Precalculus Examples

$\cos (\arcsin (\frac{1}{5}))$

Step 1

Draw a triangle in the plane with vertices $(\sqrt{1^{2} - {(\frac{1}{5})}^{2}}, \frac{1}{5})$ , $(\sqrt{1^{2} - {(\frac{1}{5})}^{2}}, 0)$ , and the origin. Then $\arcsin (\frac{1}{5})$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(\sqrt{1^{2} - {(\frac{1}{5})}^{2}}, \frac{1}{5})$ . Therefore, $\cos (\arcsin (\frac{1}{5}))$ is $\sqrt{\frac{24}{25}}$ .

$\sqrt{\frac{24}{25}}$

Step 2

Rewrite $\sqrt{\frac{24}{25}}$ as $\frac{\sqrt{24}}{\sqrt{25}}$ .

$\frac{\sqrt{24}}{\sqrt{25}}$

Step 3

Simplify the numerator.

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Step 3.1

Rewrite $24$ as $2^{2} \cdot 6$ .

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Step 3.1.1

Factor $4$ out of $24$ .

$\frac{\sqrt{4 (6)}}{\sqrt{25}}$

Step 3.1.2

Rewrite $4$ as $2^{2}$ .

$\frac{\sqrt{2^{2} \cdot 6}}{\sqrt{25}}$

Step 3.2

Pull terms out from under the radical.

$\frac{2 \sqrt{6}}{\sqrt{25}}$

Step 4

Simplify the denominator.

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Step 4.1

Rewrite $25$ as $5^{2}$ .

$\frac{2 \sqrt{6}}{\sqrt{5^{2}}}$

Step 4.2

Pull terms out from under the radical, assuming positive real numbers.

$\frac{2 \sqrt{6}}{5}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$\frac{2 \sqrt{6}}{5}$

Decimal Form:

$0.97979589 \dots$