Calculus Examples

$F (θ) = \sin (θ + \frac{π}{2})$ , $0 \leq θ \leq \frac{7 π}{4}$

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

Find the first derivative.

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Step 1.1.1.1

Differentiate using the chain rule, which states that $\frac{d}{d θ} [f (g (θ))]$ is $f' (g (θ)) g' (θ)$ where $f (θ) = \sin (θ)$ and $g (θ) = θ + \frac{π}{2}$ .

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Step 1.1.1.1.1

To apply the Chain Rule, set $u$ as $θ + \frac{π}{2}$ .

$\frac{d}{d u} [\sin (u)] \frac{d}{d θ} [θ + \frac{π}{2}]$

Step 1.1.1.1.2

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$\cos (u) \frac{d}{d θ} [θ + \frac{π}{2}]$

Step 1.1.1.1.3

Replace all occurrences of $u$ with $θ + \frac{π}{2}$ .

$\cos (θ + \frac{π}{2}) \frac{d}{d θ} [θ + \frac{π}{2}]$

Step 1.1.1.2

Differentiate.

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Step 1.1.1.2.1

By the Sum Rule, the derivative of $θ + \frac{π}{2}$ with respect to $θ$ is $\frac{d}{d θ} [θ] + \frac{d}{d θ} [\frac{π}{2}]$ .

$\cos (θ + \frac{π}{2}) (\frac{d}{d θ} [θ] + \frac{d}{d θ} [\frac{π}{2}])$

Step 1.1.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d θ} [θ^{n}]$ is $n θ^{n - 1}$ where $n = 1$ .

$\cos (θ + \frac{π}{2}) (1 + \frac{d}{d θ} [\frac{π}{2}])$

Step 1.1.1.2.3

Since $\frac{π}{2}$ is constant with respect to $θ$ , the derivative of $\frac{π}{2}$ with respect to $θ$ is $0$ .

$\cos (θ + \frac{π}{2}) (1 + 0)$

Step 1.1.1.2.4

Simplify the expression.

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Step 1.1.1.2.4.1

Add $1$ and $0$ .

$\cos (θ + \frac{π}{2}) \cdot 1$

Step 1.1.1.2.4.2

Multiply $\cos (θ + \frac{π}{2})$ by $1$ .

$f' (θ) = \cos (θ + \frac{π}{2})$

Step 1.1.2

The first derivative of $F (θ)$ with respect to $θ$ is $\cos (θ + \frac{π}{2})$ .

$\cos (θ + \frac{π}{2})$

Step 1.2

Set the first derivative equal to $0$ then solve the equation $\cos (θ + \frac{π}{2}) = 0$ .

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Step 1.2.1

Set the first derivative equal to $0$ .

$\cos (θ + \frac{π}{2}) = 0$

Step 1.2.2

Take the inverse cosine of both sides of the equation to extract $θ$ from inside the cosine.

$θ + \frac{π}{2} = \arccos (0)$

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

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

$θ + \frac{π}{2} = \frac{π}{2}$

Step 1.2.4

Move all terms not containing $θ$ to the right side of the equation.

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Step 1.2.4.1

Subtract $\frac{π}{2}$ from both sides of the equation.

$θ = \frac{π}{2} - \frac{π}{2}$

Step 1.2.4.2

Combine the numerators over the common denominator.

$θ = \frac{π - π}{2}$

Step 1.2.4.3

Subtract $π$ from $π$ .

$θ = \frac{0}{2}$

Step 1.2.4.4

Divide $0$ by $2$ .

$θ = 0$

Step 1.2.5

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$θ + \frac{π}{2} = 2 π - \frac{π}{2}$

Step 1.2.6

Solve for $θ$ .

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Step 1.2.6.1

Simplify $2 π - \frac{π}{2}$ .

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Step 1.2.6.1.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 1.2.6.1.2

Combine fractions.

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Step 1.2.6.1.2.1

Combine $2 π$ and $\frac{2}{2}$ .

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

Step 1.2.6.1.2.2

Combine the numerators over the common denominator.

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

Step 1.2.6.1.3

Simplify the numerator.

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Step 1.2.6.1.3.1

Multiply $2$ by $2$ .

$θ + \frac{π}{2} = \frac{4 π - π}{2}$

Step 1.2.6.1.3.2

Subtract $π$ from $4 π$ .

$θ + \frac{π}{2} = \frac{3 π}{2}$

Step 1.2.6.2

Move all terms not containing $θ$ to the right side of the equation.

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Step 1.2.6.2.1

Subtract $\frac{π}{2}$ from both sides of the equation.

$θ = \frac{3 π}{2} - \frac{π}{2}$

Step 1.2.6.2.2

Combine the numerators over the common denominator.

$θ = \frac{3 π - π}{2}$

Step 1.2.6.2.3

Subtract $π$ from $3 π$ .

$θ = \frac{2 π}{2}$

Step 1.2.6.2.4

Cancel the common factor of $2$ .

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Step 1.2.6.2.4.1

Cancel the common factor.

$θ = \frac{2 π}{2}$

Step 1.2.6.2.4.2

Divide $π$ by $1$ .

$θ = π$

Step 1.2.7

Find the period of $\cos (θ + \frac{π}{2})$ .

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Step 1.2.7.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 1.2.7.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 1.2.7.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 1.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 1.2.8

The period of the $\cos (θ + \frac{π}{2})$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$θ = 2 π n, π + 2 π n$ , for any integer $n$

Step 1.2.9

Consolidate the answers.

$θ = π n$ , for any integer $n$

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 $\sin (θ + \frac{π}{2})$ at each $θ$ value where the derivative is $0$ or undefined.

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Step 1.4.1

Evaluate at $θ = 0$ .

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Step 1.4.1.1

Substitute $0$ for $θ$ .

$\sin ((0) + \frac{π}{2})$

Step 1.4.1.2

Simplify.

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Step 1.4.1.2.1

Add $0$ and $\frac{π}{2}$ .

$\sin (\frac{π}{2})$

Step 1.4.1.2.2

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

$1$

Step 1.4.2

Evaluate at $θ = π$ .

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Step 1.4.2.1

Substitute $π$ for $θ$ .

$\sin ((π) + \frac{π}{2})$

Step 1.4.2.2

Simplify.

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Step 1.4.2.2.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\sin (π \cdot \frac{2}{2} + \frac{π}{2})$

Step 1.4.2.2.2

Combine fractions.

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Step 1.4.2.2.2.1

Combine $π$ and $\frac{2}{2}$ .

$\sin (\frac{π \cdot 2}{2} + \frac{π}{2})$

Step 1.4.2.2.2.2

Combine the numerators over the common denominator.

$\sin (\frac{π \cdot 2 + π}{2})$

Step 1.4.2.2.3

Simplify the numerator.

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Step 1.4.2.2.3.1

Move $2$ to the left of $π$ .

$\sin (\frac{2 \cdot π + π}{2})$

Step 1.4.2.2.3.2

Add $2 π$ and $π$ .

$\sin (\frac{3 π}{2})$

Step 1.4.2.2.4

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.

$- \sin (\frac{π}{2})$

Step 1.4.2.2.5

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

$- 1 \cdot 1$

Step 1.4.2.2.6

Multiply $- 1$ by $1$ .

$- 1$

Step 1.4.3

List all of the points.

$(0 + 2 π n, 1), (π + 2 π n, - 1)$ , for any integer $n$

Step 2

Exclude the points that are not on the interval.

$(0, 1), (π, - 1)$

Step 3

Evaluate at the included endpoints.

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Step 3.1

Evaluate at $θ = 0$ .

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Step 3.1.1

Substitute $0$ for $θ$ .

$\sin ((0) + \frac{π}{2})$

Step 3.1.2

Simplify.

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Step 3.1.2.1

Add $0$ and $\frac{π}{2}$ .

$\sin (\frac{π}{2})$

Step 3.1.2.2

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

$1$

Step 3.2

Evaluate at $θ = \frac{7 π}{4}$ .

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Step 3.2.1

Substitute $\frac{7 π}{4}$ for $θ$ .

$\sin ((\frac{7 π}{4}) + \frac{π}{2})$

Step 3.2.2

Simplify.

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Step 3.2.2.1

To write $\frac{π}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\sin (\frac{7 π}{4} + \frac{π}{2} \cdot \frac{2}{2})$

Step 3.2.2.2

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.2.2.2.1

Multiply $\frac{π}{2}$ by $\frac{2}{2}$ .

$\sin (\frac{7 π}{4} + \frac{π \cdot 2}{2 \cdot 2})$

Step 3.2.2.2.2

Multiply $2$ by $2$ .

$\sin (\frac{7 π}{4} + \frac{π \cdot 2}{4})$

Step 3.2.2.3

Combine the numerators over the common denominator.

$\sin (\frac{7 π + π \cdot 2}{4})$

Step 3.2.2.4

Simplify the numerator.

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Step 3.2.2.4.1

Move $2$ to the left of $π$ .

$\sin (\frac{7 π + 2 \cdot π}{4})$

Step 3.2.2.4.2

Add $7 π$ and $2 π$ .

$\sin (\frac{9 π}{4})$

Step 3.2.2.5

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

$\sin (\frac{π}{4})$

Step 3.2.2.6

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

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

Step 3.3

List all of the points.

$(0, 1), (\frac{7 π}{4}, \frac{\sqrt{2}}{2})$

Step 4

Compare the $F (θ)$ values found for each value of $θ$ in order to determine the absolute maximum and minimum over the given interval. The maximum will occur at the highest $F (θ)$ value and the minimum will occur at the lowest $F (θ)$ value.

Absolute Maximum: $(0, 1)$

Absolute Minimum: $(π, - 1)$

Step 5