Skills Review for Calculus of the Hyperbolic Functions

Learning Outcomes

Verify the fundamental trigonometric identities

In the Calculus of the Hyperbolic Functions section, we will learn how to differentiate and integrate hyperbolic functions. Here we will review some trigonometric formulas and how to use them. Some of these formulas are identical to those used for hyperbolic functions.

Apply Trigonometric Formulas

A General Note: Sum and Difference Formulas for Cosine

The sum and difference formulas for cosine are:

$\begin{align}\cos \left(\alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta\end{align}$

$\begin{align}\cos \left(\alpha -\beta \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta\end{align}$

A General Note: Sum and Difference Formulas for Sine

The sum and difference formulas for sine are:

$\sin \left(\alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta$

$\sin \left(\alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta$

A General Note: Sum and Difference Formulas for Tangent

The sum and difference formulas for tangent are:

$\tan \left(\alpha +\beta \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$

$\tan \left(\alpha -\beta \right)=\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }$

How To: Given an identity, verify using sum and difference formulas

Begin with the expression on the side of the equal sign that appears most complex. Rewrite that expression until it matches the other side of the equal sign. Occasionally, we might have to alter both sides, but working on only one side is the most efficient.
Look for opportunities to use the sum and difference formulas.
Rewrite sums or differences of quotients as single quotients.
If the process becomes cumbersome, rewrite the expression in terms of sines and cosines.

Example: Applying Trigonometric Formulas to Verify Trigonometric Identities

Verify the identity $\sin \left(\alpha +\beta \right)+\sin \left(\alpha -\beta \right)=2\sin \alpha \cos \beta$ .

Show Solution

We see that the left side of the equation includes the sines of the sum and the difference of angles.

$\begin{gathered}\sin \left(\alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta\\ \sin \left(\alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta \end{gathered}$

We can rewrite each using the sum and difference formulas.

$\begin{align}\sin \left(\alpha +\beta \right)+\sin \left(\alpha -\beta \right)&=\sin \alpha \cos \beta +\cos \alpha \sin \beta +\sin \alpha \cos \beta -\cos \alpha \sin \beta \\ &=2\sin \alpha \cos \beta \end{align}$

We see that the identity is verified.

Example: Applying Trigonometric Formulas to Verify Trigonometric Identities

Verify the following identity.

$\frac{\sin \left(\alpha -\beta \right)}{\cos \alpha \cos \beta }=\tan \alpha -\tan \beta$

Show Solution

We can begin by rewriting the numerator on the left side of the equation.

$\begin{align}\frac{\sin \left(\alpha -\beta \right)}{\cos \alpha \cos \beta }&=\frac{\sin \alpha \cos \beta -\cos \alpha \sin \beta }{\cos \alpha \cos \beta } \\ &=\frac{\sin \alpha \cos \beta}{\cos \alpha \cos \beta}-\frac{\cos \alpha \sin \beta }{\cos \alpha \cos \beta } && \text{Rewrite using a common denominator}. \\ &=\frac{\sin \alpha }{\cos \alpha }-\frac{\sin \beta }{\cos \beta }&& \text{Cancel}. \\ &=\tan \alpha -\tan \beta && \text{Rewrite in terms of tangent}.\end{align}$

We see that the identity is verified. In many cases, verifying tangent identities can successfully be accomplished by writing the tangent in terms of sine and cosine.

Try It

Verify the identity: $\tan \left(\pi -\theta \right)=-\tan \theta$ .

Show Solution

$\begin{align}\tan \left(\pi -\theta \right)&=\frac{\tan \left(\pi \right)-\tan \theta }{1+\tan \left(\pi \right)\tan \theta } \\ &=\frac{0-\tan \theta }{1+0\cdot \tan \theta } \\ &=-\tan \theta \end{align}$

A General Note: Double-Angle Formulas

The double-angle formulas are summarized as follows:

$\begin{align}\sin \left(2\theta \right)&=2\sin \theta \cos \theta\\\text{ }\\ \cos \left(2\theta \right)&={\cos }^{2}\theta -{\sin }^{2}\theta \\ &=1 - 2{\sin }^{2}\theta \\ &=2{\cos }^{2}\theta -1 \\\text{ }\\ \tan \left(2\theta \right)&=\frac{2\tan \theta }{1-{\tan }^{2}\theta }\end{align}$

Establishing identities using the double-angle formulas is performed using the same steps we used to establish identities using the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side.

Example: Applying Trigonometric Formulas to Verify Trigonometric Identities

Establish the following identity using double-angle formulas:

$1+\sin \left(2\theta \right)={\left(\sin \theta +\cos \theta \right)}^{2}$

Show Solution

We will work on the right side of the equal sign and rewrite the expression until it matches the left side.

$\begin{align}{\left(\sin \theta +\cos \theta \right)}^{2}&={\sin }^{2}\theta +2\sin \theta \cos \theta +{\cos }^{2}\theta \\ &=\left({\sin }^{2}\theta +{\cos }^{2}\theta \right)+2\sin \theta \cos \theta \\ &=1+2\sin \theta \cos \theta \\ &=1+\sin \left(2\theta \right)\end{align}$

Analysis of the Solution

This process is not complicated, as long as we recall the perfect square formula from algebra:

${\left(a\pm b\right)}^{2}={a}^{2}\pm 2ab+{b}^{2}$

where $a=\sin \theta$ and $b=\cos \theta$ . Part of being successful in mathematics is the ability to recognize patterns. While the terms or symbols may change, the algebra remains consistent.

Try It

Establish the identity: ${\cos }^{4}\theta -{\sin }^{4}\theta =\cos \left(2\theta \right)$ .

Show Solution

${\cos }^{4}\theta -{\sin }^{4}\theta =\left({\cos }^{2}\theta +{\sin }^{2}\theta \right)\left({\cos }^{2}\theta -{\sin }^{2}\theta \right)=\cos \left(2\theta \right)$

Example: Applying Trigonometric Formulas to Verify Trigonometric Identities

Verify the identity:

$\tan \left(2\theta \right)=\frac{2}{\cot \theta -\tan \theta }$

Show Solution

In this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation.

$\begin{align}\tan \left(2\theta \right)&=\frac{2\tan \theta }{1-{\tan }^{2}\theta }&& \text{Double-angle formula} \\ &=\frac{2\tan \theta \left(\frac{1}{\tan \theta }\right)}{\left(1-{\tan }^{2}\theta \right)\left(\frac{1}{\tan \theta }\right)}&& \text{Multiply by a term that results in desired numerator}. \\ &=\frac{2}{\frac{1}{\tan \theta }-\frac{{\tan }^{2}\theta }{\tan \theta }} \\ &=\frac{2}{\cot \theta -\tan \theta }&& \text{Use reciprocal identity for }\frac{1}{\tan \theta }.\end{align}$

Analysis of the Solution

Here is a case where the more complicated side of the initial equation appeared on the right, but we chose to work the left side. However, if we had chosen the left side to rewrite, we would have been working backwards to arrive at the equivalency. For example, suppose that we wanted to show

$\frac{2\tan \theta }{1-{\tan }^{2}\theta }=\frac{2}{\cot \theta -\tan \theta }$

Let’s work on the right side.

$\begin{align}\frac{2}{\cot \theta -\tan \theta }&=\frac{2}{\frac{1}{\tan \theta }-\tan \theta }\left(\frac{\tan \theta }{\tan \theta }\right) \\ &=\frac{2\tan \theta }{\frac{1}{\cancel{\tan \theta }}\left(\cancel{\tan \theta }\right)-\tan \theta \left(\tan \theta \right)} \\ &=\frac{2\tan \theta }{1-{\tan }^{2}\theta } \end{align}$

When using the identities to simplify a trigonometric expression or solve a trigonometric equation, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated. However, we should begin with the guidelines set forth earlier.

Try It

Verify the identity: $\cos \left(2\theta \right)\cos \theta ={\cos }^{3}\theta -\cos \theta {\sin }^{2}\theta$ .

Show Solution

$\cos \left(2\theta \right)\cos \theta =\left({\cos }^{2}\theta -{\sin }^{2}\theta \right)\cos \theta ={\cos }^{3}\theta -\cos \theta {\sin }^{2}\theta$

Applications of Integration: Prerequisite Material