Trigonometry Examples

$f (x) = - 4 \sin (x - 0.5) + 11$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $- 4 \sin (x - 0.5) + 11$ with respect to $x$ is $\frac{d}{d x} [- 4 \sin (x - 0.5)] + \frac{d}{d x} [11]$ .

$\frac{d}{d x} [- 4 \sin (x - 0.5)] + \frac{d}{d x} [11]$

Step 1.2

Evaluate $\frac{d}{d x} [- 4 \sin (x - 0.5)]$ .

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Step 1.2.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 \sin (x - 0.5)$ with respect to $x$ is $- 4 \frac{d}{d x} [\sin (x - 0.5)]$ .

$- 4 \frac{d}{d x} [\sin (x - 0.5)] + \frac{d}{d x} [11]$

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 $g (x) = x - 0.5$ .

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Step 1.2.2.1

To apply the Chain Rule, set $u$ as $x - 0.5$ .

$- 4 (\frac{d}{d u} [\sin (u)] \frac{d}{d x} [x - 0.5]) + \frac{d}{d x} [11]$

Step 1.2.2.2

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

$- 4 (\cos (u) \frac{d}{d x} [x - 0.5]) + \frac{d}{d x} [11]$

Step 1.2.2.3

Replace all occurrences of $u$ with $x - 0.5$ .

$- 4 (\cos (x - 0.5) \frac{d}{d x} [x - 0.5]) + \frac{d}{d x} [11]$

Step 1.2.3

By the Sum Rule, the derivative of $x - 0.5$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 0.5]$ .

$- 4 (\cos (x - 0.5) (\frac{d}{d x} [x] + \frac{d}{d x} [- 0.5])) + \frac{d}{d x} [11]$

Step 1.2.4

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

$- 4 (\cos (x - 0.5) (1 + \frac{d}{d x} [- 0.5])) + \frac{d}{d x} [11]$

Step 1.2.5

Since $- 0.5$ is constant with respect to $x$ , the derivative of $- 0.5$ with respect to $x$ is $0$ .

$- 4 (\cos (x - 0.5) (1 + 0)) + \frac{d}{d x} [11]$

Step 1.2.6

Add $1$ and $0$ .

$- 4 (\cos (x - 0.5) \cdot 1) + \frac{d}{d x} [11]$

Step 1.2.7

Multiply $\cos (x - 0.5)$ by $1$ .

$- 4 \cos (x - 0.5) + \frac{d}{d x} [11]$

Step 1.3

Differentiate using the Constant Rule.

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Step 1.3.1

Since $11$ is constant with respect to $x$ , the derivative of $11$ with respect to $x$ is $0$ .

$- 4 \cos (x - 0.5) + 0$

Step 1.3.2

Add $- 4 \cos (x - 0.5)$ and $0$ .

$- 4 \cos (x - 0.5)$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 \cos (x - 0.5)$ with respect to $x$ is $- 4 \frac{d}{d x} [\cos (x - 0.5)]$ .

$f'' (x) = - 4 \frac{d}{d x} \cos (x - 0.5)$

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 $g (x) = x - 0.5$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $x - 0.5$ .

$f'' (x) = - 4 (\frac{d}{d u} (\cos (u)) \frac{d}{d x} (x - 0.5))$

Step 2.2.2

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

$f'' (x) = - 4 (- \sin (u) \frac{d}{d x} (x - 0.5))$

Step 2.2.3

Replace all occurrences of $u$ with $x - 0.5$ .

$f'' (x) = - 4 (- \sin (x - 0.5) \frac{d}{d x} (x - 0.5))$

Step 2.3

Differentiate.

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Step 2.3.1

Multiply $- 1$ by $- 4$ .

$f'' (x) = 4 (\sin (x - 0.5) \frac{d}{d x} (x - 0.5))$

Step 2.3.2

By the Sum Rule, the derivative of $x - 0.5$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 0.5]$ .

$f'' (x) = 4 \sin (x - 0.5) (\frac{d}{d x} (x) + \frac{d}{d x} (- 0.5))$

Step 2.3.3

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

$f'' (x) = 4 \sin (x - 0.5) (1 + \frac{d}{d x} (- 0.5))$

Step 2.3.4

Since $- 0.5$ is constant with respect to $x$ , the derivative of $- 0.5$ with respect to $x$ is $0$ .

$f'' (x) = 4 \sin (x - 0.5) (1 + 0)$

Step 2.3.5

Simplify the expression.

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Step 2.3.5.1

Add $1$ and $0$ .

$f'' (x) = 4 \sin (x - 0.5) \cdot 1$

Step 2.3.5.2

Multiply $4$ by $1$ .

$f'' (x) = 4 \sin (x - 0.5)$

Step 3

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

$- 4 \cos (x - 0.5) = 0$

Step 4

Divide each term in $- 4 \cos (x - 0.5) = 0$ by $- 4$ and simplify.

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Step 4.1

Divide each term in $- 4 \cos (x - 0.5) = 0$ by $- 4$ .

$\frac{- 4 \cos (x - 0.5)}{- 4} = \frac{0}{- 4}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $- 4$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{- 4 \cos (x - 0.5)}{- 4} = \frac{0}{- 4}$

Step 4.2.1.2

Divide $\cos (x - 0.5)$ by $1$ .

$\cos (x - 0.5) = \frac{0}{- 4}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $- 4$ .

$\cos (x - 0.5) = 0$

Step 5

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

$x - 0.5 = \arccos (0)$

Step 6

Simplify the right side.

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Step 6.1

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

$x - 0.5 = \frac{π}{2}$

Step 7

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

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Step 7.1

Add $0.5$ to both sides of the equation.

$x = \frac{π}{2} + 0.5$

Step 7.2

Add $\frac{π}{2}$ and $0.5$ .

$x = 2.07079632$

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.

$x - 0.5 = 2 π - \frac{π}{2}$

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 $\frac{2}{2}$ .

$x - 0.5 = 2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 9.1.2

Combine fractions.

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Step 9.1.2.1

Combine $2 π$ and $\frac{2}{2}$ .

$x - 0.5 = \frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 9.1.2.2

Combine the numerators over the common denominator.

$x - 0.5 = \frac{2 π \cdot 2 - π}{2}$

Step 9.1.3

Simplify the numerator.

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Step 9.1.3.1

Multiply $2$ by $2$ .

$x - 0.5 = \frac{4 π - π}{2}$

Step 9.1.3.2

Subtract $π$ from $4 π$ .

$x - 0.5 = \frac{3 π}{2}$

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 $0.5$ to both sides of the equation.

$x = \frac{3 π}{2} + 0.5$

Step 9.2.2

Add $\frac{3 π}{2}$ and $0.5$ .

$x = 5.21238898$

Step 10

The solution to the equation $x - 0.5 = \frac{π}{2}$ .

$x = 2.07079632, 5.21238898$

Step 11

Evaluate the second derivative at $x = 2.07079632$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$4 \sin ((2.07079632) - 0.5)$

Step 12

Subtract $0.5$ from $2.07079632$ .

$4 \sin (1.57079632)$

Step 13

$x = 2.07079632$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 2.07079632$ is a local minimum

Step 14

Find the y-value when $x = 2.07079632$ .

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Step 14.1

Replace the variable $x$ with $2.07079632$ in the expression.

$f (2.07079632) = - 4 \sin ((2.07079632) - 0.5) + 11$

Step 14.2

Simplify the result.

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Step 14.2.1

Subtract $0.5$ from $2.07079632$ .

$f (2.07079632) = - 4 \sin (1.57079632) + 11$

Step 14.2.2

The final answer is $- 4 \sin (1.57079632) + 11$ .

$y = - 4 \sin (1.57079632) + 11$

Step 15

Evaluate the second derivative at $x = 5.21238898$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$4 \sin ((5.21238898) - 0.5)$

Step 16

Subtract $0.5$ from $5.21238898$ .

$4 \sin (4.71238898)$

Step 17

$x = 5.21238898$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 5.21238898$ is a local maximum

Step 18

Find the y-value when $x = 5.21238898$ .

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Step 18.1

Replace the variable $x$ with $5.21238898$ in the expression.

$f (5.21238898) = - 4 \sin ((5.21238898) - 0.5) + 11$

Step 18.2

Simplify the result.

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Step 18.2.1

Subtract $0.5$ from $5.21238898$ .

$f (5.21238898) = - 4 \sin (4.71238898) + 11$

Step 18.2.2

The final answer is $- 4 \sin (4.71238898) + 11$ .

$y = - 4 \sin (4.71238898) + 11$

Step 19

These are the local extrema for $f (x) = - 4 \sin (x - 0.5) + 11$ .

$(2.07079632, - 4 \sin (1.57079632) + 11)$ is a local minima

$(5.21238898, - 4 \sin (4.71238898) + 11)$ is a local maxima

Step 20