# 9.2 Sum and difference identities  (Page 3/6)

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## Finding the exact value of an expression involving an inverse trigonometric function

Find the exact value of $\text{\hspace{0.17em}}\mathrm{sin}\left({\mathrm{cos}}^{-1}\frac{1}{2}+{\mathrm{sin}}^{-1}\frac{3}{5}\right).\text{\hspace{0.17em}}$ Then check the answer with a graphing calculator.

The pattern displayed in this problem is $\text{\hspace{0.17em}}\mathrm{sin}\left(\alpha +\beta \right).\text{\hspace{0.17em}}$ Let $\text{\hspace{0.17em}}\alpha ={\mathrm{cos}}^{-1}\frac{1}{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\beta ={\mathrm{sin}}^{-1}\frac{3}{5}.\text{\hspace{0.17em}}$ Then we can write

$\begin{array}{ccc}\hfill \mathrm{cos}\text{\hspace{0.17em}}\alpha & =& \frac{1}{2},0\le \alpha \le \pi \hfill \\ \hfill \mathrm{sin}\text{\hspace{0.17em}}\beta & =& \frac{3}{5},-\frac{\pi }{2}\le \beta \le \frac{\pi }{2}\hfill \end{array}$

We will use the Pythagorean identities to find $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta .$

$\begin{array}{ccc}\hfill \mathrm{sin}\text{\hspace{0.17em}}\alpha & =& \sqrt{1-{\mathrm{cos}}^{2}\alpha }\hfill \\ & =& \sqrt{1-\frac{1}{4}}\hfill \\ & =& \sqrt{\frac{3}{4}}\hfill \\ & =& \frac{\sqrt{3}}{2}\hfill \\ \hfill \mathrm{cos}\text{\hspace{0.17em}}\beta & =& \sqrt{1-{\mathrm{sin}}^{2}\beta }\hfill \\ & =& \sqrt{1-\frac{9}{25}}\hfill \\ & =& \sqrt{\frac{16}{25}}\hfill \\ & =& \frac{4}{5}\hfill \end{array}$

Using the sum formula for sine,

$\begin{array}{ccc}\hfill \mathrm{sin}\left({\mathrm{cos}}^{-1}\frac{1}{2}+{\mathrm{sin}}^{-1}\frac{3}{5}\right)& =& \mathrm{sin}\left(\alpha +\beta \right)\hfill \\ & =& \mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta +\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta \hfill \\ & =& \frac{\sqrt{3}}{2}\cdot \frac{4}{5}+\frac{1}{2}\cdot \frac{3}{5}\hfill \\ & =& \frac{4\sqrt{3}+3}{10}\hfill \end{array}$

## Using the sum and difference formulas for tangent

Finding exact values for the tangent of the sum or difference of two angles is a little more complicated, but again, it is a matter of recognizing the pattern.

Finding the sum of two angles formula for tangent involves taking quotient of the sum formulas for sine and cosine and simplifying. Recall, $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x=\frac{\mathrm{sin}\text{\hspace{0.17em}}x}{\mathrm{cos}\text{\hspace{0.17em}}x},\mathrm{cos}\text{\hspace{0.17em}}x\ne 0.$

Let’s derive the sum formula for tangent.

We can derive the difference formula for tangent in a similar way.

## Sum and difference formulas for tangent

The sum and difference formulas for tangent are:

$\mathrm{tan}\left(\alpha +\beta \right)=\frac{\mathrm{tan}\text{\hspace{0.17em}}\alpha +\mathrm{tan}\text{\hspace{0.17em}}\beta }{1-\mathrm{tan}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\beta }$
$\mathrm{tan}\left(\alpha -\beta \right)=\frac{\mathrm{tan}\text{\hspace{0.17em}}\alpha -\mathrm{tan}\text{\hspace{0.17em}}\beta }{1+\mathrm{tan}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\beta }$

Given two angles, find the tangent of the sum of the angles.

1. Write the sum formula for tangent.
2. Substitute the given angles into the formula.
3. Simplify.

## Finding the exact value of an expression involving tangent

Find the exact value of $\text{\hspace{0.17em}}\mathrm{tan}\left(\frac{\pi }{6}+\frac{\pi }{4}\right).$

Let’s first write the sum formula for tangent and then substitute the given angles into the formula.

$\begin{array}{ccc}\hfill \mathrm{tan}\left(\alpha +\beta \right)& =& \frac{\mathrm{tan}\text{\hspace{0.17em}}\alpha +\mathrm{tan}\text{\hspace{0.17em}}\beta }{1-\mathrm{tan}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\beta }\hfill \\ \hfill \mathrm{tan}\left(\frac{\pi }{6}+\frac{\pi }{4}\right)& =& \frac{\mathrm{tan}\left(\frac{\pi }{6}\right)+\mathrm{tan}\left(\frac{\pi }{4}\right)}{1-\left(\mathrm{tan}\left(\frac{\pi }{6}\right)\right)\left(\mathrm{tan}\left(\frac{\pi }{4}\right)\right)}\hfill \end{array}$

Next, we determine the individual function values within the formula:

$\mathrm{tan}\left(\frac{\pi }{6}\right)=\frac{1}{\sqrt{3}},\mathrm{tan}\left(\frac{\pi }{4}\right)=1$

So we have

$\begin{array}{ccc}\hfill \mathrm{tan}\left(\frac{\pi }{6}+\frac{\pi }{4}\right)& =& \frac{\frac{1}{\sqrt{3}}+1}{1-\left(\frac{1}{\sqrt{3}}\right)\left(1\right)}\hfill \\ & =& \frac{\frac{1+\sqrt{3}}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}}\hfill \\ & =& \frac{1+\sqrt{3}}{\sqrt{3}}\left(\frac{\sqrt{3}}{\sqrt{3}-1}\right)\hfill \\ & =& \frac{\sqrt{3}+1}{\sqrt{3}-1}\hfill \end{array}$

Find the exact value of $\text{\hspace{0.17em}}\mathrm{tan}\left(\frac{2\pi }{3}+\frac{\pi }{4}\right).$

$\frac{1-\sqrt{3}}{1+\sqrt{3}}$

## Finding multiple sums and differences of angles

Given find

1. $\mathrm{sin}\left(\alpha +\beta \right)$
2. $\mathrm{cos}\left(\alpha +\beta \right)$
3. $\mathrm{tan}\left(\alpha +\beta \right)$
4. $\mathrm{tan}\left(\alpha -\beta \right)$

We can use the sum and difference formulas to identify the sum or difference of angles when the ratio of sine, cosine, or tangent is provided for each of the individual angles. To do so, we construct what is called a reference triangle to help find each component of the sum and difference formulas.

1. To find $\text{\hspace{0.17em}}\mathrm{sin}\left(\alpha +\beta \right),$ we begin with $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =\frac{3}{5}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}0<\alpha <\frac{\pi }{2}.\text{\hspace{0.17em}}$ The side opposite $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ has length 3, the hypotenuse has length 5, and $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ is in the first quadrant. See [link] . Using the Pythagorean Theorem, we can find the length of side $\text{\hspace{0.17em}}a\text{:}$
$\begin{array}{ccc}\hfill {a}^{2}+{3}^{2}& =& {5}^{2}\hfill \\ \hfill {a}^{2}& =& 16\hfill \\ \hfill a& =& 4\hfill \end{array}$

Since $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta =-\frac{5}{13}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\pi <\beta <\frac{3\pi }{2},$ the side adjacent to $\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}-5,$ the hypotenuse is 13, and $\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$ is in the third quadrant. See [link] . Again, using the Pythagorean Theorem, we have

$\begin{array}{ccc}\hfill {\left(-5\right)}^{2}+{a}^{2}& =& {13}^{2}\hfill \\ \hfill 25+{a}^{2}& =& 169\hfill \\ \hfill {a}^{2}& =& 144\hfill \\ \hfill a& =& ±12\hfill \end{array}$

Since $\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$ is in the third quadrant, $\text{\hspace{0.17em}}a=–12.$

The next step is finding the cosine of $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ and the sine of $\text{\hspace{0.17em}}\beta .\text{\hspace{0.17em}}$ The cosine of $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ is the adjacent side over the hypotenuse. We can find it from the triangle in [link] : $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\alpha =\frac{4}{5}.\text{\hspace{0.17em}}$ We can also find the sine of $\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$ from the triangle in [link] , as opposite side over the hypotenuse: $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta =-\frac{12}{13}.\text{\hspace{0.17em}}$ Now we are ready to evaluate $\text{\hspace{0.17em}}\mathrm{sin}\left(\alpha +\beta \right).$

$\begin{array}{ccc}\hfill \mathrm{sin}\left(\alpha +\beta \right)& =& \mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta +\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta \hfill \\ & =& \left(\frac{3}{5}\right)\left(-\frac{5}{13}\right)+\left(\frac{4}{5}\right)\left(-\frac{12}{13}\right)\hfill \\ & =& -\frac{15}{65}-\frac{48}{65}\hfill \\ & =& -\frac{63}{65}\hfill \end{array}$
2. We can find $\text{\hspace{0.17em}}\mathrm{cos}\left(\alpha +\beta \right)\text{\hspace{0.17em}}$ in a similar manner. We substitute the values according to the formula.
$\begin{array}{ccc}\hfill \mathrm{cos}\left(\alpha +\beta \right)& =& \mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta -\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta \hfill \\ & =& \left(\frac{4}{5}\right)\left(-\frac{5}{13}\right)-\left(\frac{3}{5}\right)\left(-\frac{12}{13}\right)\hfill \\ & =& -\frac{20}{65}+\frac{36}{65}\hfill \\ & =& \frac{16}{65}\hfill \end{array}$
3. For $\text{\hspace{0.17em}}\mathrm{tan}\left(\alpha +\beta \right),$ if $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =\frac{3}{5}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\alpha =\frac{4}{5},$ then
$\mathrm{tan}\text{\hspace{0.17em}}\alpha =\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$

If $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta =-\frac{12}{13}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta =-\frac{5}{13},$ then

$\mathrm{tan}\text{\hspace{0.17em}}\beta =\frac{\frac{-12}{13}}{\frac{-5}{13}}=\frac{12}{5}$

Then,

4. To find $\text{\hspace{0.17em}}\mathrm{tan}\left(\alpha -\beta \right),$ we have the values we need. We can substitute them in and evaluate.
$\begin{array}{ccc}\hfill \mathrm{tan}\left(\alpha -\beta \right)& =& \frac{\mathrm{tan}\text{\hspace{0.17em}}\alpha -\mathrm{tan}\text{\hspace{0.17em}}\beta }{1+\mathrm{tan}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\beta }\hfill \\ & =& \frac{\frac{3}{4}-\frac{12}{5}}{1+\frac{3}{4}\left(\frac{12}{5}\right)}\hfill \\ & =& \frac{-\frac{33}{20}}{\frac{56}{20}}\hfill \\ & =& -\frac{33}{56}\hfill \end{array}$

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