Trigonometry Examples

\tan (\cos^{-1} (2 x))

$\tan (\cos^{-1} (2 x))$

Step 1

Draw a triangle in the plane with vertices

(2 x, \sqrt{1^{2} - {(2 x)}^{2}})

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

(2 x, 0)

$(2 x, 0)$ , and the origin. Then

\cos^{-1} (2 x)

$\cos^{-1} (2 x)$ is the angle between the positive x-axis and the ray beginning at the origin and passing through

(2 x, \sqrt{1^{2} - {(2 x)}^{2}})

$(2 x, \sqrt{1^{2} - {(2 x)}^{2}})$ . Therefore,

\tan (\cos^{-1} (2 x))

$\tan (\cos^{-1} (2 x))$ is

\frac{\sqrt{1 - {(2 x)}^{2}}}{2 x}

$\frac{\sqrt{1 - {(2 x)}^{2}}}{2 x}$ .

\frac{\sqrt{1 - {(2 x)}^{2}}}{2 x}

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

Step 2

Simplify the numerator.

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Step 2.1

Rewrite

1

$1$ as

1^{2}

$1^{2}$ .

\frac{\sqrt{1^{2} - {(2 x)}^{2}}}{2 x}

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

Step 2.2

Since both terms are perfect squares, factor using the difference of squares formula,

a^{2} - b^{2} = (a + b) (a - b)

$a^{2} - b^{2} = (a + b) (a - b)$ where

a = 1

$a = 1$ and

b = 2 x

$b = 2 x$ .

\frac{\sqrt{(1 + 2 x) (1 - (2 x))}}{2 x}

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

Step 2.3

Multiply

2

$2$ by

- 1

$- 1$ .

\frac{\sqrt{(1 + 2 x) (1 - 2 x)}}{2 x}

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

\frac{\sqrt{(1 + 2 x) (1 - 2 x)}}{2 x}

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

\tan (\cos^{- 1} 2 x)

$\tan (\cos^{- 1} 2 x)$

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