Trigonometry Examples

y = sin (x) - 6

$y = \sin (x) - 6$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of

sin (x) - 6

$\sin (x) - 6$ with respect to

x

$x$ is

\frac{d}{d x} [sin (x)] + \frac{d}{d x} [- 6]

$\frac{d}{d x} [\sin (x)] + \frac{d}{d x} [- 6]$ .

\frac{d}{d x} [sin (x)] + \frac{d}{d x} [- 6]

$\frac{d}{d x} [\sin (x)] + \frac{d}{d x} [- 6]$

Step 1.2

The derivative of

sin (x)

$\sin (x)$ with respect to

x

$x$ is

cos (x)

$\cos (x)$ .

cos (x) + \frac{d}{d x} [- 6]

$\cos (x) + \frac{d}{d x} [- 6]$

Step 1.3

Differentiate using the Constant Rule.

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Step 1.3.1

Since

- 6

$- 6$ is constant with respect to

x

$x$ , the derivative of

- 6

$- 6$ with respect to

x

$x$ is

0

$0$ .

cos (x) + 0

$\cos (x) + 0$

Step 1.3.2

Add

cos (x)

$\cos (x)$ and

0

$0$ .

cos (x)

$\cos (x)$

cos (x)

$\cos (x)$

cos (x)

$\cos (x)$

Step 2

The derivative of

cos (x)

$\cos (x)$ with respect to

x

$x$ is

- sin (x)

$- \sin (x)$ .

$f'' (x) = - \sin (x)$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to

$0$ and solve.

$\cos (x) = 0$

Step 4

Take the inverse cosine of both sides of the equation to extract

$x$ from inside the cosine.

$x = \arccos (0)$

Step 5

Simplify the right side.

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Step 5.1

The exact value of

$\arccos (0)$ is

$\frac{π}{2}$ .

$x = \frac{π}{2}$

Step 6

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 7

Simplify

$2 π - \frac{π}{2}$ .

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Step 7.1

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 7.2

Combine fractions.

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Step 7.2.1

Combine

$2 π$ and

$\frac{2}{2}$ .

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

Step 7.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Step 7.3

Simplify the numerator.

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Step 7.3.1

Multiply

$2$ by

$2$ .

$x = \frac{4 π - π}{2}$

Step 7.3.2

Subtract

$π$ from

$4 π$ .

$x = \frac{3 π}{2}$

Step 8

The solution to the equation

$x = \frac{π}{2}$ .

$x = \frac{π}{2}, \frac{3 π}{2}$

Step 9

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.

$- \sin (\frac{π}{2})$

Step 10

Evaluate the second derivative.

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Step 10.1

The exact value of

$\sin (\frac{π}{2})$ is

$1$ .

$- 1 \cdot 1$

Step 10.2

Multiply

$- 1$ by

$1$ .

$- 1$

Step 11

$x = \frac{π}{2}$ 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 12

Find the y-value when

$x = \frac{π}{2}$ .

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Step 12.1

Replace the variable

$x$ with

$\frac{π}{2}$ in the expression.

$f (\frac{π}{2}) = \sin (\frac{π}{2}) - 6$

Step 12.2

Simplify the result.

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Step 12.2.1

The exact value of

$\sin (\frac{π}{2})$ is

$1$ .

$f (\frac{π}{2}) = 1 - 6$

Step 12.2.2

Subtract

$6$ from

$1$ .

$f (\frac{π}{2}) = - 5$

Step 12.2.3

The final answer is

$- 5$ .

$y = - 5$

Step 13

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.

$- \sin (\frac{3 π}{2})$

Step 14

Evaluate the second derivative.

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Step 14.1

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.

$- - \sin (\frac{π}{2})$

Step 14.2

The exact value of

$\sin (\frac{π}{2})$ is

$1$ .

$- (- 1 \cdot 1)$

Step 14.3

Multiply

$- (- 1 \cdot 1)$ .

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Step 14.3.1

Multiply

$- 1$ by

$1$ .

$- - 1$

Step 14.3.2

Multiply

$- 1$ by

$- 1$ .

$1$

Step 15

$x = \frac{3 π}{2}$ 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 16

Find the y-value when

$x = \frac{3 π}{2}$ .

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Step 16.1

Replace the variable

$x$ with

$\frac{3 π}{2}$ in the expression.

$f (\frac{3 π}{2}) = \sin (\frac{3 π}{2}) - 6$

Step 16.2

Simplify the result.

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Step 16.2.1

Simplify each term.

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Step 16.2.1.1

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.

$f (\frac{3 π}{2}) = - \sin (\frac{π}{2}) - 6$

Step 16.2.1.2

The exact value of

$\sin (\frac{π}{2})$ is

$1$ .

$f (\frac{3 π}{2}) = - 1 \cdot 1 - 6$

Step 16.2.1.3

Multiply

$- 1$ by

$1$ .

$f (\frac{3 π}{2}) = - 1 - 6$

Step 16.2.2

Subtract

$6$ from

$- 1$ .

$f (\frac{3 π}{2}) = - 7$

Step 16.2.3

The final answer is

$- 7$ .

$y = - 7$

Step 17

These are the local extrema for

$f (x) = \sin (x) - 6$ .

$(\frac{π}{2}, - 5)$ is a local maxima

$(\frac{3 π}{2}, - 7)$ is a local minima

Step 18