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Exercises Trigonometry

Example 5 from Algebra and Trigonometry, 9.2 Sum and Difference Identities

Finding the Exact Value of an Expression Involving Tangent

Find the exact value of $\tan (\frac{π}{6} + \frac{π}{4}) .$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
Let’s first write the sum formula for tangent and then substitute the given angles into the formula.
$\begin{matrix} \tan (α + β) & = & \frac{\tan α + \tan β}{1 - \tan α \tan β} \\ \tan (\frac{π}{6} + \frac{π}{4}) & = & \frac{\tan (\frac{π}{6}) + \tan (\frac{π}{4})}{1 - (\tan (\frac{π}{6})) (\tan (\frac{π}{4}))} \end{matrix}$
Next, we determine the individual function values within the formula:
$\tan (\frac{π}{6}) = \frac{1}{\sqrt{3}}, \tan (\frac{π}{4}) = 1$

So we have

$\begin{matrix} \tan (\frac{π}{6} + \frac{π}{4}) & = & \frac{\frac{1}{\sqrt{3}} + 1}{1 - (\frac{1}{\sqrt{3}}) (1)} \\ = & \frac{\frac{1 + \sqrt{3}}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} \\ = & \frac{1 + \sqrt{3}}{\sqrt{3}} (\frac{\sqrt{3}}{\sqrt{3} - 1}) \\ = & \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \end{matrix}$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Find the exact value of $\tan (\frac{2 π}{3} + \frac{π}{4}) .$

How did it go?

PreviousFinding the Exact Value of an Expression Involving an Inverse Trigonometric Function Next Finding Multiple Sums and Differences of Angles

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 9.2, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.