Trigonometry Examples

$\sin (2 \arccos (x))$

Step 1

Select a few points to graph.

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Step 1.1

Find the point at $x = - 1$ .

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Step 1.1.1

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = \sin (2 \arccos (- 1))$

Step 1.1.2

Simplify the result.

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Step 1.1.2.1

The exact value of $\arccos (- 1)$ is $π$ .

$f (- 1) = \sin (2 π)$

Step 1.1.2.2

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$f (- 1) = \sin (0)$

Step 1.1.2.3

The exact value of $\sin (0)$ is $0$ .

$f (- 1) = 0$

Step 1.1.2.4

The final answer is $0$ .

$0$

Step 1.2

Find the point at $x = - \frac{1}{2}$ .

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Step 1.2.1

Replace the variable $x$ with $- \frac{1}{2}$ in the expression.

$f (- \frac{1}{2}) = \sin (2 \arccos (- \frac{1}{2}))$

Step 1.2.2

Simplify the result.

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Step 1.2.2.1

The exact value of $\arccos (- \frac{1}{2})$ is $\frac{2 π}{3}$ .

$f (- \frac{1}{2}) = \sin (2 (\frac{2 π}{3}))$

Step 1.2.2.2

Multiply $2 \frac{2 π}{3}$ .

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Step 1.2.2.2.1

Combine $2$ and $\frac{2 π}{3}$ .

$f (- \frac{1}{2}) = \sin (\frac{2 (2 π)}{3})$

Step 1.2.2.2.2

Multiply $2$ by $2$ .

$f (- \frac{1}{2}) = \sin (\frac{4 π}{3})$

Step 1.2.2.3

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the third quadrant.

$f (- \frac{1}{2}) = - \sin (\frac{π}{3})$

Step 1.2.2.4

The exact value of $\sin (\frac{π}{3})$ is $\frac{\sqrt{3}}{2}$ .

$f (- \frac{1}{2}) = - \frac{\sqrt{3}}{2}$

Step 1.2.2.5

The final answer is $- \frac{\sqrt{3}}{2}$ .

$- \frac{\sqrt{3}}{2}$

Step 1.3

Find the point at $x = 0$ .

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Step 1.3.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = \sin (2 \arccos (0))$

Step 1.3.2

Simplify the result.

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Step 1.3.2.1

The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$f (0) = \sin (2 (\frac{π}{2}))$

Step 1.3.2.2

Cancel the common factor of $2$ .

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Step 1.3.2.2.1

Cancel the common factor.

$f (0) = \sin (2 (\frac{π}{2}))$

Step 1.3.2.2.2

Rewrite the expression.

$f (0) = \sin (π)$

Step 1.3.2.3

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$f (0) = \sin (0)$

Step 1.3.2.4

The exact value of $\sin (0)$ is $0$ .

$f (0) = 0$

Step 1.3.2.5

The final answer is $0$ .

$0$

Step 1.4

Find the point at $x = \frac{1}{2}$ .

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Step 1.4.1

Replace the variable $x$ with $\frac{1}{2}$ in the expression.

$f (\frac{1}{2}) = \sin (2 \arccos (\frac{1}{2}))$

Step 1.4.2

Simplify the result.

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Step 1.4.2.1

The exact value of $\arccos (\frac{1}{2})$ is $\frac{π}{3}$ .

$f (\frac{1}{2}) = \sin (2 (\frac{π}{3}))$

Step 1.4.2.2

Combine $2$ and $\frac{π}{3}$ .

$f (\frac{1}{2}) = \sin (\frac{2 π}{3})$

Step 1.4.2.3

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$f (\frac{1}{2}) = \sin (\frac{π}{3})$

Step 1.4.2.4

The exact value of $\sin (\frac{π}{3})$ is $\frac{\sqrt{3}}{2}$ .

$f (\frac{1}{2}) = \frac{\sqrt{3}}{2}$

Step 1.4.2.5

The final answer is $\frac{\sqrt{3}}{2}$ .

$\frac{\sqrt{3}}{2}$

Step 1.5

Find the point at $x = 1$ .

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Step 1.5.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = \sin (2 \arccos (1))$

Step 1.5.2

Simplify the result.

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Step 1.5.2.1

The exact value of $\arccos (1)$ is $0$ .

$f (1) = \sin (2 \cdot 0)$

Step 1.5.2.2

Multiply $2$ by $0$ .

$f (1) = \sin (0)$

Step 1.5.2.3

The exact value of $\sin (0)$ is $0$ .

$f (1) = 0$

Step 1.5.2.4

The final answer is $0$ .

$0$

Step 1.6

List the points in a table.

$\begin{array}{c} x & f (x) \\ - 1 & 0 \\ - \frac{1}{2} & - \frac{\sqrt{3}}{2} \\ 0 & 0 \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \\ 1 & 0 \end{array}$

Step 2

The trig function can be graphed using the points.

$\begin{array}{c} x & f (x) \\ - 1 & 0 \\ - \frac{1}{2} & - \frac{\sqrt{3}}{2} \\ 0 & 0 \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \\ 1 & 0 \end{array}$

Step 3