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3.8: Tangent Sum and Difference Formulas

Difficulty Level: At Grade Created by: CK-12
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Suppose you were given two angles and asked to find the tangent of the difference of them. For example, can you compute:

\begin{align*}\tan (120^\circ - 40^\circ)\end{align*}tan(12040)

Would you just subtract the angles and then take the tangent of the result? Or is something more complicated required to solve this problem? Keep reading, and by the end of this Concept, you'll be able to calculate trig functions like the one above.

Tangent Sum and Difference Formulas

In this Concept, we want to find a formula that will make computing the tangent of a sum of arguments or a difference of arguments easier. As first, it may seem that you should just add (or subtract) the arguments and take the tangent of the result. However, it's not quite that easy.

To find the sum formula for tangent:

\begin{align*}\tan(a + b) & = \frac{\sin(a+b)}{\cos(a+b)} && \text{Using}\ \tan \theta = \frac{\sin \theta}{\cos \theta} \\ & = \frac{\sin a \cos b + \sin b \cos a}{\cos a \cos b - \sin a \sin b} && \text{Substituting the sum formulas for sine and cosine} \\ & = \frac{\frac{\sin a \cos b + \sin b \cos a}{\cos a \cos b}}{\frac{\cos a \cos b - \sin a \sin b}{\cos a \cos b}} && \text{Divide both the numerator and the denominator by}\ \cos a \cos b \\ & = \frac{\frac{\sin a \cos b} {\cos a \cos b} + \frac{\sin b \cos a}{\cos a \cos b}}{\frac{\cos a \cos b} {\cos a \cos b} - \frac{\sin a \sin b}{\cos a \cos b}} && \text{Reduce each of the fractions} \\ & = \frac{\frac{\sin a}{\cos a} + \frac{\sin b}{\cos b}}{1- \frac{\sin a \sin b}{\cos a \cos b}} && \text{Substitute}\ \frac{\sin \theta}{\cos \theta} = \tan \theta \\ \tan(a + b) & = \frac{\tan a + \tan b}{1 - \tan a \tan b} && \text{Sum formula for tangent}\end{align*}tan(a+b)tan(a+b)=sin(a+b)cos(a+b)=sinacosb+sinbcosacosacosbsinasinb=sinacosb+sinbcosacosacosbcosacosbsinasinbcosacosb=sinacosbcosacosb+sinbcosacosacosbcosacosbcosacosbsinasinbcosacosb=sinacosa+sinbcosb1sinasinbcosacosb=tana+tanb1tanatanbUsing tanθ=sinθcosθSubstituting the sum formulas for sine and cosineDivide both the numerator and the denominator by cosacosbReduce each of the fractionsSubstitute sinθcosθ=tanθSum formula for tangent

In conclusion, \begin{align*}\tan(a+b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}\end{align*}tan(a+b)=tana+tanb1tanatanb. Substituting \begin{align*}-b\end{align*}b for \begin{align*}b\end{align*}b in the above results in the difference formula for tangent:

\begin{align*} \tan(a-b) = \frac{\tan a - \tan b}{1 + \tan a \tan b}\end{align*}tan(ab)=tanatanb1+tanatanb

Solve the following problems using the Tangent Sum and Difference Formula

Find the exact value of \begin{align*}\tan 285^\circ\end{align*}tan285.

Use the difference formula for tangent, with \begin{align*}285^\circ = 330^\circ - 45^\circ\end{align*}285=33045

\begin{align*}\tan (330^\circ - 45^\circ) & = \frac{\tan 330^\circ - \tan 45^\circ}{1 + \tan 330^\circ \tan 45^\circ} \\ & = \frac{- \frac{\sqrt{3}}{3} - 1}{1 - \frac{\sqrt{3}}{3} \cdot 1} = \frac{-3 - \sqrt{3}}{3-\sqrt{3}} \\ & = \frac{-3- \sqrt{3}}{3- \sqrt{3}} \cdot \frac{3+ \sqrt{3}}{3+ \sqrt{3}} \\ & = \frac{-9-6 \sqrt{3} -3}{9-3} \\ & = \frac{-12 - 6 \sqrt{3}}{6} \\ & = - 2 - \sqrt{3}\end{align*}tan(33045)=tan330tan451+tan330tan45=3311331=3333=33333+33+3=963393=12636=23

To verify this on the calculator, \begin{align*}\tan 285^\circ = -3.732\end{align*}tan285=3.732 and \begin{align*}-2 -\sqrt{3} = -3.732\end{align*}23=3.732.

Verify the tangent difference formula by finding \begin{align*}\tan \frac{6\pi}{6}\end{align*}tan6π6, since this should be equal to \begin{align*}\tan \pi = 0\end{align*}tanπ=0.

Use the difference formula for tangent, with \begin{align*}\tan \frac{6\pi}{6} = \tan (\frac{7 \pi}{6} - \frac{\pi}{6})\end{align*}tan6π6=tan(7π6π6)

\begin{align*}\tan (\frac{7\pi}{6} - \frac{\pi}{6}) & = \frac{\tan \frac{7 \pi}{6} - \tan \frac{\pi}{6}}{1 + \tan \frac{7 \pi}{6} \tan \frac{\pi}{6}} \\ & = \frac{\frac{\sqrt{2}}{6} - \frac{\sqrt{2}}{6}}{1 - \frac{\sqrt{2}}{6} \cdot \frac{\sqrt{2}}{6}} = \frac{0}{1-\frac{2}{36}} \\ & = \frac{0}{\frac{34}{36}} \\ & = 0 \end{align*}

Find the exact value of \begin{align*}\tan 165^\circ\end{align*}.

Use the difference formula for tangent, with \begin{align*}165^\circ = 225^\circ - 60^\circ\end{align*}

\begin{align*}\tan (225^\circ - 60^\circ) & = \frac{\tan 225^\circ - \tan 60^\circ}{1 + \tan 225^\circ \tan 60^\circ} \\ & = \frac{1 - \sqrt{3}}{1 - 1 \cdot \sqrt{3}} = 1 \\ \end{align*}

Examples

Example 1

Earlier, you were asked to find \begin{align*}\tan (120^\circ - 40^\circ)\end{align*}.

You can use the tangent difference formula:

\begin{align*} \tan(a-b) = \frac{\tan a - \tan b}{1 + \tan a \tan b}\end{align*}

to help solve this. Substituting in known quantities:

\begin{align*} \tan(120^\circ - 40^\circ) = \frac{\tan 120^\circ - \tan 40^\circ}{1 + (\tan 120^\circ)(\tan 40^\circ)} = \frac{-1.732 - .839}{1 + (-1.732)(.839)} = \frac{-2.571}{-.453148} = 5.674\end{align*}

Example 2

Find the exact value for \begin{align*}\tan 75^\circ\end{align*}

\begin{align*} \tan 75^\circ &= \tan (45^\circ + 30^\circ)\\ &= \frac{\tan 45^\circ + \tan 30^\circ}{1- \tan 45^\circ \tan 30^\circ} \\ &= \frac{1 + \frac{\sqrt{3}}{3}}{1 - 1 \cdot \frac{\sqrt{3}}{3}} = \frac{\frac{3+\sqrt{3}}{3}}{\frac{3-\sqrt{3}}{3}}\\ &= \frac{3+\sqrt{3}}{3-\sqrt{3}} \cdot \frac{3+\sqrt{3}}{3+\sqrt{3}}\\ &= \frac{9+6 \sqrt{3} + 3}{9-3} = \frac{12 + 6 \sqrt{3}}{6}\\ &= 2 + \sqrt{3}\end{align*}

Example 3

Simplify \begin{align*}\tan(\pi + \theta)\end{align*}

\begin{align*}\tan (\pi + \theta) = \frac{\tan \pi + \tan \theta}{1- \tan \pi \tan \theta} = \frac{\tan \theta}{1} = \tan \theta\end{align*}

Example 4

Find the exact value for \begin{align*}\tan 15^\circ\end{align*}

\begin{align*} \tan 15^\circ &= \tan (45^\circ - 30^\circ)\\ &= \frac{\tan 45^\circ - \tan 30^\circ}{1+ \tan 45^\circ \tan 30^\circ} \\ &= \frac{1 - \frac{\sqrt{3}}{3}}{1 + 1 \cdot \frac{\sqrt{3}}{3}} = \frac{\frac{3-\sqrt{3}}{3}}{\frac{3+\sqrt{3}}{3}}\\ &= \frac{3-\sqrt{3}}{3+\sqrt{3}} \cdot \frac{3-\sqrt{3}}{3-\sqrt{3}}\\ &= \frac{9+6 \sqrt{3} + 3}{9-3} = \frac{12 + 6 \sqrt{3}}{6}\\ &= 2 + \sqrt{3}\end{align*}

Review

Find the exact value for each tangent expression.

  1. \begin{align*}\tan\frac{5\pi}{12}\end{align*}
  2. \begin{align*}\tan\frac{11\pi}{12}\end{align*}
  3. \begin{align*}\tan-165^\circ\end{align*}
  4. \begin{align*}\tan255^\circ\end{align*}
  5. \begin{align*}\tan-15^\circ\end{align*}

Write each expression as the tangent of an angle.

  1. \begin{align*}\frac{\tan15^\circ+\tan42^\circ}{1-\tan15^\circ\tan42^\circ}\end{align*}
  2. \begin{align*}\frac{\tan65^\circ-\tan12^\circ}{1+\tan65^\circ\tan12^\circ}\end{align*}
  3. \begin{align*}\frac{\tan10^\circ+\tan50^\circ}{1-\tan10^\circ\tan50^\circ}\end{align*}
  4. \begin{align*}\frac{\tan2y+\tan4}{1-\tan2\tan4y}\end{align*}
  5. \begin{align*}\frac{\tan x-\tan3x}{1+\tan x\tan3x}\end{align*}
  6. \begin{align*}\frac{\tan2x-\tan y}{1+\tan2x\tan y}\end{align*}
  7. Prove that \begin{align*}\tan(x+\frac{\pi}{4})=\frac{1+\tan(x)}{1-\tan(x)}\end{align*}
  8. Prove that \begin{align*}\tan(x-\frac{\pi}{2})=-\cot(x)\end{align*}
  9. Prove that \begin{align*}\tan(\frac{\pi}{2}-x)=\cot(x)\end{align*}
  10. Prove that \begin{align*}\tan(x+y)\tan(x-y)=\frac{\tan^2(x)-\tan^2(y)}{1-\tan^2(x)\tan^2(y)}\end{align*}

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 3.8. 

Vocabulary

Tangent Difference Formula

Tangent Difference Formula

The tangent difference formula relates the tangent of a difference of two arguments to a set of tangent functions, each containing one argument.
Tangent Sum Formula

Tangent Sum Formula

The tangent sum formula relates the tangent of a sum of two arguments to a set of tangent functions, each containing one argument.

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Date Created:
Sep 26, 2012
Last Modified:
Mar 23, 2016
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