Calculus Examples

$- \sin^{2} (x) + \cos^{2} (x)$

Step 1

Find the first derivative of the function.

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Step 1.1

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

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

Step 1.2

Evaluate $\frac{d}{d x} [- \sin^{2} (x)]$ .

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Step 1.2.1

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

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

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

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Step 1.2.2.1

To apply the Chain Rule, set $u_{1}$ as $\sin (x)$ .

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

Step 1.2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 2$ .

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

Step 1.2.2.3

Replace all occurrences of $u_{1}$ with $\sin (x)$ .

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

Step 1.2.3

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

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

Step 1.2.4

Multiply $2$ by $- 1$ .

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

Step 1.3

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

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Step 1.3.1

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

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Step 1.3.1.1

To apply the Chain Rule, set $u_{2}$ as $\cos (x)$ .

$- 2 \sin (x) \cos (x) + \frac{d}{d u_{2}} [{u_{2}}^{2}] \frac{d}{d x} [\cos (x)]$

Step 1.3.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 2$ .

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

Step 1.3.1.3

Replace all occurrences of $u_{2}$ with $\cos (x)$ .

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

Step 1.3.2

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

$- 2 \sin (x) \cos (x) + 2 \cos (x) (- \sin (x))$

Step 1.3.3

Multiply $- 1$ by $2$ .

$- 2 \sin (x) \cos (x) - 2 \cos (x) \sin (x)$

Step 1.4

Combine terms.

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Step 1.4.1

Reorder the factors of $- 2 \sin (x) \cos (x)$ .

$- 2 \cos (x) \sin (x) - 2 \cos (x) \sin (x)$

Step 1.4.2

Subtract $2 \cos (x) \sin (x)$ from $- 2 \cos (x) \sin (x)$ .

$- 4 \cos (x) \sin (x)$

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 $- 4 \cos (x) \sin (x)$ with respect to $x$ is $- 4 \frac{d}{d x} [\cos (x) \sin (x)]$ .

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

Step 2.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = \cos (x)$ and $g (x) = \sin (x)$ .

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

Step 2.3

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

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

Step 2.4

Raise $\cos (x)$ to the power of $1$ .

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

Step 2.5

Raise $\cos (x)$ to the power of $1$ .

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

Step 2.6

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 2.7

Add $1$ and $1$ .

$f'' (x) = - 4 (\cos^{2} (x) + \sin (x) \frac{d}{d x} (\cos (x)))$

Step 2.8

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

$f'' (x) = - 4 (\cos^{2} (x) + \sin (x) (- \sin (x)))$

Step 2.9

Raise $\sin (x)$ to the power of $1$ .

$f'' (x) = - 4 (\cos^{2} (x) - (\sin (x) \sin (x)))$

Step 2.10

Raise $\sin (x)$ to the power of $1$ .

$f'' (x) = - 4 (\cos^{2} (x) - (\sin (x) \sin (x)))$

Step 2.11

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - 4 (\cos^{2} (x) - {\sin (x)}^{1 + 1})$

Step 2.12

Add $1$ and $1$ .

$f'' (x) = - 4 (\cos^{2} (x) - \sin^{2} (x))$

Step 2.13

Simplify.

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Step 2.13.1

Apply the distributive property.

$f'' (x) = - 4 \cos^{2} (x) - 4 (- \sin^{2} (x))$

Step 2.13.2

Multiply $- 1$ by $- 4$ .

$f'' (x) = - 4 \cos^{2} (x) + 4 \sin^{2} (x)$

Step 2.13.3

Rewrite $4 \sin^{2} (x)$ as ${(2 \sin (x))}^{2}$ .

$f'' (x) = - 4 \cos^{2} (x) + {(2 \sin (x))}^{2}$

Step 2.13.4

Rewrite $4 \cos^{2} (x)$ as ${(2 \cos (x))}^{2}$ .

$f'' (x) = - {(2 \cos (x))}^{2} + {(2 \sin (x))}^{2}$

Step 2.13.5

Reorder $- {(2 \cos (x))}^{2}$ and ${(2 \sin (x))}^{2}$ .

$f'' (x) = {(2 \sin (x))}^{2} - {(2 \cos (x))}^{2}$

Step 2.13.6

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 \sin (x)$ and $b = 2 \cos (x)$ .

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

Step 2.13.7

Multiply $2$ by $- 1$ .

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

Step 2.13.8

Expand $(2 \sin (x) + 2 \cos (x)) (2 \sin (x) - 2 \cos (x))$ using the FOIL Method.

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Step 2.13.8.1

Apply the distributive property.

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

Step 2.13.8.2

Apply the distributive property.

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

Step 2.13.8.3

Apply the distributive property.

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

Step 2.13.9

Combine the opposite terms in $2 \sin (x) (2 \sin (x)) + 2 \sin (x) (- 2 \cos (x)) + 2 \cos (x) (2 \sin (x)) + 2 \cos (x) (- 2 \cos (x))$ .

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Step 2.13.9.1

Reorder the factors in the terms $2 \sin (x) (- 2 \cos (x))$ and $2 \cos (x) (2 \sin (x))$ .

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

Step 2.13.9.2

Add $- 2 \cdot 2 \cos (x) \sin (x)$ and $2 \cdot 2 \cos (x) \sin (x)$ .

$f'' (x) = 2 \sin (x) (2 \sin (x)) + 0 + 2 \cos (x) (- 2 \cos (x))$

Step 2.13.9.3

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

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

Step 2.13.10

Simplify each term.

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Step 2.13.10.1

Multiply $2 \sin (x) (2 \sin (x))$ .

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Step 2.13.10.1.1

Multiply $2$ by $2$ .

$f'' (x) = 4 \sin (x) \sin (x) + 2 \cos (x) (- 2 \cos (x))$

Step 2.13.10.1.2

Raise $\sin (x)$ to the power of $1$ .

$f'' (x) = 4 (\sin (x) \sin (x)) + 2 \cos (x) (- 2 \cos (x))$

Step 2.13.10.1.3

Raise $\sin (x)$ to the power of $1$ .

$f'' (x) = 4 (\sin (x) \sin (x)) + 2 \cos (x) (- 2 \cos (x))$

Step 2.13.10.1.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = 4 {\sin (x)}^{1 + 1} + 2 \cos (x) (- 2 \cos (x))$

Step 2.13.10.1.5

Add $1$ and $1$ .

$f'' (x) = 4 \sin^{2} (x) + 2 \cos (x) (- 2 \cos (x))$

Step 2.13.10.2

Multiply $2 \cos (x) (- 2 \cos (x))$ .

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Step 2.13.10.2.1

Multiply $- 2$ by $2$ .

$f'' (x) = 4 \sin^{2} (x) - 4 \cos (x) \cos (x)$

Step 2.13.10.2.2

Raise $\cos (x)$ to the power of $1$ .

$f'' (x) = 4 \sin^{2} (x) - 4 (\cos (x) \cos (x))$

Step 2.13.10.2.3

Raise $\cos (x)$ to the power of $1$ .

$f'' (x) = 4 \sin^{2} (x) - 4 (\cos (x) \cos (x))$

Step 2.13.10.2.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = 4 \sin^{2} (x) - 4 {\cos (x)}^{1 + 1}$

Step 2.13.10.2.5

Add $1$ and $1$ .

$f'' (x) = 4 \sin^{2} (x) - 4 \cos^{2} (x)$

Step 3

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

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

Step 4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$\cos (x) = 0$

$\sin (x) = 0$

Step 5

Set $\cos (x)$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $\cos (x)$ equal to $0$ .

$\cos (x) = 0$

Step 5.2

Solve $\cos (x) = 0$ for $x$ .

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Step 5.2.1

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

$x = \arccos (0)$

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

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

$x = \frac{π}{2}$

Step 5.2.3

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.

$x = 2 π - \frac{π}{2}$

Step 5.2.4

Simplify $2 π - \frac{π}{2}$ .

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Step 5.2.4.1

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

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

Step 5.2.4.2

Combine fractions.

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Step 5.2.4.2.1

Combine $2 π$ and $\frac{2}{2}$ .

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

Step 5.2.4.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Step 5.2.4.3

Simplify the numerator.

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Step 5.2.4.3.1

Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Step 5.2.4.3.2

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

Step 5.2.5

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

$x = \frac{π}{2}, \frac{3 π}{2}$

Step 6

Set $\sin (x)$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $\sin (x)$ equal to $0$ .

$\sin (x) = 0$

Step 6.2

Solve $\sin (x) = 0$ for $x$ .

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Step 6.2.1

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

$x = \arcsin (0)$

Step 6.2.2

Simplify the right side.

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Step 6.2.2.1

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

$x = 0$

Step 6.2.3

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.

$x = π - 0$

Step 6.2.4

Subtract $0$ from $π$ .

$x = π$

Step 6.2.5

The solution to the equation $x = 0$ .

$x = 0, π$

Step 7

The final solution is all the values that make $- 4 \cos (x) \sin (x) = 0$ true.

$x = \frac{π}{2}, \frac{3 π}{2}, 0, π$

Step 8

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.

$4 \sin^{2} (\frac{π}{2}) - 4 \cos^{2} (\frac{π}{2})$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify each term.

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Step 9.1.1

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

$4 \cdot 1^{2} - 4 \cos^{2} (\frac{π}{2})$

Step 9.1.2

One to any power is one.

$4 \cdot 1 - 4 \cos^{2} (\frac{π}{2})$

Step 9.1.3

Multiply $4$ by $1$ .

$4 - 4 \cos^{2} (\frac{π}{2})$

Step 9.1.4

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

$4 - 4 \cdot 0^{2}$

Step 9.1.5

Raising $0$ to any positive power yields $0$ .

$4 - 4 \cdot 0$

Step 9.1.6

Multiply $- 4$ by $0$ .

$4 + 0$

Step 9.2

Add $4$ and $0$ .

$4$

Step 10

$x = \frac{π}{2}$ 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 11

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

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Step 11.1

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

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

Step 11.2

Simplify the result.

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Step 11.2.1

Reorder $- \sin^{2} (\frac{π}{2})$ and $\cos^{2} (\frac{π}{2})$ .

$f (\frac{π}{2}) = \cos^{2} (\frac{π}{2}) - \sin^{2} (\frac{π}{2})$

Step 11.2.2

Apply the cosine double-angle identity.

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

Step 11.2.3

Cancel the common factor of $2$ .

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Step 11.2.3.1

Cancel the common factor.

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

Step 11.2.3.2

Rewrite the expression.

$f (\frac{π}{2}) = \cos (π)$

Step 11.2.4

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.

$f (\frac{π}{2}) = - \cos (0)$

Step 11.2.5

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

$f (\frac{π}{2}) = - 1 \cdot 1$

Step 11.2.6

Multiply $- 1$ by $1$ .

$f (\frac{π}{2}) = - 1$

Step 11.2.7

The final answer is $- 1$ .

$y = - 1$

Step 12

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.

$4 \sin^{2} (\frac{3 π}{2}) - 4 \cos^{2} (\frac{3 π}{2})$

Step 13

Evaluate the second derivative.

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Step 13.1

Simplify each term.

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Step 13.1.1

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.

$4 {(- \sin (\frac{π}{2}))}^{2} - 4 \cos^{2} (\frac{3 π}{2})$

Step 13.1.2

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

$4 {(- 1 \cdot 1)}^{2} - 4 \cos^{2} (\frac{3 π}{2})$

Step 13.1.3

Multiply $- 1$ by $1$ .

$4 {(- 1)}^{2} - 4 \cos^{2} (\frac{3 π}{2})$

Step 13.1.4

Raise $- 1$ to the power of $2$ .

$4 \cdot 1 - 4 \cos^{2} (\frac{3 π}{2})$

Step 13.1.5

Multiply $4$ by $1$ .

$4 - 4 \cos^{2} (\frac{3 π}{2})$

Step 13.1.6

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$4 - 4 \cos^{2} (\frac{π}{2})$

Step 13.1.7

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

$4 - 4 \cdot 0^{2}$

Step 13.1.8

Raising $0$ to any positive power yields $0$ .

$4 - 4 \cdot 0$

Step 13.1.9

Multiply $- 4$ by $0$ .

$4 + 0$

Step 13.2

Add $4$ and $0$ .

$4$

Step 14

$x = \frac{3 π}{2}$ 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 15

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

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Step 15.1

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

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

Step 15.2

Simplify the result.

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Step 15.2.1

Reorder $- \sin^{2} (\frac{3 π}{2})$ and $\cos^{2} (\frac{3 π}{2})$ .

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

Step 15.2.2

Apply the cosine double-angle identity.

$f (\frac{3 π}{2}) = \cos (2 (\frac{3 π}{2}))$

Step 15.2.3

Cancel the common factor of $2$ .

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Step 15.2.3.1

Cancel the common factor.

$f (\frac{3 π}{2}) = \cos (2 (\frac{3 π}{2}))$

Step 15.2.3.2

Rewrite the expression.

$f (\frac{3 π}{2}) = \cos (3 π)$

Step 15.2.4

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

$f (\frac{3 π}{2}) = \cos (π)$

Step 15.2.5

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.

$f (\frac{3 π}{2}) = - \cos (0)$

Step 15.2.6

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

$f (\frac{3 π}{2}) = - 1 \cdot 1$

Step 15.2.7

Multiply $- 1$ by $1$ .

$f (\frac{3 π}{2}) = - 1$

Step 15.2.8

The final answer is $- 1$ .

$y = - 1$

Step 16

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.

$4 \sin^{2} (0) - 4 \cos^{2} (0)$

Step 17

Evaluate the second derivative.

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Step 17.1

Simplify each term.

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Step 17.1.1

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

$4 \cdot 0^{2} - 4 \cos^{2} (0)$

Step 17.1.2

Raising $0$ to any positive power yields $0$ .

$4 \cdot 0 - 4 \cos^{2} (0)$

Step 17.1.3

Multiply $4$ by $0$ .

$0 - 4 \cos^{2} (0)$

Step 17.1.4

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

$0 - 4 \cdot 1^{2}$

Step 17.1.5

One to any power is one.

$0 - 4 \cdot 1$

Step 17.1.6

Multiply $- 4$ by $1$ .

$0 - 4$

Step 17.2

Subtract $4$ from $0$ .

$- 4$

Step 18

$x = 0$ 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 19

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

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Step 19.1

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

$f (0) = - \sin^{2} (0) + \cos^{2} (0)$

Step 19.2

Simplify the result.

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Step 19.2.1

Reorder $- \sin^{2} (0)$ and $\cos^{2} (0)$ .

$f (0) = \cos^{2} (0) - \sin^{2} (0)$

Step 19.2.2

Apply the cosine double-angle identity.

$f (0) = \cos (2 \cdot 0)$

Step 19.2.3

Multiply $2$ by $0$ .

$f (0) = \cos (0)$

Step 19.2.4

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

$f (0) = 1$

Step 19.2.5

The final answer is $1$ .

$y = 1$

Step 20

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.

$4 \sin^{2} (π) - 4 \cos^{2} (π)$

Step 21

Evaluate the second derivative.

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Step 21.1

Simplify each term.

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Step 21.1.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$4 \sin^{2} (0) - 4 \cos^{2} (π)$

Step 21.1.2

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

$4 \cdot 0^{2} - 4 \cos^{2} (π)$

Step 21.1.3

Raising $0$ to any positive power yields $0$ .

$4 \cdot 0 - 4 \cos^{2} (π)$

Step 21.1.4

Multiply $4$ by $0$ .

$0 - 4 \cos^{2} (π)$

Step 21.1.5

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.

$0 - 4 {(- \cos (0))}^{2}$

Step 21.1.6

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

$0 - 4 {(- 1 \cdot 1)}^{2}$

Step 21.1.7

Multiply $- 1$ by $1$ .

$0 - 4 {(- 1)}^{2}$

Step 21.1.8

Raise $- 1$ to the power of $2$ .

$0 - 4 \cdot 1$

Step 21.1.9

Multiply $- 4$ by $1$ .

$0 - 4$

Step 21.2

Subtract $4$ from $0$ .

$- 4$

Step 22

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

$x = π$ is a local maximum

Step 23

Find the y-value when $x = π$ .

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Step 23.1

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

$f (π) = - \sin^{2} (π) + \cos^{2} (π)$

Step 23.2

Simplify the result.

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Step 23.2.1

Reorder $- \sin^{2} (π)$ and $\cos^{2} (π)$ .

$f (π) = \cos^{2} (π) - \sin^{2} (π)$

Step 23.2.2

Apply the cosine double-angle identity.

$f (π) = \cos (2 π)$

Step 23.2.3

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

$f (π) = \cos (0)$

Step 23.2.4

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

$f (π) = 1$

Step 23.2.5

The final answer is $1$ .

$y = 1$

Step 24

These are the local extrema for $f (x) = - \sin^{2} (x) + \cos^{2} (x)$ .

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

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

$(0, 1)$ is a local maxima

$(π, 1)$ is a local maxima

Step 25