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Q. 102

Expert-verified
Found in: Page 498

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# If $x=2\mathrm{tan}\theta$, express $\mathrm{cos}\left(2\theta \right)$ as a function of x.

On expressing $\mathrm{cos}\left(2\theta \right)$ as a function of x is $\mathrm{cos}\left(2\theta \right)=\frac{4-{x}^{2}}{4+{x}^{2}}$.

See the step by step solution

## Step 1. Given information.

Consider the given question,

$x=2\mathrm{tan}\theta$

From the double-angle formula,

$\mathrm{cos}\left(2\theta \right)=2{\mathrm{cos}}^{2}\theta -1$

From inverse trigonometry,

$\theta ={\mathrm{tan}}^{-1}\frac{x}{2}\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left(\frac{x}{\sqrt{4+{x}^{2}}}\right)\phantom{\rule{0ex}{0ex}}=co{s}^{-1}\left(\frac{2}{\sqrt{4+{x}^{2}}}\right)$

## Step 2. Use inverse trigonometry and solve the equation.

From inverse trigonometry,

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

Then,

$\theta ={\mathrm{tan}}^{-1}\frac{x}{2}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\left(2\theta \right)=2{\mathrm{cos}}^{2}\left({\mathrm{tan}}^{-1}\frac{x}{2}\right)-1\phantom{\rule{0ex}{0ex}}\mathrm{cos}\left(2\theta \right)=2{\mathrm{cos}}^{2}\left({\mathrm{cos}}^{-1}\left(\frac{2}{\sqrt{4+{x}^{2}}}\right)\right)-1\phantom{\rule{0ex}{0ex}}\mathrm{cos}\left(2\theta \right)=2{\left(\frac{2}{\sqrt{4+{x}^{2}}}\right)}^{2}-1\phantom{\rule{0ex}{0ex}}\mathrm{cos}\left(2\theta \right)=\frac{4-{x}^{2}}{4+{x}^{2}}$