Calculus Examples

$2 \cos (t) + \sin (2 t)$

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

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

$f' (t) = \frac{d}{d t} (2 \cos (t)) + \frac{d}{d t} (\sin (2 t))$

Step 1.1.2

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

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Step 1.1.2.1

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

$f' (t) = 2 \frac{d}{d t} (\cos (t)) + \frac{d}{d t} (\sin (2 t))$

Step 1.1.2.2

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

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

Step 1.1.2.3

Multiply $- 1$ by $2$ .

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

Step 1.1.3

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

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Step 1.1.3.1

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \sin (t)$ and $g (t) = 2 t$ .

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Step 1.1.3.1.1

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

$f' (t) = - 2 \sin (t) + \frac{d}{d u} (\sin (u)) \frac{d}{d t} (2 t)$

Step 1.1.3.1.2

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

$f' (t) = - 2 \sin (t) + \cos (u) \frac{d}{d t} (2 t)$

Step 1.1.3.1.3

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

$f' (t) = - 2 \sin (t) + \cos (2 t) \frac{d}{d t} (2 t)$

Step 1.1.3.2

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

$f' (t) = - 2 \sin (t) + \cos (2 t) (2 \frac{d}{d t} (t))$

Step 1.1.3.3

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

$f' (t) = - 2 \sin (t) + \cos (2 t) (2 \cdot 1)$

Step 1.1.3.4

Multiply $2$ by $1$ .

$f' (t) = - 2 \sin (t) + \cos (2 t) \cdot 2$

Step 1.1.3.5

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

$f' (t) = - 2 \sin (t) + 2 \cos (2 t)$

Step 1.2

The first derivative of $f (t)$ with respect to $t$ is $- 2 \sin (t) + 2 \cos (2 t)$ .

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

Step 2

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

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Step 2.1

Set the first derivative equal to $0$ .

$- 2 \sin (t) + 2 \cos (2 t) = 0$

Step 2.2

Simplify each term.

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Step 2.2.1

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

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

Step 2.2.2

Apply the distributive property.

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

Step 2.2.3

Multiply $2$ by $1$ .

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

Step 2.2.4

Multiply $- 2$ by $2$ .

$- 2 \sin (t) + 2 - 4 \sin^{2} (t) = 0$

Step 2.3

Factor $- 2 \sin (t) + 2 - 4 \sin^{2} (t)$ .

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Step 2.3.1

Factor $2$ out of $- 2 \sin (t) + 2 - 4 \sin^{2} (t)$ .

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Step 2.3.1.1

Factor $2$ out of $- 2 \sin (t)$ .

$2 (- \sin (t)) + 2 - 4 \sin^{2} (t) = 0$

Step 2.3.1.2

Factor $2$ out of $2$ .

$2 (- \sin (t)) + 2 (1) - 4 \sin^{2} (t) = 0$

Step 2.3.1.3

Factor $2$ out of $- 4 \sin^{2} (t)$ .

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

Step 2.3.1.4

Factor $2$ out of $2 (- \sin (t)) + 2 (1)$ .

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

Step 2.3.1.5

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

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

Step 2.3.2

Factor.

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Step 2.3.2.1

Factor by grouping.

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Step 2.3.2.1.1

Reorder terms.

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

Step 2.3.2.1.2

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 2.3.2.1.2.1

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

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

Step 2.3.2.1.2.2

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

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

Step 2.3.2.1.2.3

Apply the distributive property.

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

Step 2.3.2.1.2.4

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

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

Step 2.3.2.1.3

Factor out the greatest common factor from each group.

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Step 2.3.2.1.3.1

Group the first two terms and the last two terms.

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

Step 2.3.2.1.3.2

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

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

Step 2.3.2.1.4

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

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

Step 2.3.2.2

Remove unnecessary parentheses.

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

Step 2.4

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

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

$\sin (t) + 1 = 0$

Step 2.5

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

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Step 2.5.1

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

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

Step 2.5.2

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

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Step 2.5.2.1

Subtract $1$ from both sides of the equation.

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

Step 2.5.2.2

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

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Step 2.5.2.2.1

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

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

Step 2.5.2.2.2

Simplify the left side.

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Step 2.5.2.2.2.1

Cancel the common factor of $- 2$ .

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Step 2.5.2.2.2.1.1

Cancel the common factor.

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

Step 2.5.2.2.2.1.2

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

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

Step 2.5.2.2.3

Simplify the right side.

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Step 2.5.2.2.3.1

Dividing two negative values results in a positive value.

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

Step 2.5.2.3

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

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

Step 2.5.2.4

Simplify the right side.

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Step 2.5.2.4.1

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

$t = \frac{π}{6}$

Step 2.5.2.5

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.

$t = π - \frac{π}{6}$

Step 2.5.2.6

Simplify $π - \frac{π}{6}$ .

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Step 2.5.2.6.1

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

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

Step 2.5.2.6.2

Combine fractions.

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Step 2.5.2.6.2.1

Combine $π$ and $\frac{6}{6}$ .

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

Step 2.5.2.6.2.2

Combine the numerators over the common denominator.

$t = \frac{π \cdot 6 - π}{6}$

Step 2.5.2.6.3

Simplify the numerator.

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Step 2.5.2.6.3.1

Move $6$ to the left of $π$ .

$t = \frac{6 \cdot π - π}{6}$

Step 2.5.2.6.3.2

Subtract $π$ from $6 π$ .

$t = \frac{5 π}{6}$

Step 2.5.2.7

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

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Step 2.5.2.7.1

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

$\frac{2 π}{| b |}$

Step 2.5.2.7.2

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

$\frac{2 π}{| 1 |}$

Step 2.5.2.7.3

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

$\frac{2 π}{1}$

Step 2.5.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 2.5.2.8

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

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

Step 2.6

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

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Step 2.6.1

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

$\sin (t) + 1 = 0$

Step 2.6.2

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

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Step 2.6.2.1

Subtract $1$ from both sides of the equation.

$\sin (t) = - 1$

Step 2.6.2.2

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

$t = \arcsin (- 1)$

Step 2.6.2.3

Simplify the right side.

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Step 2.6.2.3.1

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

$t = - \frac{π}{2}$

Step 2.6.2.4

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.

$t = 2 π + \frac{π}{2} + π$

Step 2.6.2.5

Simplify the expression to find the second solution.

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Step 2.6.2.5.1

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

$t = 2 π + \frac{π}{2} + π - 2 π$

Step 2.6.2.5.2

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

$t = \frac{3 π}{2}$

Step 2.6.2.6

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

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Step 2.6.2.6.1

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

$\frac{2 π}{| b |}$

Step 2.6.2.6.2

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

$\frac{2 π}{| 1 |}$

Step 2.6.2.6.3

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

$\frac{2 π}{1}$

Step 2.6.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 2.6.2.7

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

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Step 2.6.2.7.1

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

$- \frac{π}{2} + 2 π$

Step 2.6.2.7.2

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 2.6.2.7.3

Combine fractions.

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Step 2.6.2.7.3.1

Combine $2 π$ and $\frac{2}{2}$ .

$\frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 2.6.2.7.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 2 - π}{2}$

Step 2.6.2.7.4

Simplify the numerator.

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Step 2.6.2.7.4.1

Multiply $2$ by $2$ .

$\frac{4 π - π}{2}$

Step 2.6.2.7.4.2

Subtract $π$ from $4 π$ .

$\frac{3 π}{2}$

Step 2.6.2.7.5

List the new angles.

$t = \frac{3 π}{2}$

Step 2.6.2.8

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

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

Step 2.7

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

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

Step 2.8

Consolidate the answers.

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

Step 3

Find the values where the derivative is undefined.

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Step 3.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 4

Evaluate $2 \cos (t) + \sin (2 t)$ at each $t$ value where the derivative is $0$ or undefined.

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Step 4.1

Evaluate at $t = \frac{π}{6}$ .

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Step 4.1.1

Substitute $\frac{π}{6}$ for $t$ .

$2 \cos (\frac{π}{6}) + \sin (2 (\frac{π}{6}))$

Step 4.1.2

Simplify.

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Step 4.1.2.1

Simplify each term.

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Step 4.1.2.1.1

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

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

Step 4.1.2.1.2

Cancel the common factor of $2$ .

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Step 4.1.2.1.2.1

Cancel the common factor.

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

Step 4.1.2.1.2.2

Rewrite the expression.

$\sqrt{3} + \sin (2 (\frac{π}{6}))$

Step 4.1.2.1.3

Cancel the common factor of $2$ .

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Step 4.1.2.1.3.1

Factor $2$ out of $6$ .

$\sqrt{3} + \sin (2 \frac{π}{2 (3)})$

Step 4.1.2.1.3.2

Cancel the common factor.

$\sqrt{3} + \sin (2 \frac{π}{2 \cdot 3})$

Step 4.1.2.1.3.3

Rewrite the expression.

$\sqrt{3} + \sin (\frac{π}{3})$

Step 4.1.2.1.4

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

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

Step 4.1.2.2

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

$\sqrt{3} \cdot \frac{2}{2} + \frac{\sqrt{3}}{2}$

Step 4.1.2.3

Combine fractions.

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Step 4.1.2.3.1

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

$\frac{\sqrt{3} \cdot 2}{2} + \frac{\sqrt{3}}{2}$

Step 4.1.2.3.2

Combine the numerators over the common denominator.

$\frac{\sqrt{3} \cdot 2 + \sqrt{3}}{2}$

Step 4.1.2.4

Simplify the numerator.

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Step 4.1.2.4.1

Move $2$ to the left of $\sqrt{3}$ .

$\frac{2 \cdot \sqrt{3} + \sqrt{3}}{2}$

Step 4.1.2.4.2

Add $2 \sqrt{3}$ and $\sqrt{3}$ .

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

Step 4.2

Evaluate at $t = \frac{5 π}{6}$ .

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Step 4.2.1

Substitute $\frac{5 π}{6}$ for $t$ .

$2 \cos (\frac{5 π}{6}) + \sin (2 (\frac{5 π}{6}))$

Step 4.2.2

Simplify.

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Step 4.2.2.1

Simplify each term.

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Step 4.2.2.1.1

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.

$2 (- \cos (\frac{π}{6})) + \sin (2 (\frac{5 π}{6}))$

Step 4.2.2.1.2

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

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

Step 4.2.2.1.3

Cancel the common factor of $2$ .

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Step 4.2.2.1.3.1

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

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

Step 4.2.2.1.3.2

Cancel the common factor.

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

Step 4.2.2.1.3.3

Rewrite the expression.

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

Step 4.2.2.1.4

Cancel the common factor of $2$ .

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Step 4.2.2.1.4.1

Factor $2$ out of $6$ .

$- \sqrt{3} + \sin (2 \frac{5 π}{2 (3)})$

Step 4.2.2.1.4.2

Cancel the common factor.

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

Step 4.2.2.1.4.3

Rewrite the expression.

$- \sqrt{3} + \sin (\frac{5 π}{3})$

Step 4.2.2.1.5

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.

$- \sqrt{3} - \sin (\frac{π}{3})$

Step 4.2.2.1.6

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

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

Step 4.2.2.2

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

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

Step 4.2.2.3

Combine fractions.

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Step 4.2.2.3.1

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

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

Step 4.2.2.3.2

Combine the numerators over the common denominator.

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

Step 4.2.2.4

Simplify the numerator.

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Step 4.2.2.4.1

Multiply $2$ by $- 1$ .

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

Step 4.2.2.4.2

Subtract $\sqrt{3}$ from $- 2 \sqrt{3}$ .

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

Step 4.2.2.5

Move the negative in front of the fraction.

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

Step 4.3

Evaluate at $t = \frac{3 π}{2}$ .

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Step 4.3.1

Substitute $\frac{3 π}{2}$ for $t$ .

$2 \cos (\frac{3 π}{2}) + \sin (2 (\frac{3 π}{2}))$

Step 4.3.2

Simplify.

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Step 4.3.2.1

Simplify each term.

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Step 4.3.2.1.1

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

$2 \cos (\frac{π}{2}) + \sin (2 (\frac{3 π}{2}))$

Step 4.3.2.1.2

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

$2 \cdot 0 + \sin (2 (\frac{3 π}{2}))$

Step 4.3.2.1.3

Multiply $2$ by $0$ .

$0 + \sin (2 (\frac{3 π}{2}))$

Step 4.3.2.1.4

Cancel the common factor of $2$ .

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Step 4.3.2.1.4.1

Cancel the common factor.

$0 + \sin (2 \frac{3 π}{2})$

Step 4.3.2.1.4.2

Rewrite the expression.

$0 + \sin (3 π)$

Step 4.3.2.1.5

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

$0 + \sin (π)$

Step 4.3.2.1.6

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

$0 + \sin (0)$

Step 4.3.2.1.7

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

$0 + 0$

Step 4.3.2.2

Add $0$ and $0$ .

$0$

Step 4.4

List all of the points.

$(\frac{π}{6} + 2 π n, \frac{3 \sqrt{3}}{2}), (\frac{5 π}{6} + 2 π n, - \frac{3 \sqrt{3}}{2}), (\frac{3 π}{2} + 2 π n, 0)$ , for any integer $n$

Step 5