Trigonometry Examples

$\sin (\arccos (- \frac{1}{3}))$

Step 1

Draw a triangle in the plane with vertices $(- \frac{1}{3}, \sqrt{1^{2} - {(- \frac{1}{3})}^{2}})$ , $(- \frac{1}{3}, 0)$ , and the origin. Then $\arccos (- \frac{1}{3})$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(- \frac{1}{3}, \sqrt{1^{2} - {(- \frac{1}{3})}^{2}})$ . Therefore, $\sin (\arccos (- \frac{1}{3}))$ is $\sqrt{\frac{8}{9}}$ .

$\sqrt{\frac{8}{9}}$

Step 2

Rewrite $\sqrt{\frac{8}{9}}$ as $\frac{\sqrt{8}}{\sqrt{9}}$ .

$\frac{\sqrt{8}}{\sqrt{9}}$

Step 3

Simplify the numerator.

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Step 3.1

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 3.1.1

Factor $4$ out of $8$ .

$\frac{\sqrt{4 (2)}}{\sqrt{9}}$

Step 3.1.2

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

$\frac{\sqrt{2^{2} \cdot 2}}{\sqrt{9}}$

Step 3.2

Pull terms out from under the radical.

$\frac{2 \sqrt{2}}{\sqrt{9}}$

Step 4

Simplify the denominator.

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Step 4.1

Rewrite $9$ as $3^{2}$ .

$\frac{2 \sqrt{2}}{\sqrt{3^{2}}}$

Step 4.2

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

$\frac{2 \sqrt{2}}{3}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$\frac{2 \sqrt{2}}{3}$

Decimal Form:

$0.94280904 \dots$