Precalculus Examples

$\cos (\arctan (\frac{3}{2}) - \arcsin (\frac{4}{5}))$

Use the difference formula for cosine to simplify the expression. The formula states that $\cos (A - B) = \cos (A) \cos (B) + \sin (A) \sin (B)$ .

$\cos (\arctan (\frac{3}{2})) \cdot \cos (\arcsin (\frac{4}{5})) + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

Remove parentheses.

$\cos (\arctan (\frac{3}{2})) \cdot \cos (\arcsin (\frac{4}{5})) + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

Simplify each term.

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Draw a triangle in the plane with vertices $(1, \frac{3}{2})$ , $(1, 0)$ , and the origin. Then $\arctan (\frac{3}{2})$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(1, \frac{3}{2})$ . Therefore, $\cos (\arctan (\frac{3}{2}))$ is $\frac{2 \sqrt{13}}{13}$ .

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Multiply $\frac{2}{\sqrt{13}}$ by $\frac{\sqrt{13}}{\sqrt{13}}$ .

$\frac{2 \sqrt{13}}{\sqrt{13} \sqrt{13}} \cdot \cos (\arcsin (\frac{4}{5})) + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

Raise $\sqrt{13}$ to the power of $1$ .

$\frac{2 \sqrt{13}}{{\sqrt{13}}^{1} \sqrt{13}} \cdot \cos (\arcsin (\frac{4}{5})) + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

Raise $\sqrt{13}$ to the power of $1$ .

$\frac{2 \sqrt{13}}{{\sqrt{13}}^{1} {\sqrt{13}}^{1}} \cdot \cos (\arcsin (\frac{4}{5})) + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{2 \sqrt{13}}{{\sqrt{13}}^{1 + 1}} \cdot \cos (\arcsin (\frac{4}{5})) + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

Add $1$ and $1$ .

$\frac{2 \sqrt{13}}{{\sqrt{13}}^{2}} \cdot \cos (\arcsin (\frac{4}{5})) + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

Rewrite ${\sqrt{13}}^{2}$ as $13$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{13}$ as $13^{\frac{1}{2}}$ .

$\frac{2 \sqrt{13}}{{(13^{\frac{1}{2}})}^{2}} \cdot \cos (\arcsin (\frac{4}{5})) + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{2 \sqrt{13}}{13^{\frac{1}{2} \cdot 2}} \cdot \cos (\arcsin (\frac{4}{5})) + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

Combine $\frac{1}{2}$ and $2$ .

$\frac{2 \sqrt{13}}{13^{\frac{2}{2}}} \cdot \cos (\arcsin (\frac{4}{5})) + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 \sqrt{13}}{13^{\frac{2}{2}}} \cdot \cos (\arcsin (\frac{4}{5})) + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

Rewrite the expression.

$\frac{2 \sqrt{13}}{13^{1}} \cdot \cos (\arcsin (\frac{4}{5})) + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

Evaluate the exponent.

$\frac{2 \sqrt{13}}{13} \cdot \cos (\arcsin (\frac{4}{5})) + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

Draw a triangle in the plane with vertices $(\sqrt{1^{2} - {(\frac{4}{5})}^{2}}, \frac{4}{5})$ , $(\sqrt{1^{2} - {(\frac{4}{5})}^{2}}, 0)$ , and the origin. Then $\arcsin (\frac{4}{5})$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(\sqrt{1^{2} - {(\frac{4}{5})}^{2}}, \frac{4}{5})$ . Therefore, $\cos (\arcsin (\frac{4}{5}))$ is $\sqrt{\frac{9}{25}}$ .

$\frac{2 \sqrt{13}}{13} \cdot \sqrt{\frac{9}{25}} + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

Rewrite $\sqrt{\frac{9}{25}}$ as $\frac{\sqrt{9}}{\sqrt{25}}$ .

$\frac{2 \sqrt{13}}{13} \cdot \frac{\sqrt{9}}{\sqrt{25}} + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

Simplify the numerator.

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

$\frac{2 \sqrt{13}}{13} \cdot \frac{\sqrt{3^{2}}}{\sqrt{25}} + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

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

$\frac{2 \sqrt{13}}{13} \cdot \frac{3}{\sqrt{25}} + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

Simplify the denominator.

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

$\frac{2 \sqrt{13}}{13} \cdot \frac{3}{\sqrt{5^{2}}} + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

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

$\frac{2 \sqrt{13}}{13} \cdot \frac{3}{5} + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

Multiply $\frac{2 \sqrt{13}}{13} \cdot \frac{3}{5}$ .

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Multiply $\frac{2 \sqrt{13}}{13}$ by $\frac{3}{5}$ .

$\frac{2 \sqrt{13} \cdot 3}{13 \cdot 5} + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

Multiply $3$ by $2$ .

$\frac{6 \sqrt{13}}{13 \cdot 5} + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

Multiply $13$ by $5$ .

$\frac{6 \sqrt{13}}{65} + \sin (\arctan (\frac{3}{2})) \cdot \sin (\arcsin (\frac{4}{5}))$

Draw a triangle in the plane with vertices $(1, \frac{3}{2})$ , $(1, 0)$ , and the origin. Then $\arctan (\frac{3}{2})$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(1, \frac{3}{2})$ . Therefore, $\sin (\arctan (\frac{3}{2}))$ is $\frac{3 \sqrt{13}}{13}$ .

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Multiply $\frac{3}{\sqrt{13}}$ by $\frac{\sqrt{13}}{\sqrt{13}}$ .

$\frac{6 \sqrt{13}}{65} + \frac{3 \sqrt{13}}{\sqrt{13} \sqrt{13}} \cdot \sin (\arcsin (\frac{4}{5}))$

Raise $\sqrt{13}$ to the power of $1$ .

$\frac{6 \sqrt{13}}{65} + \frac{3 \sqrt{13}}{{\sqrt{13}}^{1} \sqrt{13}} \cdot \sin (\arcsin (\frac{4}{5}))$

Raise $\sqrt{13}$ to the power of $1$ .

$\frac{6 \sqrt{13}}{65} + \frac{3 \sqrt{13}}{{\sqrt{13}}^{1} {\sqrt{13}}^{1}} \cdot \sin (\arcsin (\frac{4}{5}))$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{6 \sqrt{13}}{65} + \frac{3 \sqrt{13}}{{\sqrt{13}}^{1 + 1}} \cdot \sin (\arcsin (\frac{4}{5}))$

Add $1$ and $1$ .

$\frac{6 \sqrt{13}}{65} + \frac{3 \sqrt{13}}{{\sqrt{13}}^{2}} \cdot \sin (\arcsin (\frac{4}{5}))$

Rewrite ${\sqrt{13}}^{2}$ as $13$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{13}$ as $13^{\frac{1}{2}}$ .

$\frac{6 \sqrt{13}}{65} + \frac{3 \sqrt{13}}{{(13^{\frac{1}{2}})}^{2}} \cdot \sin (\arcsin (\frac{4}{5}))$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{6 \sqrt{13}}{65} + \frac{3 \sqrt{13}}{13^{\frac{1}{2} \cdot 2}} \cdot \sin (\arcsin (\frac{4}{5}))$

Combine $\frac{1}{2}$ and $2$ .

$\frac{6 \sqrt{13}}{65} + \frac{3 \sqrt{13}}{13^{\frac{2}{2}}} \cdot \sin (\arcsin (\frac{4}{5}))$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{6 \sqrt{13}}{65} + \frac{3 \sqrt{13}}{13^{\frac{2}{2}}} \cdot \sin (\arcsin (\frac{4}{5}))$

Rewrite the expression.

$\frac{6 \sqrt{13}}{65} + \frac{3 \sqrt{13}}{13^{1}} \cdot \sin (\arcsin (\frac{4}{5}))$

Evaluate the exponent.

$\frac{6 \sqrt{13}}{65} + \frac{3 \sqrt{13}}{13} \cdot \sin (\arcsin (\frac{4}{5}))$

The functions sine and arcsine are inverses.

$\frac{6 \sqrt{13}}{65} + \frac{3 \sqrt{13}}{13} \cdot \frac{4}{5}$

Multiply $\frac{3 \sqrt{13}}{13} \cdot \frac{4}{5}$ .

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Multiply $\frac{3 \sqrt{13}}{13}$ by $\frac{4}{5}$ .

$\frac{6 \sqrt{13}}{65} + \frac{3 \sqrt{13} \cdot 4}{13 \cdot 5}$

Multiply $4$ by $3$ .

$\frac{6 \sqrt{13}}{65} + \frac{12 \sqrt{13}}{13 \cdot 5}$

Multiply $13$ by $5$ .

$\frac{6 \sqrt{13}}{65} + \frac{12 \sqrt{13}}{65}$

Simplify terms.

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Combine the numerators over the common denominator.

$\frac{6 \sqrt{13} + 12 \sqrt{13}}{65}$

Add $6 \sqrt{13}$ and $12 \sqrt{13}$ .

$\frac{18 \sqrt{13}}{65}$

The result can be shown in multiple forms.

Exact Form:

$\frac{18 \sqrt{13}}{65}$

Decimal Form:

$0.99846035 \dots$