Trigonometry Examples

$\arctan (x) = \arccos (\frac{5}{13})$

Step 1

Take the inverse arctangent of both sides of the equation to extract $x$ from inside the arctangent.

$x = \tan (\arccos (\frac{5}{13}))$

Step 2

Simplify the right side.

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Step 2.1

Draw a triangle in the plane with vertices $(\frac{5}{13}, \sqrt{1^{2} - {(\frac{5}{13})}^{2}})$ , $(\frac{5}{13}, 0)$ , and the origin. Then $\arccos (\frac{5}{13})$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(\frac{5}{13}, \sqrt{1^{2} - {(\frac{5}{13})}^{2}})$ . Therefore, $\tan (\arccos (\frac{5}{13}))$ is $\frac{12}{5}$ .

$x = \frac{12}{5}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = \frac{12}{5}$

Decimal Form:

$x = 2.4$

Mixed Number Form:

$x = 2 \frac{2}{5}$