Find the Exact Value cos(arcsin(2x))

$\cos (\arcsin (2 x))$

Step 1

Write the expression using exponents.

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

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

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

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

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 $b = 2 x$ .

$\sqrt{(1 + 2 x) (1 - (2 x))}$

Step 3

Multiply $2$ by $- 1$ .

$\sqrt{(1 + 2 x) (1 - 2 x)}$

Find the Exact Value cos(arcsin(2x))

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