Trigonometry Examples

$f (x) = \frac{1}{2} \cos (2 x) - 1$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $\frac{1}{2} \cdot \cos (2 x) - 1$ with respect to $x$ is $\frac{d}{d x} [\frac{1}{2} \cdot \cos (2 x)] + \frac{d}{d x} [- 1]$ .

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

Step 1.2

Evaluate $\frac{d}{d x} [\frac{1}{2} \cdot \cos (2 x)]$ .

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Step 1.2.1

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{1}{2} \cdot \cos (2 x)$ with respect to $x$ is $\frac{1}{2} \frac{d}{d x} [\cos (2 x)]$ .

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

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 $f (x) = \cos (x)$ and $g (x) = 2 x$ .

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Step 1.2.2.1

To apply the Chain Rule, set $u$ as $2 x$ .

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

Step 1.2.2.2

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

$\frac{1}{2} (- \sin (u) \frac{d}{d x} [2 x]) + \frac{d}{d x} [- 1]$

Step 1.2.2.3

Replace all occurrences of $u$ with $2 x$ .

$\frac{1}{2} (- \sin (2 x) \frac{d}{d x} [2 x]) + \frac{d}{d x} [- 1]$

Step 1.2.3

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$\frac{1}{2} (- \sin (2 x) (2 \frac{d}{d x} [x])) + \frac{d}{d x} [- 1]$

Step 1.2.4

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

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

Step 1.2.5

Multiply $2$ by $1$ .

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

Step 1.2.6

Multiply $2$ by $- 1$ .

$\frac{1}{2} (- 2 \sin (2 x)) + \frac{d}{d x} [- 1]$

Step 1.2.7

Combine $- 2$ and $\frac{1}{2}$ .

$\frac{- 2}{2} \sin (2 x) + \frac{d}{d x} [- 1]$

Step 1.2.8

Combine $\frac{- 2}{2}$ and $\sin (2 x)$ .

$\frac{- 2 \sin (2 x)}{2} + \frac{d}{d x} [- 1]$

Step 1.2.9

Cancel the common factor of $- 2$ and $2$ .

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Step 1.2.9.1

Factor $2$ out of $- 2 \sin (2 x)$ .

$\frac{2 (- \sin (2 x))}{2} + \frac{d}{d x} [- 1]$

Step 1.2.9.2

Cancel the common factors.

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Step 1.2.9.2.1

Factor $2$ out of $2$ .

$\frac{2 (- \sin (2 x))}{2 (1)} + \frac{d}{d x} [- 1]$

Step 1.2.9.2.2

Cancel the common factor.

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

Step 1.2.9.2.3

Rewrite the expression.

$\frac{- \sin (2 x)}{1} + \frac{d}{d x} [- 1]$

Step 1.2.9.2.4

Divide $- \sin (2 x)$ by $1$ .

$- \sin (2 x) + \frac{d}{d x} [- 1]$

Step 1.3

Differentiate using the Constant Rule.

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Step 1.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$- \sin (2 x) + 0$

Step 1.3.2

Add $- \sin (2 x)$ and $0$ .

$- \sin (2 x)$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \sin (2 x)$ with respect to $x$ is $- \frac{d}{d x} [\sin (2 x)]$ .

$f'' (x) = - \frac{d}{d x} \sin (2 x)$

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 $f (x) = \sin (x)$ and $g (x) = 2 x$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $2 x$ .

$f'' (x) = - (\frac{d}{d u} (\sin (u)) \frac{d}{d x} (2 x))$

Step 2.2.2

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

$f'' (x) = - (\cos (u) \frac{d}{d x} (2 x))$

Step 2.2.3

Replace all occurrences of $u$ with $2 x$ .

$f'' (x) = - (\cos (2 x) \frac{d}{d x} (2 x))$

Step 2.3

Differentiate.

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Step 2.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$f'' (x) = - \cos (2 x) (2 \frac{d}{d x} (x))$

Step 2.3.2

Multiply $2$ by $- 1$ .

$f'' (x) = - 2 \cos (2 x) \frac{d}{d x} x$

Step 2.3.3

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

$f'' (x) = - 2 \cos (2 x) \cdot 1$

Step 2.3.4

Multiply $- 2$ by $1$ .

$f'' (x) = - 2 \cos (2 x)$

Step 3

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

$- \sin (2 x) = 0$

Step 4

Divide each term in $- \sin (2 x) = 0$ by $- 1$ and simplify.

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Step 4.1

Divide each term in $- \sin (2 x) = 0$ by $- 1$ .

$\frac{- \sin (2 x)}{- 1} = \frac{0}{- 1}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Dividing two negative values results in a positive value.

$\frac{\sin (2 x)}{1} = \frac{0}{- 1}$

Step 4.2.2

Divide $\sin (2 x)$ by $1$ .

$\sin (2 x) = \frac{0}{- 1}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $- 1$ .

$\sin (2 x) = 0$

Step 5

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

$2 x = \arcsin (0)$

Step 6

Simplify the right side.

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Step 6.1

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

$2 x = 0$

Step 7

Divide each term in $2 x = 0$ by $2$ and simplify.

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Step 7.1

Divide each term in $2 x = 0$ by $2$ .

$\frac{2 x}{2} = \frac{0}{2}$

Step 7.2

Simplify the left side.

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Step 7.2.1

Cancel the common factor of $2$ .

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Step 7.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{0}{2}$

Step 7.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{2}$

Step 7.3

Simplify the right side.

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Step 7.3.1

Divide $0$ by $2$ .

$x = 0$

Step 8

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

$2 x = π - 0$

Step 9

Solve for $x$ .

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Step 9.1

Simplify.

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Step 9.1.1

Multiply $- 1$ by $0$ .

$2 x = π + 0$

Step 9.1.2

Add $π$ and $0$ .

$2 x = π$

Step 9.2

Divide each term in $2 x = π$ by $2$ and simplify.

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Step 9.2.1

Divide each term in $2 x = π$ by $2$ .

$\frac{2 x}{2} = \frac{π}{2}$

Step 9.2.2

Simplify the left side.

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Step 9.2.2.1

Cancel the common factor of $2$ .

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Step 9.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{π}{2}$

Step 9.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{π}{2}$

Step 10

The solution to the equation $2 x = 0$ .

$x = 0, \frac{π}{2}$

Step 11

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 2 \cos (2 (0))$

Step 12

Evaluate the second derivative.

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Step 12.1

Multiply $2$ by $0$ .

$- 2 \cos (0)$

Step 12.2

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

$- 2 \cdot 1$

Step 12.3

Multiply $- 2$ by $1$ .

$- 2$

Step 13

$x = 0$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 0$ is a local maximum

Step 14

Find the y-value when $x = 0$ .

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Step 14.1

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

$f (0) = \frac{1}{2} \cdot \cos (2 (0)) - 1$

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

Multiply $2$ by $0$ .

$f (0) = \frac{1}{2} \cdot \cos (0) - 1$

Step 14.2.1.2

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

$f (0) = \frac{1}{2} \cdot 1 - 1$

Step 14.2.1.3

Multiply $\frac{1}{2}$ by $1$ .

$f (0) = \frac{1}{2} - 1$

Step 14.2.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (0) = \frac{1}{2} - 1 \cdot \frac{2}{2}$

Step 14.2.3

Combine $- 1$ and $\frac{2}{2}$ .

$f (0) = \frac{1}{2} + \frac{- 1 \cdot 2}{2}$

Step 14.2.4

Combine the numerators over the common denominator.

$f (0) = \frac{1 - 1 \cdot 2}{2}$

Step 14.2.5

Simplify the numerator.

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Step 14.2.5.1

Multiply $- 1$ by $2$ .

$f (0) = \frac{1 - 2}{2}$

Step 14.2.5.2

Subtract $2$ from $1$ .

$f (0) = \frac{- 1}{2}$

Step 14.2.6

Move the negative in front of the fraction.

$f (0) = - \frac{1}{2}$

Step 14.2.7

The final answer is $- \frac{1}{2}$ .

$y = - \frac{1}{2}$

Step 15

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

$- 2 \cos (2 (\frac{π}{2}))$

Step 16

Evaluate the second derivative.

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Step 16.1

Cancel the common factor of $2$ .

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Step 16.1.1

Cancel the common factor.

$- 2 \cos (2 \frac{π}{2})$

Step 16.1.2

Rewrite the expression.

$- 2 \cos (π)$

Step 16.2

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

$- 2 (- \cos (0))$

Step 16.3

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

$- 2 (- 1 \cdot 1)$

Step 16.4

Multiply $- 2 (- 1 \cdot 1)$ .

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Step 16.4.1

Multiply $- 1$ by $1$ .

$- 2 \cdot - 1$

Step 16.4.2

Multiply $- 2$ by $- 1$ .

$2$

Step 17

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

$x = \frac{π}{2}$ is a local minimum

Step 18

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

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Step 18.1

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

$f (\frac{π}{2}) = \frac{1}{2} \cdot \cos (2 (\frac{π}{2})) - 1$

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

Cancel the common factor of $2$ .

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Step 18.2.1.1.1

Cancel the common factor.

$f (\frac{π}{2}) = \frac{1}{2} \cdot \cos (2 (\frac{π}{2})) - 1$

Step 18.2.1.1.2

Rewrite the expression.

$f (\frac{π}{2}) = \frac{1}{2} \cdot \cos (π) - 1$

Step 18.2.1.2

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

$f (\frac{π}{2}) = \frac{1}{2} \cdot (- \cos (0)) - 1$

Step 18.2.1.3

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

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

Step 18.2.1.4

Multiply $- 1$ by $1$ .

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

Step 18.2.1.5

Combine $\frac{1}{2}$ and $- 1$ .

$f (\frac{π}{2}) = \frac{- 1}{2} - 1$

Step 18.2.1.6

Move the negative in front of the fraction.

$f (\frac{π}{2}) = - \frac{1}{2} - 1$

Step 18.2.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 18.2.3

Combine $- 1$ and $\frac{2}{2}$ .

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

Step 18.2.4

Combine the numerators over the common denominator.

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

Step 18.2.5

Simplify the numerator.

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Step 18.2.5.1

Multiply $- 1$ by $2$ .

$f (\frac{π}{2}) = \frac{- 1 - 2}{2}$

Step 18.2.5.2

Subtract $2$ from $- 1$ .

$f (\frac{π}{2}) = \frac{- 3}{2}$

Step 18.2.6

Move the negative in front of the fraction.

$f (\frac{π}{2}) = - \frac{3}{2}$

Step 18.2.7

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

$y = - \frac{3}{2}$

Step 19

These are the local extrema for $f (x) = \frac{1}{2} \cdot \cos (2 x) - 1$ .

$(0, - \frac{1}{2})$ is a local maxima

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

Step 20