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Expert-verified Found in: Page 544 ### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508 # Determining Identities GraphicallyGraph $f$ and $g$ in the same viewing rectangle. Do the graphs suggest that the equation $\mathbit{f}\mathbf{\left(}\mathbit{x}\mathbf{\right)}\mathbf{=}\mathbit{g}\mathbf{\left(}\mathbit{x}\mathbf{\right)}$ is an identity? Prove your answer$\mathbit{f}\mathbf{\left(}\mathbit{x}\mathbf{\right)}\mathbf{=}\mathbit{c}\mathbit{o}{\mathbit{s}}^{\mathbf{2}}\mathbit{x}\mathbf{-}\mathbit{s}\mathbit{i}{\mathbit{n}}^{\mathbf{2}}\mathbit{x}\mathbf{,}\mathbf{\text{ }}\mathbit{g}\mathbf{\left(}\mathbit{x}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{2}\mathbit{s}\mathbit{i}{\mathbit{n}}^{\mathbf{2}}\mathbit{x}$

The expression $f\left(x\right)=g\left(x\right)$ is an identity.

See the step by step solution

## Step 1. Given information

Two functions as-

$\begin{array}{l}f\left(x\right)={\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x\\ \text{ }g\left(x\right)=1-2{\mathrm{sin}}^{2}x\end{array}$

## Step 2. Concept used

Create distinct graphs for both functions on the same graph. Both functions are equivalent if both graphs coincide with one other. As a result, combining them into a single equation yields an identity.

## Step 3. Calculation

Now, draw graphs of both functions: Since both graphs coincide each other. Therefore $f\left(x\right)=g\left(x\right)$ is an identity ### Want to see more solutions like these? 