Calculus Examples

$18 \cos (x) + 9 \sin (2 x)$

Step 1

Find the derivative.

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Step 1.1

By the Sum Rule, the derivative of $18 \cos (x) + 9 \sin (2 x)$ with respect to $x$ is $\frac{d}{d x} [18 \cos (x)] + \frac{d}{d x} [9 \sin (2 x)]$ .

$\frac{d}{d x} [18 \cos (x)] + \frac{d}{d x} [9 \sin (2 x)]$

Step 1.2

Evaluate $\frac{d}{d x} [18 \cos (x)]$ .

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Step 1.2.1

Since $18$ is constant with respect to $x$ , the derivative of $18 \cos (x)$ with respect to $x$ is $18 \frac{d}{d x} [\cos (x)]$ .

$18 \frac{d}{d x} [\cos (x)] + \frac{d}{d x} [9 \sin (2 x)]$

Step 1.2.2

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

$18 (- \sin (x)) + \frac{d}{d x} [9 \sin (2 x)]$

Step 1.2.3

Multiply $- 1$ by $18$ .

$- 18 \sin (x) + \frac{d}{d x} [9 \sin (2 x)]$

Step 1.3

Evaluate $\frac{d}{d x} [9 \sin (2 x)]$ .

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Step 1.3.1

Since $9$ is constant with respect to $x$ , the derivative of $9 \sin (2 x)$ with respect to $x$ is $9 \frac{d}{d x} [\sin (2 x)]$ .

$- 18 \sin (x) + 9 \frac{d}{d x} [\sin (2 x)]$

Step 1.3.2

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 $g (x) = 2 x$ .

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Step 1.3.2.1

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

$- 18 \sin (x) + 9 (\frac{d}{d u} [\sin (u)] \frac{d}{d x} [2 x])$

Step 1.3.2.2

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

$- 18 \sin (x) + 9 (\cos (u) \frac{d}{d x} [2 x])$

Step 1.3.2.3

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

$- 18 \sin (x) + 9 (\cos (2 x) \frac{d}{d x} [2 x])$

Step 1.3.3

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

$- 18 \sin (x) + 9 (\cos (2 x) (2 \frac{d}{d x} [x]))$

Step 1.3.4

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

$- 18 \sin (x) + 9 (\cos (2 x) (2 \cdot 1))$

Step 1.3.5

Multiply $2$ by $1$ .

$- 18 \sin (x) + 9 (\cos (2 x) \cdot 2)$

Step 1.3.6

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

$- 18 \sin (x) + 9 (2 \cdot \cos (2 x))$

Step 1.3.7

Multiply $2$ by $9$ .

$- 18 \sin (x) + 18 \cos (2 x)$

Step 2

Set the derivative equal to $0$ then solve the equation $- 18 \sin (x) + 18 \cos (2 x) = 0$ .

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Step 2.1

Simplify each term.

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Step 2.1.1

Use the double-angle identity to transform $\cos (2 x)$ to $1 - 2 \sin^{2} (x)$ .

$- 18 \sin (x) + 18 (1 - 2 \sin^{2} (x)) = 0$

Step 2.1.2

Apply the distributive property.

$- 18 \sin (x) + 18 \cdot 1 + 18 (- 2 \sin^{2} (x)) = 0$

Step 2.1.3

Multiply $18$ by $1$ .

$- 18 \sin (x) + 18 + 18 (- 2 \sin^{2} (x)) = 0$

Step 2.1.4

Multiply $- 2$ by $18$ .

$- 18 \sin (x) + 18 - 36 \sin^{2} (x) = 0$

Step 2.2

Factor $- 18 \sin (x) + 18 - 36 \sin^{2} (x)$ .

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Step 2.2.1

Factor $18$ out of $- 18 \sin (x) + 18 - 36 \sin^{2} (x)$ .

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Step 2.2.1.1

Factor $18$ out of $- 18 \sin (x)$ .

$18 (- \sin (x)) + 18 - 36 \sin^{2} (x) = 0$

Step 2.2.1.2

Factor $18$ out of $18$ .

$18 (- \sin (x)) + 18 (1) - 36 \sin^{2} (x) = 0$

Step 2.2.1.3

Factor $18$ out of $- 36 \sin^{2} (x)$ .

$18 (- \sin (x)) + 18 (1) + 18 (- 2 \sin^{2} (x)) = 0$

Step 2.2.1.4

Factor $18$ out of $18 (- \sin (x)) + 18 (1)$ .

$18 (- \sin (x) + 1) + 18 (- 2 \sin^{2} (x)) = 0$

Step 2.2.1.5

Factor $18$ out of $18 (- \sin (x) + 1) + 18 (- 2 \sin^{2} (x))$ .

$18 (- \sin (x) + 1 - 2 \sin^{2} (x)) = 0$

Step 2.2.2

Factor.

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Step 2.2.2.1

Factor by grouping.

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Step 2.2.2.1.1

Reorder terms.

$18 (- 2 \sin^{2} (x) - \sin (x) + 1) = 0$

Step 2.2.2.1.2

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 2 \cdot 1 = - 2$ and whose sum is $b = - 1$ .

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Step 2.2.2.1.2.1

Factor $- 1$ out of $- \sin (x)$ .

$18 (- 2 \sin^{2} (x) - \sin (x) + 1) = 0$

Step 2.2.2.1.2.2

Rewrite $- 1$ as $1$ plus $- 2$

$18 (- 2 \sin^{2} (x) + (1 - 2) \sin (x) + 1) = 0$

Step 2.2.2.1.2.3

Apply the distributive property.

$18 (- 2 \sin^{2} (x) + 1 \sin (x) - 2 \sin (x) + 1) = 0$

Step 2.2.2.1.2.4

Multiply $\sin (x)$ by $1$ .

$18 (- 2 \sin^{2} (x) + \sin (x) - 2 \sin (x) + 1) = 0$

Step 2.2.2.1.3

Factor out the greatest common factor from each group.

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Step 2.2.2.1.3.1

Group the first two terms and the last two terms.

$18 (- 2 \sin^{2} (x) + \sin (x) - 2 \sin (x) + 1) = 0$

Step 2.2.2.1.3.2

Factor out the greatest common factor (GCF) from each group.

$18 (\sin (x) (- 2 \sin (x) + 1) + 1 (- 2 \sin (x) + 1)) = 0$

Step 2.2.2.1.4

Factor the polynomial by factoring out the greatest common factor, $- 2 \sin (x) + 1$ .

$18 ((- 2 \sin (x) + 1) (\sin (x) + 1)) = 0$

Step 2.2.2.2

Remove unnecessary parentheses.

$18 (- 2 \sin (x) + 1) (\sin (x) + 1) = 0$

Step 2.3

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

$- 2 \sin (x) + 1 = 0$

$\sin (x) + 1 = 0$

Step 2.4

Set $- 2 \sin (x) + 1$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $- 2 \sin (x) + 1$ equal to $0$ .

$- 2 \sin (x) + 1 = 0$

Step 2.4.2

Solve $- 2 \sin (x) + 1 = 0$ for $x$ .

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Step 2.4.2.1

Subtract $1$ from both sides of the equation.

$- 2 \sin (x) = - 1$

Step 2.4.2.2

Divide each term in $- 2 \sin (x) = - 1$ by $- 2$ and simplify.

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Step 2.4.2.2.1

Divide each term in $- 2 \sin (x) = - 1$ by $- 2$ .

$\frac{- 2 \sin (x)}{- 2} = \frac{- 1}{- 2}$

Step 2.4.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.1

Cancel the common factor of $- 2$ .

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Step 2.4.2.2.2.1.1

Cancel the common factor.

$\frac{- 2 \sin (x)}{- 2} = \frac{- 1}{- 2}$

Step 2.4.2.2.2.1.2

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

$\sin (x) = \frac{- 1}{- 2}$

Step 2.4.2.2.3

Simplify the right side.

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Step 2.4.2.2.3.1

Dividing two negative values results in a positive value.

$\sin (x) = \frac{1}{2}$

Step 2.4.2.3

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

$x = \arcsin (\frac{1}{2})$

Step 2.4.2.4

Simplify the right side.

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Step 2.4.2.4.1

The exact value of $\arcsin (\frac{1}{2})$ is $\frac{π}{6}$ .

$x = \frac{π}{6}$

Step 2.4.2.5

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.

$x = π - \frac{π}{6}$

Step 2.4.2.6

Simplify $π - \frac{π}{6}$ .

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Step 2.4.2.6.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$x = π \cdot \frac{6}{6} - \frac{π}{6}$

Step 2.4.2.6.2

Combine fractions.

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Step 2.4.2.6.2.1

Combine $π$ and $\frac{6}{6}$ .

$x = \frac{π \cdot 6}{6} - \frac{π}{6}$

Step 2.4.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 - π}{6}$

Step 2.4.2.6.3

Simplify the numerator.

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Step 2.4.2.6.3.1

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π - π}{6}$

Step 2.4.2.6.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{6}$

Step 2.4.2.7

Find the period of $\sin (x)$ .

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Step 2.4.2.7.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 2.4.2.7.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 2.4.2.7.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 2.4.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 2.4.2.8

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer $n$

Step 2.5

Set $\sin (x) + 1$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $\sin (x) + 1$ equal to $0$ .

$\sin (x) + 1 = 0$

Step 2.5.2

Solve $\sin (x) + 1 = 0$ for $x$ .

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Step 2.5.2.1

Subtract $1$ from both sides of the equation.

$\sin (x) = - 1$

Step 2.5.2.2

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

$x = \arcsin (- 1)$

Step 2.5.2.3

Simplify the right side.

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Step 2.5.2.3.1

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

$x = - \frac{π}{2}$

Step 2.5.2.4

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$x = 2 π + \frac{π}{2} + π$

Step 2.5.2.5

Simplify the expression to find the second solution.

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Step 2.5.2.5.1

Subtract $2 π$ from $2 π + \frac{π}{2} + π$ .

$x = 2 π + \frac{π}{2} + π - 2 π$

Step 2.5.2.5.2

The resulting angle of $\frac{3 π}{2}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{2} + π$ .

$x = \frac{3 π}{2}$

Step 2.5.2.6

Find the period of $\sin (x)$ .

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Step 2.5.2.6.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 2.5.2.6.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 2.5.2.6.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 2.5.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 2.5.2.7

Add $2 π$ to every negative angle to get positive angles.

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Step 2.5.2.7.1

Add $2 π$ to $- \frac{π}{2}$ to find the positive angle.

$- \frac{π}{2} + 2 π$

Step 2.5.2.7.2

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

$2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 2.5.2.7.3

Combine fractions.

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Step 2.5.2.7.3.1

Combine $2 π$ and $\frac{2}{2}$ .

$\frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 2.5.2.7.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 2 - π}{2}$

Step 2.5.2.7.4

Simplify the numerator.

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Step 2.5.2.7.4.1

Multiply $2$ by $2$ .

$\frac{4 π - π}{2}$

Step 2.5.2.7.4.2

Subtract $π$ from $4 π$ .

$\frac{3 π}{2}$

Step 2.5.2.7.5

List the new angles.

$x = \frac{3 π}{2}$

Step 2.5.2.8

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{3 π}{2} + 2 π n, \frac{3 π}{2} + 2 π n$ , for any integer $n$

Step 2.6

The final solution is all the values that make $18 (- 2 \sin (x) + 1) (\sin (x) + 1) = 0$ true.

$x = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n, \frac{3 π}{2} + 2 π n$ , for any integer $n$

Step 2.7

Consolidate the answers.

$x = \frac{π}{6} + \frac{2 π n}{3}$ , for any integer $n$

Step 3

Solve the original function $f (x) = 18 \cos (x) + 9 \sin (2 x)$ at $x = \frac{π}{6}$ .

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Step 3.1

Replace the variable $x$ with $\frac{π}{6}$ in the expression.

$f (\frac{π}{6}) = 18 \cos (\frac{π}{6}) + 9 \sin (2 (\frac{π}{6}))$

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$f (\frac{π}{6}) = 18 (\frac{\sqrt{3}}{2}) + 9 \sin (2 (\frac{π}{6}))$

Step 3.2.1.2

Cancel the common factor of $2$ .

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Step 3.2.1.2.1

Factor $2$ out of $18$ .

$f (\frac{π}{6}) = 2 (9) (\frac{\sqrt{3}}{2}) + 9 \sin (2 (\frac{π}{6}))$

Step 3.2.1.2.2

Cancel the common factor.

$f (\frac{π}{6}) = 2 \cdot (9 (\frac{\sqrt{3}}{2})) + 9 \sin (2 (\frac{π}{6}))$

Step 3.2.1.2.3

Rewrite the expression.

$f (\frac{π}{6}) = 9 \sqrt{3} + 9 \sin (2 (\frac{π}{6}))$

Step 3.2.1.3

Cancel the common factor of $2$ .

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Step 3.2.1.3.1

Factor $2$ out of $6$ .

$f (\frac{π}{6}) = 9 \sqrt{3} + 9 \sin (2 (\frac{π}{2 (3)}))$

Step 3.2.1.3.2

Cancel the common factor.

$f (\frac{π}{6}) = 9 \sqrt{3} + 9 \sin (2 (\frac{π}{2 \cdot 3}))$

Step 3.2.1.3.3

Rewrite the expression.

$f (\frac{π}{6}) = 9 \sqrt{3} + 9 \sin (\frac{π}{3})$

Step 3.2.1.4

The exact value of $\sin (\frac{π}{3})$ is $\frac{\sqrt{3}}{2}$ .

$f (\frac{π}{6}) = 9 \sqrt{3} + 9 (\frac{\sqrt{3}}{2})$

Step 3.2.1.5

Combine $9$ and $\frac{\sqrt{3}}{2}$ .

$f (\frac{π}{6}) = 9 \sqrt{3} + \frac{9 \sqrt{3}}{2}$

Step 3.2.2

To write $9 \sqrt{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (\frac{π}{6}) = 9 \sqrt{3} \cdot \frac{2}{2} + \frac{9 \sqrt{3}}{2}$

Step 3.2.3

Combine fractions.

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Step 3.2.3.1

Combine $9 \sqrt{3}$ and $\frac{2}{2}$ .

$f (\frac{π}{6}) = \frac{9 \sqrt{3} \cdot 2}{2} + \frac{9 \sqrt{3}}{2}$

Step 3.2.3.2

Combine the numerators over the common denominator.

$f (\frac{π}{6}) = \frac{9 \sqrt{3} \cdot 2 + 9 \sqrt{3}}{2}$

Step 3.2.4

Simplify the numerator.

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Step 3.2.4.1

Multiply $2$ by $9$ .

$f (\frac{π}{6}) = \frac{18 \sqrt{3} + 9 \sqrt{3}}{2}$

Step 3.2.4.2

Add $18 \sqrt{3}$ and $9 \sqrt{3}$ .

$f (\frac{π}{6}) = \frac{27 \sqrt{3}}{2}$

Step 3.2.5

The final answer is $\frac{27 \sqrt{3}}{2}$ .

$\frac{27 \sqrt{3}}{2}$

Step 4

Solve the original function $f (x) = 18 \cos (x) + 9 \sin (2 x)$ at $x = \frac{π}{6} + \frac{2 π}{3}$ .

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Step 4.1

Replace the variable $x$ with $\frac{π}{6} + \frac{2 π}{3}$ in the expression.

$f (\frac{π}{6} + \frac{2 π}{3}) = 18 \cos (\frac{π}{6} + \frac{2 π}{3}) + 9 \sin (2 (\frac{π}{6} + \frac{2 π}{3}))$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

To write $\frac{2 π}{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (\frac{π}{6} + \frac{2 π}{3}) = 18 \cos (\frac{π}{6} + \frac{2 π}{3} \cdot \frac{2}{2}) + 9 \sin (2 (\frac{π}{6} + \frac{2 π}{3}))$

Step 4.2.1.2

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Step 4.2.1.2.1

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

$f (\frac{π}{6} + \frac{2 π}{3}) = 18 \cos (\frac{π}{6} + \frac{2 π \cdot 2}{3 \cdot 2}) + 9 \sin (2 (\frac{π}{6} + \frac{2 π}{3}))$

Step 4.2.1.2.2

Multiply $3$ by $2$ .

$f (\frac{π}{6} + \frac{2 π}{3}) = 18 \cos (\frac{π}{6} + \frac{2 π \cdot 2}{6}) + 9 \sin (2 (\frac{π}{6} + \frac{2 π}{3}))$

Step 4.2.1.3

Combine the numerators over the common denominator.

$f (\frac{π}{6} + \frac{2 π}{3}) = 18 \cos (\frac{π + 2 π \cdot 2}{6}) + 9 \sin (2 (\frac{π}{6} + \frac{2 π}{3}))$

Step 4.2.1.4

Simplify the numerator.

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Step 4.2.1.4.1

Multiply $2$ by $2$ .

$f (\frac{π}{6} + \frac{2 π}{3}) = 18 \cos (\frac{π + 4 π}{6}) + 9 \sin (2 (\frac{π}{6} + \frac{2 π}{3}))$

Step 4.2.1.4.2

Add $π$ and $4 π$ .

$f (\frac{π}{6} + \frac{2 π}{3}) = 18 \cos (\frac{5 π}{6}) + 9 \sin (2 (\frac{π}{6} + \frac{2 π}{3}))$

Step 4.2.1.5

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.

$f (\frac{π}{6} + \frac{2 π}{3}) = 18 (- \cos (\frac{π}{6})) + 9 \sin (2 (\frac{π}{6} + \frac{2 π}{3}))$

Step 4.2.1.6

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$f (\frac{π}{6} + \frac{2 π}{3}) = 18 (- \frac{\sqrt{3}}{2}) + 9 \sin (2 (\frac{π}{6} + \frac{2 π}{3}))$

Step 4.2.1.7

Cancel the common factor of $2$ .

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Step 4.2.1.7.1

Move the leading negative in $- \frac{\sqrt{3}}{2}$ into the numerator.

$f (\frac{π}{6} + \frac{2 π}{3}) = 18 (\frac{- \sqrt{3}}{2}) + 9 \sin (2 (\frac{π}{6} + \frac{2 π}{3}))$

Step 4.2.1.7.2

Factor $2$ out of $18$ .

$f (\frac{π}{6} + \frac{2 π}{3}) = 2 (9) (\frac{- \sqrt{3}}{2}) + 9 \sin (2 (\frac{π}{6} + \frac{2 π}{3}))$

Step 4.2.1.7.3

Cancel the common factor.

$f (\frac{π}{6} + \frac{2 π}{3}) = 2 \cdot (9 (\frac{- \sqrt{3}}{2})) + 9 \sin (2 (\frac{π}{6} + \frac{2 π}{3}))$

Step 4.2.1.7.4

Rewrite the expression.

$f (\frac{π}{6} + \frac{2 π}{3}) = 9 (- \sqrt{3}) + 9 \sin (2 (\frac{π}{6} + \frac{2 π}{3}))$

Step 4.2.1.8

Multiply $- 1$ by $9$ .

$f (\frac{π}{6} + \frac{2 π}{3}) = - 9 \sqrt{3} + 9 \sin (2 (\frac{π}{6} + \frac{2 π}{3}))$

Step 4.2.1.9

To write $\frac{2 π}{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (\frac{π}{6} + \frac{2 π}{3}) = - 9 \sqrt{3} + 9 \sin (2 (\frac{π}{6} + \frac{2 π}{3} \cdot \frac{2}{2}))$

Step 4.2.1.10

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Step 4.2.1.10.1

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

$f (\frac{π}{6} + \frac{2 π}{3}) = - 9 \sqrt{3} + 9 \sin (2 (\frac{π}{6} + \frac{2 π \cdot 2}{3 \cdot 2}))$

Step 4.2.1.10.2

Multiply $3$ by $2$ .

$f (\frac{π}{6} + \frac{2 π}{3}) = - 9 \sqrt{3} + 9 \sin (2 (\frac{π}{6} + \frac{2 π \cdot 2}{6}))$

Step 4.2.1.11

Combine the numerators over the common denominator.

$f (\frac{π}{6} + \frac{2 π}{3}) = - 9 \sqrt{3} + 9 \sin (2 (\frac{π + 2 π \cdot 2}{6}))$

Step 4.2.1.12

Simplify the numerator.

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Step 4.2.1.12.1

Multiply $2$ by $2$ .

$f (\frac{π}{6} + \frac{2 π}{3}) = - 9 \sqrt{3} + 9 \sin (2 (\frac{π + 4 π}{6}))$

Step 4.2.1.12.2

Add $π$ and $4 π$ .

$f (\frac{π}{6} + \frac{2 π}{3}) = - 9 \sqrt{3} + 9 \sin (2 (\frac{5 π}{6}))$

Step 4.2.1.13

Cancel the common factor of $2$ .

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Step 4.2.1.13.1

Factor $2$ out of $6$ .

$f (\frac{π}{6} + \frac{2 π}{3}) = - 9 \sqrt{3} + 9 \sin (2 (\frac{5 π}{2 (3)}))$

Step 4.2.1.13.2

Cancel the common factor.

$f (\frac{π}{6} + \frac{2 π}{3}) = - 9 \sqrt{3} + 9 \sin (2 (\frac{5 π}{2 \cdot 3}))$

Step 4.2.1.13.3

Rewrite the expression.

$f (\frac{π}{6} + \frac{2 π}{3}) = - 9 \sqrt{3} + 9 \sin (\frac{5 π}{3})$

Step 4.2.1.14

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.

$f (\frac{π}{6} + \frac{2 π}{3}) = - 9 \sqrt{3} + 9 (- \sin (\frac{π}{3}))$

Step 4.2.1.15

The exact value of $\sin (\frac{π}{3})$ is $\frac{\sqrt{3}}{2}$ .

$f (\frac{π}{6} + \frac{2 π}{3}) = - 9 \sqrt{3} + 9 (- \frac{\sqrt{3}}{2})$

Step 4.2.1.16

Multiply $9 (- \frac{\sqrt{3}}{2})$ .

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Step 4.2.1.16.1

Multiply $- 1$ by $9$ .

$f (\frac{π}{6} + \frac{2 π}{3}) = - 9 \sqrt{3} - 9 \frac{\sqrt{3}}{2}$

Step 4.2.1.16.2

Combine $- 9$ and $\frac{\sqrt{3}}{2}$ .

$f (\frac{π}{6} + \frac{2 π}{3}) = - 9 \sqrt{3} + \frac{- 9 \sqrt{3}}{2}$

Step 4.2.1.17

Move the negative in front of the fraction.

$f (\frac{π}{6} + \frac{2 π}{3}) = - 9 \sqrt{3} - \frac{9 \sqrt{3}}{2}$

Step 4.2.2

To write $- 9 \sqrt{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (\frac{π}{6} + \frac{2 π}{3}) = - 9 \sqrt{3} \cdot \frac{2}{2} - \frac{9 \sqrt{3}}{2}$

Step 4.2.3

Combine fractions.

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Step 4.2.3.1

Combine $- 9 \sqrt{3}$ and $\frac{2}{2}$ .

$f (\frac{π}{6} + \frac{2 π}{3}) = \frac{- 9 \sqrt{3} \cdot 2}{2} - \frac{9 \sqrt{3}}{2}$

Step 4.2.3.2

Combine the numerators over the common denominator.

$f (\frac{π}{6} + \frac{2 π}{3}) = \frac{- 9 \sqrt{3} \cdot 2 - 9 \sqrt{3}}{2}$

Step 4.2.4

Simplify the numerator.

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Step 4.2.4.1

Multiply $2$ by $- 9$ .

$f (\frac{π}{6} + \frac{2 π}{3}) = \frac{- 18 \sqrt{3} - 9 \sqrt{3}}{2}$

Step 4.2.4.2

Subtract $9 \sqrt{3}$ from $- 18 \sqrt{3}$ .

$f (\frac{π}{6} + \frac{2 π}{3}) = \frac{- 27 \sqrt{3}}{2}$

Step 4.2.5

Move the negative in front of the fraction.

$f (\frac{π}{6} + \frac{2 π}{3}) = - \frac{27 \sqrt{3}}{2}$

Step 4.2.6

The final answer is $- \frac{27 \sqrt{3}}{2}$ .

$- \frac{27 \sqrt{3}}{2}$

Step 5

The horizontal tangent line on function $f (x) = 18 \cos (x) + 9 \sin (2 x)$ is $y = \frac{27 \sqrt{3}}{2}, y = - \frac{27 \sqrt{3}}{2}$ .

$y = \frac{27 \sqrt{3}}{2}, y = - \frac{27 \sqrt{3}}{2}$

Step 6