Trigonometry Examples

$y = 3 \cos (2 θ)$

Step 1

Find the first derivative of the function.

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Step 1.1

Since $3$ is constant with respect to $θ$ , the derivative of $3 \cos (2 θ)$ with respect to $θ$ is $3 \frac{d}{d θ} [\cos (2 θ)]$ .

$3 \frac{d}{d θ} [\cos (2 θ)]$

Step 1.2

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

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Step 1.2.1

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

$3 (\frac{d}{d u} [\cos (u)] \frac{d}{d θ} [2 θ])$

Step 1.2.2

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

$3 (- \sin (u) \frac{d}{d θ} [2 θ])$

Step 1.2.3

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

$3 (- \sin (2 θ) \frac{d}{d θ} [2 θ])$

Step 1.3

Differentiate.

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Step 1.3.1

Multiply $- 1$ by $3$ .

$- 3 (\sin (2 θ) \frac{d}{d θ} [2 θ])$

Step 1.3.2

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

$- 3 \sin (2 θ) (2 \frac{d}{d θ} [θ])$

Step 1.3.3

Multiply $2$ by $- 3$ .

$- 6 \sin (2 θ) \frac{d}{d θ} [θ]$

Step 1.3.4

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

$- 6 \sin (2 θ) \cdot 1$

Step 1.3.5

Multiply $- 6$ by $1$ .

$- 6 \sin (2 θ)$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $- 6$ is constant with respect to $θ$ , the derivative of $- 6 \sin (2 θ)$ with respect to $θ$ is $- 6 \frac{d}{d θ} [\sin (2 θ)]$ .

$f'' (θ) = - 6 \frac{d}{d θ} \sin (2 θ)$

Step 2.2

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

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Step 2.2.1

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

$f'' (θ) = - 6 (\frac{d}{d u} (\sin (u)) \frac{d}{d θ} (2 θ))$

Step 2.2.2

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

$f'' (θ) = - 6 (\cos (u) \frac{d}{d θ} (2 θ))$

Step 2.2.3

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

$f'' (θ) = - 6 (\cos (2 θ) \frac{d}{d θ} (2 θ))$

Step 2.3

Differentiate.

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Step 2.3.1

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

$f'' (θ) = - 6 \cos (2 θ) (2 \frac{d}{d θ} (θ))$

Step 2.3.2

Multiply $2$ by $- 6$ .

$f'' (θ) = - 12 \cos (2 θ) \frac{d}{d θ} θ$

Step 2.3.3

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

$f'' (θ) = - 12 \cos (2 θ) \cdot 1$

Step 2.3.4

Multiply $- 12$ by $1$ .

$f'' (θ) = - 12 \cos (2 θ)$

Step 3

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

$- 6 \sin (2 θ) = 0$

Step 4

Divide each term in $- 6 \sin (2 θ) = 0$ by $- 6$ and simplify.

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Step 4.1

Divide each term in $- 6 \sin (2 θ) = 0$ by $- 6$ .

$\frac{- 6 \sin (2 θ)}{- 6} = \frac{0}{- 6}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $- 6$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{- 6 \sin (2 θ)}{- 6} = \frac{0}{- 6}$

Step 4.2.1.2

Divide $\sin (2 θ)$ by $1$ .

$\sin (2 θ) = \frac{0}{- 6}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $- 6$ .

$\sin (2 θ) = 0$

Step 5

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

$2 θ = \arcsin (0)$

Step 6

Simplify the right side.

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Step 6.1

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

$2 θ = 0$

Step 7

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

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Step 7.1

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

$\frac{2 θ}{2} = \frac{0}{2}$

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.

$\frac{2 θ}{2} = \frac{0}{2}$

Step 7.2.1.2

Divide $θ$ by $1$ .

$θ = \frac{0}{2}$

Step 7.3

Simplify the right side.

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Step 7.3.1

Divide $0$ by $2$ .

$θ = 0$

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.

$2 θ = π - 0$

Step 9

Solve for $θ$ .

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Step 9.1

Simplify.

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Step 9.1.1

Multiply $- 1$ by $0$ .

$2 θ = π + 0$

Step 9.1.2

Add $π$ and $0$ .

$2 θ = π$

Step 9.2

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

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Step 9.2.1

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

$\frac{2 θ}{2} = \frac{π}{2}$

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.

$\frac{2 θ}{2} = \frac{π}{2}$

Step 9.2.2.1.2

Divide $θ$ by $1$ .

$θ = \frac{π}{2}$

Step 10

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

$θ = 0, \frac{π}{2}$

Step 11

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

$- 12 \cos (2 (0))$

Step 12

Evaluate the second derivative.

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Step 12.1

Multiply $2$ by $0$ .

$- 12 \cos (0)$

Step 12.2

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

$- 12 \cdot 1$

Step 12.3

Multiply $- 12$ by $1$ .

$- 12$

Step 13

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

$θ = 0$ is a local maximum

Step 14

Find the y-value when $θ = 0$ .

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Step 14.1

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

$f (0) = 3 \cos (2 (0))$

Step 14.2

Simplify the result.

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Step 14.2.1

Multiply $2$ by $0$ .

$f (0) = 3 \cos (0)$

Step 14.2.2

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

$f (0) = 3 \cdot 1$

Step 14.2.3

Multiply $3$ by $1$ .

$f (0) = 3$

Step 14.2.4

The final answer is $3$ .

$y = 3$

Step 15

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

$- 12 \cos (2 (\frac{π}{2}))$

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.

$- 12 \cos (2 \frac{π}{2})$

Step 16.1.2

Rewrite the expression.

$- 12 \cos (π)$

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.

$- 12 (- \cos (0))$

Step 16.3

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

$- 12 (- 1 \cdot 1)$

Step 16.4

Multiply $- 12 (- 1 \cdot 1)$ .

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Step 16.4.1

Multiply $- 1$ by $1$ .

$- 12 \cdot - 1$

Step 16.4.2

Multiply $- 12$ by $- 1$ .

$12$

Step 17

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

$θ = \frac{π}{2}$ is a local minimum

Step 18

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

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Step 18.1

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

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

Step 18.2

Simplify the result.

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Step 18.2.1

Cancel the common factor of $2$ .

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Step 18.2.1.1

Cancel the common factor.

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

Step 18.2.1.2

Rewrite the expression.

$f (\frac{π}{2}) = 3 \cos (π)$

Step 18.2.2

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.

$f (\frac{π}{2}) = 3 (- \cos (0))$

Step 18.2.3

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

$f (\frac{π}{2}) = 3 (- 1 \cdot 1)$

Step 18.2.4

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

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Step 18.2.4.1

Multiply $- 1$ by $1$ .

$f (\frac{π}{2}) = 3 \cdot - 1$

Step 18.2.4.2

Multiply $3$ by $- 1$ .

$f (\frac{π}{2}) = - 3$

Step 18.2.5

The final answer is $- 3$ .

$y = - 3$

Step 19

These are the local extrema for $f (θ) = 3 \cos (2 θ)$ .

$(0, 3)$ is a local maxima

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

Step 20