Find the Exact Value cos(arcsin(6x))

$\cos (\arcsin (6 x))$

Step 1

Write the expression using exponents.

Draw a triangle in the plane with vertices $(\sqrt{1^{2} - {(6 x)}^{2}}, 6 x)$ , $(\sqrt{1^{2} - {(6 x)}^{2}}, 0)$ , and the origin. Then $\arcsin (6 x)$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(\sqrt{1^{2} - {(6 x)}^{2}}, 6 x)$ . Therefore, $\cos (\arcsin (6 x))$ is $\sqrt{1 - {(6 x)}^{2}}$ .

$\sqrt{1 - {(6 x)}^{2}}$

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

$\sqrt{1^{2} - {(6 x)}^{2}}$

Step 2

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 $b = 6 x$ .

$\sqrt{(1 + 6 x) (1 - (6 x))}$

Step 3

Multiply $6$ by $- 1$ .

$\sqrt{(1 + 6 x) (1 - 6 x)}$

Find the Exact Value cos(arcsin(6x))

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