The inverse of the function sin(x) + cos(x) is...?

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Nov 30, 2016

Explanation:

Here is the graph:

This looks like a cosine function that has been multiplied by an amplitude and phase shift to the right:

$f \left(x\right) = A \cos \left(x - \phi\right) = \sin \left(x\right) + \cos \left(x\right)$

My graphing tool allows me to obtain values of points and I can tell you that $A = \sqrt{2}$

$\cos \left(x - \phi\right) = \frac{1}{\sqrt{2}} \sin \left(x\right) + \frac{1}{\sqrt{2}} \cos \left(x\right)$

This fits the trigonometric identity:

$\cos \left(x - \phi\right) = \sin \left(x\right) \sin \left(\phi\right) + \cos \left(x\right) \cos \left(\phi\right)$

where $\cos \left(\phi\right) = \sin \left(\phi\right) = \frac{1}{\sqrt{2}}$

This happens at $\phi = \frac{\pi}{4}$

The graphing tool confirms that the x coordinates are shift by $\frac{\pi}{4}$.

$f \left(x\right) = \sqrt{2} \cos \left(x - \frac{\pi}{4}\right)$

Substitute ${f}^{-} 1 \left(x\right)$ for every x:

$f \left({f}^{-} 1 \left(x\right)\right) = \sqrt{2} \cos \left({f}^{-} 1 \left(x\right) - \frac{\pi}{4}\right)$

The left side becomes x by definition:

$x = \sqrt{2} \cos \left({f}^{-} 1 \left(x\right) - \frac{\pi}{4}\right)$

Make the $\sqrt{2}$ on the right disappear by multiplying both sides by $\frac{\sqrt{2}}{2}$:

$\frac{\sqrt{2}}{2} x = \cos \left({f}^{-} 1 \left(x\right) - \frac{\pi}{4}\right)$

Use the inverse cosine on both sides:

${\cos}^{-} 1 \left(\frac{\sqrt{2}}{2} x\right) = {f}^{-} 1 \left(x\right) - \frac{\pi}{4}$

Solve for ${f}^{-} 1 \left(x\right)$:

${f}^{-} 1 \left(x\right) = {\cos}^{-} 1 \left(\frac{\sqrt{2}}{2} x\right) + \frac{\pi}{4}$

To confirm that this is truly an inverse, verify that $f \left({f}^{-} 1 \left(x\right)\right) = {f}^{-} 1 \left(f \left(x\right)\right) = x$

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