Expand the Trigonometric Expression sin(arccos(u)+arctan(v))

$\sin (\arccos (u) + \arctan (v))$

Apply the sum of angles identity.

$\sin (\arccos (u)) \cos (\arctan (v)) + \cos (\arccos (u)) \sin (\arctan (v))$

Simplify terms.

Simplify each term.

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

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

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

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

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 $b = u$ .

$\sqrt{(1 + u) (1 - u)} \cos (\arctan (v)) + \cos (\arccos (u)) \sin (\arctan (v))$

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

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

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

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

Combine and simplify the denominator.

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

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

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

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

Combine using the product rule for radicals.

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

The functions cosine and arccosine are inverses.

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

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

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

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

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

Combine and simplify the denominator.

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

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

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

Combine the numerators over the common denominator.

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

Expand the Trigonometric Expression sin(arccos(u)+arctan(v))

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