Find the Exact Value tan(arccos(4x))

$\tan (\arccos (4 x))$

Step 1

Draw a triangle in the plane with vertices $(4 x, \sqrt{1^{2} - {(4 x)}^{2}})$ , $(4 x, 0)$ , and the origin. Then $\arccos (4 x)$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(4 x, \sqrt{1^{2} - {(4 x)}^{2}})$ . Therefore, $\tan (\arccos (4 x))$ is $\frac{\sqrt{1 - {(4 x)}^{2}}}{4 x}$ .

$\frac{\sqrt{1 - {(4 x)}^{2}}}{4 x}$

Step 2

Simplify the numerator.

Rewrite $1$ as $1^{2}$ .

$\frac{\sqrt{1^{2} - {(4 x)}^{2}}}{4 x}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = 4 x$ .

$\frac{\sqrt{(1 + 4 x) (1 - (4 x))}}{4 x}$

Multiply $4$ by $- 1$ .

$\frac{\sqrt{(1 + 4 x) (1 - 4 x)}}{4 x}$

Find the Exact Value tan(arccos(4x))

Related Questions

Support LCMGCF.com by disabling your adblocker.