Calculus Examples

f (θ) = \sin (θ)

$f (θ) = \sin (θ)$ ,

- \frac{π}{2} \leq θ \leq \frac{5 π}{6}

$- \frac{π}{2} \leq θ \leq \frac{5 π}{6}$

Step 1

Find the critical points.

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Step 1.1

Find the first derivative.

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Step 1.1.1

The derivative of

\sin (θ)

$\sin (θ)$ with respect to

θ

$θ$ is

\cos (θ)

$\cos (θ)$ .

f' (θ) = \cos (θ)

$f' (θ) = \cos (θ)$

Step 1.1.2

The first derivative of

f (θ)

$f (θ)$ with respect to

θ

$θ$ is

\cos (θ)

$\cos (θ)$ .

\cos (θ)

$\cos (θ)$

\cos (θ)

$\cos (θ)$

Step 1.2

Set the first derivative equal to

0

$0$ then solve the equation

\cos (θ) = 0

$\cos (θ) = 0$ .

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Step 1.2.1

Set the first derivative equal to

0

$0$ .

\cos (θ) = 0

$\cos (θ) = 0$

Step 1.2.2

Take the inverse cosine of both sides of the equation to extract

θ

$θ$ from inside the cosine.

θ = \arccos (0)

$θ = \arccos (0)$

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

The exact value of

\arccos (0)

$\arccos (0)$ is

\frac{π}{2}

$\frac{π}{2}$ .

θ = \frac{π}{2}

$θ = \frac{π}{2}$

θ = \frac{π}{2}

$θ = \frac{π}{2}$

Step 1.2.4

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from

2 π

$2 π$ to find the solution in the fourth quadrant.

θ = 2 π - \frac{π}{2}

$θ = 2 π - \frac{π}{2}$

Step 1.2.5

Simplify

2 π - \frac{π}{2}

$2 π - \frac{π}{2}$ .

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Step 1.2.5.1

To write

2 π

$2 π$ as a fraction with a common denominator, multiply by

\frac{2}{2}

$\frac{2}{2}$ .

θ = 2 π \cdot \frac{2}{2} - \frac{π}{2}

$θ = 2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 1.2.5.2

Combine fractions.

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Step 1.2.5.2.1

Combine

2 π

$2 π$ and

\frac{2}{2}

$\frac{2}{2}$ .

θ = \frac{2 π \cdot 2}{2} - \frac{π}{2}

$θ = \frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 1.2.5.2.2

Combine the numerators over the common denominator.

θ = \frac{2 π \cdot 2 - π}{2}

$θ = \frac{2 π \cdot 2 - π}{2}$

θ = \frac{2 π \cdot 2 - π}{2}

$θ = \frac{2 π \cdot 2 - π}{2}$

Step 1.2.5.3

Simplify the numerator.

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Step 1.2.5.3.1

Multiply

2

$2$ by

2

$2$ .

θ = \frac{4 π - π}{2}

$θ = \frac{4 π - π}{2}$

Step 1.2.5.3.2

Subtract

π

$π$ from

4 π

$4 π$ .

θ = \frac{3 π}{2}

$θ = \frac{3 π}{2}$

θ = \frac{3 π}{2}

$θ = \frac{3 π}{2}$

θ = \frac{3 π}{2}

$θ = \frac{3 π}{2}$

Step 1.2.6

Find the period of

\cos (θ)

$\cos (θ)$ .

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Step 1.2.6.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 1.2.6.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 1.2.6.3

The absolute value is the distance between a number and zero. The distance between

0

$0$ and

1

$1$ is

1

$1$ .

\frac{2 π}{1}

$\frac{2 π}{1}$

Step 1.2.6.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 1.2.7

The period of the

\cos (θ)

$\cos (θ)$ function is

2 π

$2 π$ so values will repeat every

2 π

$2 π$ radians in both directions.

θ = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n

$θ = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n$ , for any integer

n

$n$

Step 1.2.8

Consolidate the answers.

θ = \frac{π}{2} + π n

$θ = \frac{π}{2} + π n$ , for any integer

n

$n$

θ = \frac{π}{2} + π n

$θ = \frac{π}{2} + π n$ , for any integer

n

$n$

Step 1.3

Find the values where the derivative is undefined.

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Step 1.3.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 1.4

Evaluate

\sin (θ)

$\sin (θ)$ at each

θ

$θ$ value where the derivative is

0

$0$ or undefined.

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Step 1.4.1

Evaluate at

θ = \frac{π}{2}

$θ = \frac{π}{2}$ .

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Step 1.4.1.1

Substitute

\frac{π}{2}

$\frac{π}{2}$ for

θ

$θ$ .

\sin (\frac{π}{2})

$\sin (\frac{π}{2})$

Step 1.4.1.2

The exact value of

\sin (\frac{π}{2})

$\sin (\frac{π}{2})$ is

1

$1$ .

1

$1$

1

$1$

Step 1.4.2

Evaluate at

θ = \frac{3 π}{2}

$θ = \frac{3 π}{2}$ .

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Step 1.4.2.1

Substitute

\frac{3 π}{2}

$\frac{3 π}{2}$ for

θ

$θ$ .

\sin (\frac{3 π}{2})

$\sin (\frac{3 π}{2})$

Step 1.4.2.2

Simplify.

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Step 1.4.2.2.1

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 fourth quadrant.

- \sin (\frac{π}{2})

$- \sin (\frac{π}{2})$

Step 1.4.2.2.2

The exact value of

\sin (\frac{π}{2})

$\sin (\frac{π}{2})$ is

1

$1$ .

- 1 \cdot 1

$- 1 \cdot 1$

Step 1.4.2.2.3

Multiply

- 1

$- 1$ by

1

$1$ .

- 1

$- 1$

- 1

$- 1$

- 1

$- 1$

Step 1.4.3

List all of the points.

(\frac{π}{2} + 2 π n, 1), (\frac{3 π}{2} + 2 π n, - 1)

$(\frac{π}{2} + 2 π n, 1), (\frac{3 π}{2} + 2 π n, - 1)$ , for any integer

n

$n$

(\frac{π}{2} + 2 π n, 1), (\frac{3 π}{2} + 2 π n, - 1)

$(\frac{π}{2} + 2 π n, 1), (\frac{3 π}{2} + 2 π n, - 1)$ , for any integer

n

$n$

(\frac{π}{2} + 2 π n, 1), (\frac{3 π}{2} + 2 π n, - 1)

$(\frac{π}{2} + 2 π n, 1), (\frac{3 π}{2} + 2 π n, - 1)$ , for any integer

n

$n$

Step 2

Exclude the points that are not on the interval.

(\frac{π}{2}, 1), (- \frac{π}{2}, - 1)

$(\frac{π}{2}, 1), (- \frac{π}{2}, - 1)$

Step 3

Evaluate at the included endpoints.

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Step 3.1

Evaluate at

θ = - \frac{π}{2}

$θ = - \frac{π}{2}$ .

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Step 3.1.1

Substitute

- \frac{π}{2}

$- \frac{π}{2}$ for

θ

$θ$ .

\sin (- \frac{π}{2})

$\sin (- \frac{π}{2})$

Step 3.1.2

Simplify.

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Step 3.1.2.1

Add full rotations of

2 π

$2 π$ until the angle is greater than or equal to

0

$0$ and less than

2 π

$2 π$ .

\sin (\frac{3 π}{2})

$\sin (\frac{3 π}{2})$

Step 3.1.2.2

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 fourth quadrant.

- \sin (\frac{π}{2})

$- \sin (\frac{π}{2})$

Step 3.1.2.3

The exact value of

\sin (\frac{π}{2})

$\sin (\frac{π}{2})$ is

1

$1$ .

- 1 \cdot 1

$- 1 \cdot 1$

Step 3.1.2.4

Multiply

- 1

$- 1$ by

1

$1$ .

- 1

$- 1$

- 1

$- 1$

- 1

$- 1$

Step 3.2

Evaluate at

θ = \frac{5 π}{6}

$θ = \frac{5 π}{6}$ .

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Step 3.2.1

Substitute

\frac{5 π}{6}

$\frac{5 π}{6}$ for

θ

$θ$ .

\sin (\frac{5 π}{6})

$\sin (\frac{5 π}{6})$

Step 3.2.2

Simplify.

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Step 3.2.2.1

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

\sin (\frac{π}{6})

$\sin (\frac{π}{6})$

Step 3.2.2.2

The exact value of

\sin (\frac{π}{6})

$\sin (\frac{π}{6})$ is

\frac{1}{2}

$\frac{1}{2}$ .

\frac{1}{2}

$\frac{1}{2}$

\frac{1}{2}

$\frac{1}{2}$

\frac{1}{2}

$\frac{1}{2}$

Step 3.3

List all of the points.

(- \frac{π}{2}, - 1), (\frac{5 π}{6}, \frac{1}{2})

$(- \frac{π}{2}, - 1), (\frac{5 π}{6}, \frac{1}{2})$

(- \frac{π}{2}, - 1), (\frac{5 π}{6}, \frac{1}{2})

$(- \frac{π}{2}, - 1), (\frac{5 π}{6}, \frac{1}{2})$

Step 4

Compare the

f (θ)

$f (θ)$ values found for each value of

θ

$θ$ in order to determine the absolute maximum and minimum over the given interval. The maximum will occur at the highest

f (θ)

$f (θ)$ value and the minimum will occur at the lowest

f (θ)

$f (θ)$ value.

Absolute Maximum:

(\frac{π}{2}, 1)

$(\frac{π}{2}, 1)$

Absolute Minimum:

(- \frac{π}{2}, - 1)

$(- \frac{π}{2}, - 1)$

Step 5