advantages:

• It allows us to use smaller angle measures.
• It helps us identify and combine identical values within expressions, simplifying them.

Let us now practice using the supplementary angle identity and consider these advantages in more detail.

2 Practice Problems on the Identities

First, note that \frac{5 \pi}{6} corresponds to 150^\circ, which is an obtuse angle. Here, \frac{5 \pi}{6} is expressed in radians, while 150^\circ is in degrees. If you are not familiar with the difference, please refer to this explanation of degrees and radians.

Step 1: Constructing \pi

Since \sin\frac{5 \pi}{6} involves an obtuse angle, let us consider whether the supplementary angle identity, which uses \pi, can help simplify it.

We express:

\frac{5 \pi}{6} = \pi - \frac{\pi}{6}

Therefore,

\sin \frac{5 \pi}{6} = \sin \left( \pi - \frac{\pi}{6} \right) \tag{4}

holds true.

Step 2: Apply Equation 1

Because the sine function is involved, we apply the identity Equation 1:

\sin{(\pi - \theta)} = \sin{\theta}

Here, setting \theta = \frac{\pi}{6} gives:

\sin{\left(\pi - \frac{\pi}{6}\right)} = \sin{\frac{\pi}{6}}.

Hence, from Equation 4, we conclude:

\begin{aligned} \sin \frac{5 \pi}{6} &= \sin \left( \pi - \frac{\pi}{6} \right)\\ &= \sin{\frac{\pi}{6}}. \end{aligned}

Since \frac{\pi}{6} is an acute angle, this completes the problem. Expressed in degrees, this reads:

\sin 150^\circ = \sin 30^\circ.

As you can see, the supplementary angle identity enables us to express a trigonometric function value originally given by a large angle, 150^\circ, in terms of a smaller, acute angle, 30^\circ.

We proceed similarly to Exercise 1.

Step 1: Construct \pi

Note that \frac{6 \pi}{7} is an obtuse angle. Observe that:

\frac{6 \pi}{7} = \pi - \frac{\pi}{7}

Therefore,

\cos \frac{6 \pi}{7} = \cos \left(\pi - \frac{\pi}{7}\right).

Step 2: Apply Equation 2

Letting \theta = \frac{\pi}{7}, the identity Equation 2 tells us:

\begin{aligned} \cos{(\pi - \theta)} &= \cos{\left( \pi - \frac{\pi}{7} \right)}\\ &= - \cos \frac{\pi}{7}. \end{aligned}

Since \frac{\pi}{7} is acute, we have solved the problem.

We follow the same steps as in Exercise 1 and Exercise 2.

Step 1: Construct \pi

Because \frac{3 \pi}{5} is an obtuse angle, we use the supplementary angle identity to express it with an acute angle:

\frac{3 \pi}{5} = \pi - \frac{2 \pi}{5}.

Therefore,

\tan \frac{3 \pi}{5} = \tan \left( \pi - \frac{2 \pi}{5} \right).

Step 2: Apply Equation 3

Applying Equation 3, we have:

\begin{aligned} \tan \frac{3 \pi}{5} &= \tan \left( \pi - \frac{2 \pi}{5} \right)\\ &= -\tan \frac{2 \pi}{5}. \end{aligned}

Since \frac{2 \pi}{5} is acute, the problem is solved.

Initially glance, this expression may not seem easy to simplify. However, note that 170^\circ is obtuse, while 10^\circ is acute. We can use the supplementary angle identity to express \sin 170^\circ in terms of an acute angle:

\begin{aligned} \sin 170^\circ &= \sin (180^\circ - 10^\circ)\\ &= \sin 10^\circ. \quad (\because \sin (\pi - \theta) = \sin \theta) \end{aligned}

Substituting this back into the original expression gives:

\begin{aligned} \sin 170^\circ - \sin 10^\circ &= \sin 10^\circ - \sin 10^\circ\\ &= 0. \end{aligned}

Thus, the expression simplifies neatly to zero. This example shows that when an expression contains a mix of obtuse and acute angles, rewriting the parts involving obtuse angles in terms of acute angles can help simplify it.

3 Notes on Using the Theorem

As explained in Section 1, these identities allow us to use smaller angles instead of larger ones. Indeed, in Section 2, we expressed trigonometric functions with obtuse angles in terms of acute angles.

As shown in Exercise 4, unifying angles to acute ones helps us identify common values within expressions, enabling simplification.

Whenever you encounter obtuse angles in trigonometric problems, it is worthwhile to consider applying the supplementary angle identities.

Please remember that, unlike Equation 1, the identity Equation 2 and Equation 3 include a negative sign on the right-hand side; this detail is important to keep in mind.

4 Summary

In this article, we have discussed the supplementary angle identity in trigonometry. We reviewed the concept of supplementary angles and several examples showing how to use it effectively.