Trigonometry Examples

$f (x) = - 3 \sin (x - π) + 2$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $- 3 \sin (x - π) + 2$ with respect to $x$ is $\frac{d}{d x} [- 3 \sin (x - π)] + \frac{d}{d x} [2]$ .

$\frac{d}{d x} [- 3 \sin (x - π)] + \frac{d}{d x} [2]$

Step 1.2

Evaluate $\frac{d}{d x} [- 3 \sin (x - π)]$ .

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Step 1.2.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 \sin (x - π)$ with respect to $x$ is $- 3 \frac{d}{d x} [\sin (x - π)]$ .

$- 3 \frac{d}{d x} [\sin (x - π)] + \frac{d}{d x} [2]$

Step 1.2.2

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

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Step 1.2.2.1

To apply the Chain Rule, set $u$ as $x - π$ .

$- 3 (\frac{d}{d u} [\sin (u)] \frac{d}{d x} [x - π]) + \frac{d}{d x} [2]$

Step 1.2.2.2

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

$- 3 (\cos (u) \frac{d}{d x} [x - π]) + \frac{d}{d x} [2]$

Step 1.2.2.3

Replace all occurrences of $u$ with $x - π$ .

$- 3 (\cos (x - π) \frac{d}{d x} [x - π]) + \frac{d}{d x} [2]$

Step 1.2.3

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

$- 3 (\cos (x - π) (\frac{d}{d x} [x] + \frac{d}{d x} [- π])) + \frac{d}{d x} [2]$

Step 1.2.4

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

$- 3 (\cos (x - π) (1 + \frac{d}{d x} [- π])) + \frac{d}{d x} [2]$

Step 1.2.5

Since $- π$ is constant with respect to $x$ , the derivative of $- π$ with respect to $x$ is $0$ .

$- 3 (\cos (x - π) (1 + 0)) + \frac{d}{d x} [2]$

Step 1.2.6

Add $1$ and $0$ .

$- 3 (\cos (x - π) \cdot 1) + \frac{d}{d x} [2]$

Step 1.2.7

Multiply $\cos (x - π)$ by $1$ .

$- 3 \cos (x - π) + \frac{d}{d x} [2]$

Step 1.3

Differentiate using the Constant Rule.

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Step 1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$- 3 \cos (x - π) + 0$

Step 1.3.2

Add $- 3 \cos (x - π)$ and $0$ .

$- 3 \cos (x - π)$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 \cos (x - π)$ with respect to $x$ is $- 3 \frac{d}{d x} [\cos (x - π)]$ .

$f'' (x) = - 3 \frac{d}{d x} \cos (x - π)$

Step 2.2

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

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Step 2.2.1

To apply the Chain Rule, set $u$ as $x - π$ .

$f'' (x) = - 3 (\frac{d}{d u} (\cos (u)) \frac{d}{d x} (x - π))$

Step 2.2.2

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

$f'' (x) = - 3 (- \sin (u) \frac{d}{d x} (x - π))$

Step 2.2.3

Replace all occurrences of $u$ with $x - π$ .

$f'' (x) = - 3 (- \sin (x - π) \frac{d}{d x} (x - π))$

Step 2.3

Differentiate.

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Step 2.3.1

Multiply $- 1$ by $- 3$ .

$f'' (x) = 3 (\sin (x - π) \frac{d}{d x} (x - π))$

Step 2.3.2

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

$f'' (x) = 3 \sin (x - π) (\frac{d}{d x} (x) + \frac{d}{d x} (- π))$

Step 2.3.3

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

$f'' (x) = 3 \sin (x - π) (1 + \frac{d}{d x} (- π))$

Step 2.3.4

Since $- π$ is constant with respect to $x$ , the derivative of $- π$ with respect to $x$ is $0$ .

$f'' (x) = 3 \sin (x - π) (1 + 0)$

Step 2.3.5

Simplify the expression.

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Step 2.3.5.1

Add $1$ and $0$ .

$f'' (x) = 3 \sin (x - π) \cdot 1$

Step 2.3.5.2

Multiply $3$ by $1$ .

$f'' (x) = 3 \sin (x - π)$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$- 3 \cos (x - π) = 0$

Step 4

Divide each term in $- 3 \cos (x - π) = 0$ by $- 3$ and simplify.

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Step 4.1

Divide each term in $- 3 \cos (x - π) = 0$ by $- 3$ .

$\frac{- 3 \cos (x - π)}{- 3} = \frac{0}{- 3}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $- 3$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{- 3 \cos (x - π)}{- 3} = \frac{0}{- 3}$

Step 4.2.1.2

Divide $\cos (x - π)$ by $1$ .

$\cos (x - π) = \frac{0}{- 3}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $- 3$ .

$\cos (x - π) = 0$

Step 5

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

$x - π = \arccos (0)$

Step 6

Simplify the right side.

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Step 6.1

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

$x - π = \frac{π}{2}$

Step 7

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

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Step 7.1

Add $π$ to both sides of the equation.

$x = \frac{π}{2} + π$

Step 7.2

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

$x = \frac{π}{2} + π \cdot \frac{2}{2}$

Step 7.3

Combine $π$ and $\frac{2}{2}$ .

$x = \frac{π}{2} + \frac{π \cdot 2}{2}$

Step 7.4

Combine the numerators over the common denominator.

$x = \frac{π + π \cdot 2}{2}$

Step 7.5

Simplify the numerator.

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Step 7.5.1

Move $2$ to the left of $π$ .

$x = \frac{π + 2 \cdot π}{2}$

Step 7.5.2

Add $π$ and $2 π$ .

$x = \frac{3 π}{2}$

Step 8

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.

$x - π = 2 π - \frac{π}{2}$

Step 9

Solve for $x$ .

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Step 9.1

Simplify $2 π - \frac{π}{2}$ .

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Step 9.1.1

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

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

Step 9.1.2

Combine fractions.

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Step 9.1.2.1

Combine $2 π$ and $\frac{2}{2}$ .

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

Step 9.1.2.2

Combine the numerators over the common denominator.

$x - π = \frac{2 π \cdot 2 - π}{2}$

Step 9.1.3

Simplify the numerator.

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Step 9.1.3.1

Multiply $2$ by $2$ .

$x - π = \frac{4 π - π}{2}$

Step 9.1.3.2

Subtract $π$ from $4 π$ .

$x - π = \frac{3 π}{2}$

Step 9.2

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

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Step 9.2.1

Add $π$ to both sides of the equation.

$x = \frac{3 π}{2} + π$

Step 9.2.2

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

$x = \frac{3 π}{2} + π \cdot \frac{2}{2}$

Step 9.2.3

Combine $π$ and $\frac{2}{2}$ .

$x = \frac{3 π}{2} + \frac{π \cdot 2}{2}$

Step 9.2.4

Combine the numerators over the common denominator.

$x = \frac{3 π + π \cdot 2}{2}$

Step 9.2.5

Simplify the numerator.

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Step 9.2.5.1

Move $2$ to the left of $π$ .

$x = \frac{3 π + 2 \cdot π}{2}$

Step 9.2.5.2

Add $3 π$ and $2 π$ .

$x = \frac{5 π}{2}$

Step 10

The solution to the equation $x - π = \frac{π}{2}$ .

$x = \frac{3 π}{2}, \frac{5 π}{2}$

Step 11

Evaluate the second derivative at $x = \frac{3 π}{2}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$3 \sin ((\frac{3 π}{2}) - π)$

Step 12

Evaluate the second derivative.

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Step 12.1

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

$3 \sin (\frac{3 π}{2} - π \cdot \frac{2}{2})$

Step 12.2

Combine fractions.

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Step 12.2.1

Combine $- π$ and $\frac{2}{2}$ .

$3 \sin (\frac{3 π}{2} + \frac{- π \cdot 2}{2})$

Step 12.2.2

Combine the numerators over the common denominator.

$3 \sin (\frac{3 π - π \cdot 2}{2})$

Step 12.3

Simplify the numerator.

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Step 12.3.1

Multiply $2$ by $- 1$ .

$3 \sin (\frac{3 π - 2 π}{2})$

Step 12.3.2

Subtract $2 π$ from $3 π$ .

$3 \sin (\frac{π}{2})$

Step 12.4

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

$3 \cdot 1$

Step 12.5

Multiply $3$ by $1$ .

$3$

Step 13

$x = \frac{3 π}{2}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{3 π}{2}$ is a local minimum

Step 14

Find the y-value when $x = \frac{3 π}{2}$ .

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Step 14.1

Replace the variable $x$ with $\frac{3 π}{2}$ in the expression.

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

Step 14.2

Simplify the result.

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Step 14.2.1

Simplify each term.

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Step 14.2.1.1

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

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

Step 14.2.1.2

Combine $- π$ and $\frac{2}{2}$ .

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

Step 14.2.1.3

Combine the numerators over the common denominator.

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

Step 14.2.1.4

Simplify the numerator.

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Step 14.2.1.4.1

Multiply $2$ by $- 1$ .

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

Step 14.2.1.4.2

Subtract $2 π$ from $3 π$ .

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

Step 14.2.1.5

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

$f (\frac{3 π}{2}) = - 3 \cdot 1 + 2$

Step 14.2.1.6

Multiply $- 3$ by $1$ .

$f (\frac{3 π}{2}) = - 3 + 2$

Step 14.2.2

Add $- 3$ and $2$ .

$f (\frac{3 π}{2}) = - 1$

Step 14.2.3

The final answer is $- 1$ .

$y = - 1$

Step 15

Evaluate the second derivative at $x = \frac{5 π}{2}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$3 \sin ((\frac{5 π}{2}) - π)$

Step 16

Evaluate the second derivative.

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Step 16.1

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

$3 \sin (\frac{5 π}{2} - π \cdot \frac{2}{2})$

Step 16.2

Combine fractions.

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Step 16.2.1

Combine $- π$ and $\frac{2}{2}$ .

$3 \sin (\frac{5 π}{2} + \frac{- π \cdot 2}{2})$

Step 16.2.2

Combine the numerators over the common denominator.

$3 \sin (\frac{5 π - π \cdot 2}{2})$

Step 16.3

Simplify the numerator.

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Step 16.3.1

Multiply $2$ by $- 1$ .

$3 \sin (\frac{5 π - 2 π}{2})$

Step 16.3.2

Subtract $2 π$ from $5 π$ .

$3 \sin (\frac{3 π}{2})$

Step 16.4

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.

$3 (- \sin (\frac{π}{2}))$

Step 16.5

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

$3 (- 1 \cdot 1)$

Step 16.6

Multiply $3 (- 1 \cdot 1)$ .

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Step 16.6.1

Multiply $- 1$ by $1$ .

$3 \cdot - 1$

Step 16.6.2

Multiply $3$ by $- 1$ .

$- 3$

Step 17

$x = \frac{5 π}{2}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{5 π}{2}$ is a local maximum

Step 18

Find the y-value when $x = \frac{5 π}{2}$ .

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Step 18.1

Replace the variable $x$ with $\frac{5 π}{2}$ in the expression.

$f (\frac{5 π}{2}) = - 3 \sin ((\frac{5 π}{2}) - π) + 2$

Step 18.2

Simplify the result.

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Step 18.2.1

Simplify each term.

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Step 18.2.1.1

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

$f (\frac{5 π}{2}) = - 3 \sin (\frac{5 π}{2} - π \cdot \frac{2}{2}) + 2$

Step 18.2.1.2

Combine $- π$ and $\frac{2}{2}$ .

$f (\frac{5 π}{2}) = - 3 \sin (\frac{5 π}{2} + \frac{- π \cdot 2}{2}) + 2$

Step 18.2.1.3

Combine the numerators over the common denominator.

$f (\frac{5 π}{2}) = - 3 \sin (\frac{5 π - π \cdot 2}{2}) + 2$

Step 18.2.1.4

Simplify the numerator.

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Step 18.2.1.4.1

Multiply $2$ by $- 1$ .

$f (\frac{5 π}{2}) = - 3 \sin (\frac{5 π - 2 π}{2}) + 2$

Step 18.2.1.4.2

Subtract $2 π$ from $5 π$ .

$f (\frac{5 π}{2}) = - 3 \sin (\frac{3 π}{2}) + 2$

Step 18.2.1.5

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.

$f (\frac{5 π}{2}) = - 3 (- \sin (\frac{π}{2})) + 2$

Step 18.2.1.6

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

$f (\frac{5 π}{2}) = - 3 (- 1 \cdot 1) + 2$

Step 18.2.1.7

Multiply $- 3 (- 1 \cdot 1)$ .

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Step 18.2.1.7.1

Multiply $- 1$ by $1$ .

$f (\frac{5 π}{2}) = - 3 \cdot - 1 + 2$

Step 18.2.1.7.2

Multiply $- 3$ by $- 1$ .

$f (\frac{5 π}{2}) = 3 + 2$

Step 18.2.2

Add $3$ and $2$ .

$f (\frac{5 π}{2}) = 5$

Step 18.2.3

The final answer is $5$ .

$y = 5$

Step 19

These are the local extrema for $f (x) = - 3 \sin (x - π) + 2$ .

$(\frac{3 π}{2}, - 1)$ is a local minima

$(\frac{5 π}{2}, 5)$ is a local maxima

Step 20