Trigonometry Examples

$\sin (\arcsin (x) + \arccos (y))$

Step 1

Apply the sum of angles identity.

$\sin (\arcsin (x)) \cos (\arccos (y)) + \cos (\arcsin (x)) \sin (\arccos (y))$

Step 2

Simplify each term.

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Step 2.1

The functions sine and arcsine are inverses.

$x \cos (\arccos (y)) + \cos (\arcsin (x)) \sin (\arccos (y))$

Step 2.2

The functions cosine and arccosine are inverses.

$x y + \cos (\arcsin (x)) \sin (\arccos (y))$

Step 2.3

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

$x y + \sqrt{1 - x^{2}} \sin (\arccos (y))$

Step 2.4

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

$x y + \sqrt{1^{2} - x^{2}} \sin (\arccos (y))$

Step 2.5

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 $b = x$ .

$x y + \sqrt{(1 + x) (1 - x)} \sin (\arccos (y))$

Step 2.6

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

$x y + \sqrt{(1 + x) (1 - x)} \sqrt{1 - y^{2}}$

Step 2.7

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

$x y + \sqrt{(1 + x) (1 - x)} \sqrt{1^{2} - y^{2}}$

Step 2.8

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 $b = y$ .

$x y + \sqrt{(1 + x) (1 - x)} \sqrt{(1 + y) (1 - y)}$

Step 2.9

Combine using the product rule for radicals.

$x y + \sqrt{(1 + x) (1 - x) (1 + y) (1 - y)}$