Trigonometry Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

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

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

Step 1.2

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

$\frac{2}{3} (- \sin (t))$

Step 1.3

Combine $\frac{2}{3}$ and $\sin (t)$ .

$- \frac{2 \sin (t)}{3}$

Step 2

Find the second derivative of the function.

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Step 2.1

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

$f'' (t) = - \frac{2}{3} \cdot \frac{d}{d t} \sin (t)$

Step 2.2

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

$f'' (t) = - \frac{2}{3} \cdot \cos (t)$

Step 2.3

Simplify.

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Step 2.3.1

Combine $\cos (t)$ and $\frac{2}{3}$ .

$f'' (t) = - \frac{\cos (t) \cdot 2}{3}$

Step 2.3.2

Move $2$ to the left of $\cos (t)$ .

$f'' (t) = - \frac{2 \cos (t)}{3}$

Step 3

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

$- \frac{2 \sin (t)}{3} = 0$

Step 4

Set the numerator equal to zero.

$2 \sin (t) = 0$

Step 5

Solve the equation for $t$ .

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Step 5.1

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

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Step 5.1.1

Divide each term in $2 \sin (t) = 0$ by $2$ .

$\frac{2 \sin (t)}{2} = \frac{0}{2}$

Step 5.1.2

Simplify the left side.

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Step 5.1.2.1

Cancel the common factor of $2$ .

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Step 5.1.2.1.1

Cancel the common factor.

$\frac{2 \sin (t)}{2} = \frac{0}{2}$

Step 5.1.2.1.2

Divide $\sin (t)$ by $1$ .

$\sin (t) = \frac{0}{2}$

Step 5.1.3

Simplify the right side.

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Step 5.1.3.1

Divide $0$ by $2$ .

$\sin (t) = 0$

Step 5.2

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

$t = \arcsin (0)$

Step 5.3

Simplify the right side.

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Step 5.3.1

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

$t = 0$

Step 5.4

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.

$t = π - 0$

Step 5.5

Subtract $0$ from $π$ .

$t = π$

Step 5.6

The solution to the equation $t = 0$ .

$t = 0, π$

Step 6

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

$- \frac{2 \cos (0)}{3}$

Step 7

Evaluate the second derivative.

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Step 7.1

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

$- \frac{2 \cdot 1}{3}$

Step 7.2

Multiply $2$ by $1$ .

$- \frac{2}{3}$

Step 8

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

$t = 0$ is a local maximum

Step 9

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

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Step 9.1

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

$f (0) = \frac{2}{3} \cdot \cos (0)$

Step 9.2

Simplify the result.

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Step 9.2.1

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

$f (0) = \frac{2}{3} \cdot 1$

Step 9.2.2

Multiply $\frac{2}{3}$ by $1$ .

$f (0) = \frac{2}{3}$

Step 9.2.3

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

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

Step 10

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

$- \frac{2 \cos (π)}{3}$

Step 11

Evaluate the second derivative.

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Step 11.1

Simplify the numerator.

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Step 11.1.1

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.

$- \frac{2 (- \cos (0))}{3}$

Step 11.1.2

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

$- \frac{2 (- 1 \cdot 1)}{3}$

Step 11.1.3

Multiply $- 1$ by $1$ .

$- \frac{2 \cdot - 1}{3}$

Step 11.2

Simplify the expression.

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Step 11.2.1

Multiply $2$ by $- 1$ .

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

Step 11.2.2

Move the negative in front of the fraction.

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

Step 11.3

Multiply $- - \frac{2}{3}$ .

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Step 11.3.1

Multiply $- 1$ by $- 1$ .

$1 (\frac{2}{3})$

Step 11.3.2

Multiply $\frac{2}{3}$ by $1$ .

$\frac{2}{3}$

Step 12

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

$t = π$ is a local minimum

Step 13

Find the y-value when $t = π$ .

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Step 13.1

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

$f (π) = \frac{2}{3} \cdot \cos (π)$

Step 13.2

Simplify the result.

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Step 13.2.1

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.

$f (π) = \frac{2}{3} \cdot (- \cos (0))$

Step 13.2.2

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

$f (π) = \frac{2}{3} \cdot (- 1 \cdot 1)$

Step 13.2.3

Multiply $- 1$ by $1$ .

$f (π) = \frac{2}{3} \cdot - 1$

Step 13.2.4

Multiply $\frac{2}{3} \cdot - 1$ .

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Step 13.2.4.1

Combine $\frac{2}{3}$ and $- 1$ .

$f (π) = \frac{2 \cdot - 1}{3}$

Step 13.2.4.2

Multiply $2$ by $- 1$ .

$f (π) = \frac{- 2}{3}$

Step 13.2.5

Move the negative in front of the fraction.

$f (π) = - \frac{2}{3}$

Step 13.2.6

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

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

Step 14

These are the local extrema for $f (t) = \frac{2}{3} \cdot \cos (t)$ .

$(0, \frac{2}{3})$ is a local maxima

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

Step 15