Precalculus Examples

$\sin (\sin^{-1} (u) - \tan^{-1} (v))$

Step 1

Apply the difference of angles identity.

$\sin (\sin^{-1} (u)) \cos (\tan^{-1} (v)) - \cos (\sin^{-1} (u)) \sin (\tan^{-1} (v))$

Step 2

Simplify terms.

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Step 2.1

Simplify each term.

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Step 2.1.1

The functions sine and arcsine are inverses.

$u \cos (\tan^{-1} (v)) - \cos (\sin^{-1} (u)) \sin (\tan^{-1} (v))$

Step 2.1.2

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

$u \frac{1}{\sqrt{1 + v^{2}}} - \cos (\sin^{-1} (u)) \sin (\tan^{-1} (v))$

Step 2.1.3

Multiply $\frac{1}{\sqrt{1 + v^{2}}}$ by $\frac{\sqrt{1 + v^{2}}}{\sqrt{1 + v^{2}}}$ .

$u (\frac{1}{\sqrt{1 + v^{2}}} \cdot \frac{\sqrt{1 + v^{2}}}{\sqrt{1 + v^{2}}}) - \cos (\sin^{-1} (u)) \sin (\tan^{-1} (v))$

Step 2.1.4

Combine and simplify the denominator.

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Step 2.1.4.1

Multiply $\frac{1}{\sqrt{1 + v^{2}}}$ by $\frac{\sqrt{1 + v^{2}}}{\sqrt{1 + v^{2}}}$ .

$u \frac{\sqrt{1 + v^{2}}}{\sqrt{1 + v^{2}} \sqrt{1 + v^{2}}} - \cos (\sin^{-1} (u)) \sin (\tan^{-1} (v))$

Step 2.1.4.2

Raise $\sqrt{1 + v^{2}}$ to the power of $1$ .

$u \frac{\sqrt{1 + v^{2}}}{{\sqrt{1 + v^{2}}}^{1} \sqrt{1 + v^{2}}} - \cos (\sin^{-1} (u)) \sin (\tan^{-1} (v))$

Step 2.1.4.3

Raise $\sqrt{1 + v^{2}}$ to the power of $1$ .

$u \frac{\sqrt{1 + v^{2}}}{{\sqrt{1 + v^{2}}}^{1} {\sqrt{1 + v^{2}}}^{1}} - \cos (\sin^{-1} (u)) \sin (\tan^{-1} (v))$

Step 2.1.4.4

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

$u \frac{\sqrt{1 + v^{2}}}{{\sqrt{1 + v^{2}}}^{1 + 1}} - \cos (\sin^{-1} (u)) \sin (\tan^{-1} (v))$

Step 2.1.4.5

Add $1$ and $1$ .

$u \frac{\sqrt{1 + v^{2}}}{{\sqrt{1 + v^{2}}}^{2}} - \cos (\sin^{-1} (u)) \sin (\tan^{-1} (v))$

Step 2.1.4.6

Rewrite ${\sqrt{1 + v^{2}}}^{2}$ as $1 + v^{2}$ .

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Step 2.1.4.6.1

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

$u \frac{\sqrt{1 + v^{2}}}{{({(1 + v^{2})}^{\frac{1}{2}})}^{2}} - \cos (\sin^{-1} (u)) \sin (\tan^{-1} (v))$

Step 2.1.4.6.2

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

$u \frac{\sqrt{1 + v^{2}}}{{(1 + v^{2})}^{\frac{1}{2} \cdot 2}} - \cos (\sin^{-1} (u)) \sin (\tan^{-1} (v))$

Step 2.1.4.6.3

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

$u \frac{\sqrt{1 + v^{2}}}{{(1 + v^{2})}^{\frac{2}{2}}} - \cos (\sin^{-1} (u)) \sin (\tan^{-1} (v))$

Step 2.1.4.6.4

Cancel the common factor of $2$ .

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Step 2.1.4.6.4.1

Cancel the common factor.

$u \frac{\sqrt{1 + v^{2}}}{{(1 + v^{2})}^{\frac{2}{2}}} - \cos (\sin^{-1} (u)) \sin (\tan^{-1} (v))$

Step 2.1.4.6.4.2

Rewrite the expression.

$u \frac{\sqrt{1 + v^{2}}}{{(1 + v^{2})}^{1}} - \cos (\sin^{-1} (u)) \sin (\tan^{-1} (v))$

Step 2.1.4.6.5

Simplify.

$u \frac{\sqrt{1 + v^{2}}}{1 + v^{2}} - \cos (\sin^{-1} (u)) \sin (\tan^{-1} (v))$

Step 2.1.5

Combine $u$ and $\frac{\sqrt{1 + v^{2}}}{1 + v^{2}}$ .

$\frac{u \sqrt{1 + v^{2}}}{1 + v^{2}} - \cos (\sin^{-1} (u)) \sin (\tan^{-1} (v))$

Step 2.1.6

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

$\frac{u \sqrt{1 + v^{2}}}{1 + v^{2}} - \sqrt{1 - u^{2}} \sin (\tan^{-1} (v))$

Step 2.1.7

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

$\frac{u \sqrt{1 + v^{2}}}{1 + v^{2}} - \sqrt{1^{2} - u^{2}} \sin (\tan^{-1} (v))$

Step 2.1.8

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 $b = u$ .

$\frac{u \sqrt{1 + v^{2}}}{1 + v^{2}} - \sqrt{(1 + u) (1 - u)} \sin (\tan^{-1} (v))$

Step 2.1.9

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

$\frac{u \sqrt{1 + v^{2}}}{1 + v^{2}} - \sqrt{(1 + u) (1 - u)} \frac{v}{\sqrt{1 + v^{2}}}$

Step 2.1.10

Multiply $\frac{v}{\sqrt{1 + v^{2}}}$ by $\frac{\sqrt{1 + v^{2}}}{\sqrt{1 + v^{2}}}$ .

$\frac{u \sqrt{1 + v^{2}}}{1 + v^{2}} - \sqrt{(1 + u) (1 - u)} (\frac{v}{\sqrt{1 + v^{2}}} \cdot \frac{\sqrt{1 + v^{2}}}{\sqrt{1 + v^{2}}})$

Step 2.1.11

Combine and simplify the denominator.

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Step 2.1.11.1

Multiply $\frac{v}{\sqrt{1 + v^{2}}}$ by $\frac{\sqrt{1 + v^{2}}}{\sqrt{1 + v^{2}}}$ .

$\frac{u \sqrt{1 + v^{2}}}{1 + v^{2}} - \sqrt{(1 + u) (1 - u)} \frac{v \sqrt{1 + v^{2}}}{\sqrt{1 + v^{2}} \sqrt{1 + v^{2}}}$

Step 2.1.11.2

Raise $\sqrt{1 + v^{2}}$ to the power of $1$ .

$\frac{u \sqrt{1 + v^{2}}}{1 + v^{2}} - \sqrt{(1 + u) (1 - u)} \frac{v \sqrt{1 + v^{2}}}{{\sqrt{1 + v^{2}}}^{1} \sqrt{1 + v^{2}}}$

Step 2.1.11.3

Raise $\sqrt{1 + v^{2}}$ to the power of $1$ .

$\frac{u \sqrt{1 + v^{2}}}{1 + v^{2}} - \sqrt{(1 + u) (1 - u)} \frac{v \sqrt{1 + v^{2}}}{{\sqrt{1 + v^{2}}}^{1} {\sqrt{1 + v^{2}}}^{1}}$

Step 2.1.11.4

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

$\frac{u \sqrt{1 + v^{2}}}{1 + v^{2}} - \sqrt{(1 + u) (1 - u)} \frac{v \sqrt{1 + v^{2}}}{{\sqrt{1 + v^{2}}}^{1 + 1}}$

Step 2.1.11.5

Add $1$ and $1$ .

$\frac{u \sqrt{1 + v^{2}}}{1 + v^{2}} - \sqrt{(1 + u) (1 - u)} \frac{v \sqrt{1 + v^{2}}}{{\sqrt{1 + v^{2}}}^{2}}$

Step 2.1.11.6

Rewrite ${\sqrt{1 + v^{2}}}^{2}$ as $1 + v^{2}$ .

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Step 2.1.11.6.1

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

$\frac{u \sqrt{1 + v^{2}}}{1 + v^{2}} - \sqrt{(1 + u) (1 - u)} \frac{v \sqrt{1 + v^{2}}}{{({(1 + v^{2})}^{\frac{1}{2}})}^{2}}$

Step 2.1.11.6.2

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

$\frac{u \sqrt{1 + v^{2}}}{1 + v^{2}} - \sqrt{(1 + u) (1 - u)} \frac{v \sqrt{1 + v^{2}}}{{(1 + v^{2})}^{\frac{1}{2} \cdot 2}}$

Step 2.1.11.6.3

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

$\frac{u \sqrt{1 + v^{2}}}{1 + v^{2}} - \sqrt{(1 + u) (1 - u)} \frac{v \sqrt{1 + v^{2}}}{{(1 + v^{2})}^{\frac{2}{2}}}$

Step 2.1.11.6.4

Cancel the common factor of $2$ .

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Step 2.1.11.6.4.1

Cancel the common factor.

$\frac{u \sqrt{1 + v^{2}}}{1 + v^{2}} - \sqrt{(1 + u) (1 - u)} \frac{v \sqrt{1 + v^{2}}}{{(1 + v^{2})}^{\frac{2}{2}}}$

Step 2.1.11.6.4.2

Rewrite the expression.

$\frac{u \sqrt{1 + v^{2}}}{1 + v^{2}} - \sqrt{(1 + u) (1 - u)} \frac{v \sqrt{1 + v^{2}}}{{(1 + v^{2})}^{1}}$

Step 2.1.11.6.5

Simplify.

$\frac{u \sqrt{1 + v^{2}}}{1 + v^{2}} - \sqrt{(1 + u) (1 - u)} \frac{v \sqrt{1 + v^{2}}}{1 + v^{2}}$

Step 2.1.12

Multiply $- \sqrt{(1 + u) (1 - u)} \frac{v \sqrt{1 + v^{2}}}{1 + v^{2}}$ .

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Step 2.1.12.1

Combine $\frac{v \sqrt{1 + v^{2}}}{1 + v^{2}}$ and $\sqrt{(1 + u) (1 - u)}$ .

$\frac{u \sqrt{1 + v^{2}}}{1 + v^{2}} - \frac{v \sqrt{1 + v^{2}} \sqrt{(1 + u) (1 - u)}}{1 + v^{2}}$

Step 2.1.12.2

Combine using the product rule for radicals.

$\frac{u \sqrt{1 + v^{2}}}{1 + v^{2}} - \frac{v \sqrt{(1 + u) (1 - u) (1 + v^{2})}}{1 + v^{2}}$

Step 2.2

Combine the numerators over the common denominator.

$\frac{u \sqrt{1 + v^{2}} - v \sqrt{(1 + u) (1 - u) (1 + v^{2})}}{1 + v^{2}}$