Trigonometry Examples

$g (x) = 4 \cos (2 x - π) - 2$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $4 \cos (2 x - π) - 2$ with respect to $x$ is $\frac{d}{d x} [4 \cos (2 x - π)] + \frac{d}{d x} [- 2]$ .

$\frac{d}{d x} [4 \cos (2 x - π)] + \frac{d}{d x} [- 2]$

Step 1.2

Evaluate $\frac{d}{d x} [4 \cos (2 x - π)]$ .

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Step 1.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 \cos (2 x - π)$ with respect to $x$ is $4 \frac{d}{d x} [\cos (2 x - π)]$ .

$4 \frac{d}{d x} [\cos (2 x - π)] + \frac{d}{d x} [- 2]$

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

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Step 1.2.2.1

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

$4 (\frac{d}{d u} [\cos (u)] \frac{d}{d x} [2 x - π]) + \frac{d}{d x} [- 2]$

Step 1.2.2.2

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

$4 (- \sin (u) \frac{d}{d x} [2 x - π]) + \frac{d}{d x} [- 2]$

Step 1.2.2.3

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

$4 (- \sin (2 x - π) \frac{d}{d x} [2 x - π]) + \frac{d}{d x} [- 2]$

Step 1.2.3

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

$4 (- \sin (2 x - π) (\frac{d}{d x} [2 x] + \frac{d}{d x} [- π])) + \frac{d}{d x} [- 2]$

Step 1.2.4

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

$4 (- \sin (2 x - π) (2 \frac{d}{d x} [x] + \frac{d}{d x} [- π])) + \frac{d}{d x} [- 2]$

Step 1.2.5

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

$4 (- \sin (2 x - π) (2 \cdot 1 + \frac{d}{d x} [- π])) + \frac{d}{d x} [- 2]$

Step 1.2.6

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

$4 (- \sin (2 x - π) (2 \cdot 1 + 0)) + \frac{d}{d x} [- 2]$

Step 1.2.7

Multiply $2$ by $1$ .

$4 (- \sin (2 x - π) (2 + 0)) + \frac{d}{d x} [- 2]$

Step 1.2.8

Add $2$ and $0$ .

$4 (- \sin (2 x - π) \cdot 2) + \frac{d}{d x} [- 2]$

Step 1.2.9

Multiply $2$ by $- 1$ .

$4 (- 2 \sin (2 x - π)) + \frac{d}{d x} [- 2]$

Step 1.2.10

Multiply $- 2$ by $4$ .

$- 8 \sin (2 x - π) + \frac{d}{d x} [- 2]$

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 $0$ .

$- 8 \sin (2 x - π) + 0$

Step 1.3.2

Add $- 8 \sin (2 x - π)$ and $0$ .

$- 8 \sin (2 x - π)$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 \sin (2 x - π)$ with respect to $x$ is $- 8 \frac{d}{d x} [\sin (2 x - π)]$ .

$g'' (x) = - 8 \frac{d}{d x} \sin (2 x - π)$

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

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Step 2.2.1

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

$g'' (x) = - 8 (\frac{d}{d u} (\sin (u)) \frac{d}{d x} (2 x - π))$

Step 2.2.2

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

$g'' (x) = - 8 (\cos (u) \frac{d}{d x} (2 x - π))$

Step 2.2.3

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

$g'' (x) = - 8 (\cos (2 x - π) \frac{d}{d x} (2 x - π))$

Step 2.3

Differentiate.

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Step 2.3.1

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

$g'' (x) = - 8 \cos (2 x - π) (\frac{d}{d x} (2 x) + \frac{d}{d x} (- π))$

Step 2.3.2

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

$g'' (x) = - 8 \cos (2 x - π) (2 \frac{d}{d x} (x) + \frac{d}{d x} (- π))$

Step 2.3.3

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

$g'' (x) = - 8 \cos (2 x - π) (2 \cdot 1 + \frac{d}{d x} (- π))$

Step 2.3.4

Multiply $2$ by $1$ .

$g'' (x) = - 8 \cos (2 x - π) (2 + \frac{d}{d x} (- π))$

Step 2.3.5

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

$g'' (x) = - 8 \cos (2 x - π) (2 + 0)$

Step 2.3.6

Simplify the expression.

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Step 2.3.6.1

Add $2$ and $0$ .

$g'' (x) = - 8 \cos (2 x - π) \cdot 2$

Step 2.3.6.2

Multiply $2$ by $- 8$ .

$g'' (x) = - 16 \cos (2 x - π)$

Step 3

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

$- 8 \sin (2 x - π) = 0$

Step 4

Divide each term in $- 8 \sin (2 x - π) = 0$ by $- 8$ and simplify.

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Step 4.1

Divide each term in $- 8 \sin (2 x - π) = 0$ by $- 8$ .

$\frac{- 8 \sin (2 x - π)}{- 8} = \frac{0}{- 8}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $- 8$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{- 8 \sin (2 x - π)}{- 8} = \frac{0}{- 8}$

Step 4.2.1.2

Divide $\sin (2 x - π)$ by $1$ .

$\sin (2 x - π) = \frac{0}{- 8}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $- 8$ .

$\sin (2 x - π) = 0$

Step 5

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

$2 x - π = \arcsin (0)$

Step 6

Simplify the right side.

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Step 6.1

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

$2 x - π = 0$

Step 7

Add $π$ to both sides of the equation.

$2 x = π$

Step 8

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

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Step 8.1

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

$\frac{2 x}{2} = \frac{π}{2}$

Step 8.2

Simplify the left side.

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Step 8.2.1

Cancel the common factor of $2$ .

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Step 8.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{π}{2}$

Step 8.2.1.2

Divide $x$ by $1$ .

$x = \frac{π}{2}$

Step 9

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.

$2 x - π = π - 0$

Step 10

Solve for $x$ .

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Step 10.1

Subtract $0$ from $π$ .

$2 x - π = π$

Step 10.2

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

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Step 10.2.1

Add $π$ to both sides of the equation.

$2 x = π + π$

Step 10.2.2

Add $π$ and $π$ .

$2 x = 2 π$

Step 10.3

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

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Step 10.3.1

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

$\frac{2 x}{2} = \frac{2 π}{2}$

Step 10.3.2

Simplify the left side.

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Step 10.3.2.1

Cancel the common factor of $2$ .

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Step 10.3.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{2 π}{2}$

Step 10.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{2 π}{2}$

Step 10.3.3

Simplify the right side.

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Step 10.3.3.1

Cancel the common factor of $2$ .

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Step 10.3.3.1.1

Cancel the common factor.

$x = \frac{2 π}{2}$

Step 10.3.3.1.2

Divide $π$ by $1$ .

$x = π$

Step 11

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

$x = \frac{π}{2}, π$

Step 12

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.

$- 16 \cos (2 (\frac{π}{2}) - π)$

Step 13

Evaluate the second derivative.

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Step 13.1

Cancel the common factor of $2$ .

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Step 13.1.1

Cancel the common factor.

$- 16 \cos (2 \frac{π}{2} - π)$

Step 13.1.2

Rewrite the expression.

$- 16 \cos (π - π)$

Step 13.2

Subtract $π$ from $π$ .

$- 16 \cos (0)$

Step 13.3

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

$- 16 \cdot 1$

Step 13.4

Multiply $- 16$ by $1$ .

$- 16$

Step 14

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

$x = \frac{π}{2}$ is a local maximum

Step 15

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

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Step 15.1

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

$g (\frac{π}{2}) = 4 \cos (2 (\frac{π}{2}) - π) - 2$

Step 15.2

Simplify the result.

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Step 15.2.1

Simplify each term.

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Step 15.2.1.1

Cancel the common factor of $2$ .

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Step 15.2.1.1.1

Cancel the common factor.

$g (\frac{π}{2}) = 4 \cos (2 (\frac{π}{2}) - π) - 2$

Step 15.2.1.1.2

Rewrite the expression.

$g (\frac{π}{2}) = 4 \cos (π - π) - 2$

Step 15.2.1.2

Subtract $π$ from $π$ .

$g (\frac{π}{2}) = 4 \cos (0) - 2$

Step 15.2.1.3

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

$g (\frac{π}{2}) = 4 \cdot 1 - 2$

Step 15.2.1.4

Multiply $4$ by $1$ .

$g (\frac{π}{2}) = 4 - 2$

Step 15.2.2

Subtract $2$ from $4$ .

$g (\frac{π}{2}) = 2$

Step 15.2.3

The final answer is $2$ .

$y = 2$

Step 16

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

$- 16 \cos (2 (π) - π)$

Step 17

Evaluate the second derivative.

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Step 17.1

Subtract $π$ from $2 π$ .

$- 16 \cos (π)$

Step 17.2

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.

$- 16 (- \cos (0))$

Step 17.3

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

$- 16 (- 1 \cdot 1)$

Step 17.4

Multiply $- 16 (- 1 \cdot 1)$ .

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Step 17.4.1

Multiply $- 1$ by $1$ .

$- 16 \cdot - 1$

Step 17.4.2

Multiply $- 16$ by $- 1$ .

$16$

Step 18

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

$x = π$ is a local minimum

Step 19

Find the y-value when $x = π$ .

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Step 19.1

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

$g (π) = 4 \cos (2 (π) - π) - 2$

Step 19.2

Simplify the result.

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Step 19.2.1

Simplify each term.

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Step 19.2.1.1

Subtract $π$ from $2 π$ .

$g (π) = 4 \cos (π) - 2$

Step 19.2.1.2

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.

$g (π) = 4 (- \cos (0)) - 2$

Step 19.2.1.3

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

$g (π) = 4 (- 1 \cdot 1) - 2$

Step 19.2.1.4

Multiply $4 (- 1 \cdot 1)$ .

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Step 19.2.1.4.1

Multiply $- 1$ by $1$ .

$g (π) = 4 \cdot - 1 - 2$

Step 19.2.1.4.2

Multiply $4$ by $- 1$ .

$g (π) = - 4 - 2$

Step 19.2.2

Subtract $2$ from $- 4$ .

$g (π) = - 6$

Step 19.2.3

The final answer is $- 6$ .

$y = - 6$

Step 20

These are the local extrema for $g (x) = 4 \cos (2 x - π) - 2$ .

$(\frac{π}{2}, 2)$ is a local maxima

$(π, - 6)$ is a local minima

Step 21