Trigonometry Examples

\cos (θ) = \frac{\sqrt{3}}{2}

$\cos (θ) = \frac{\sqrt{3}}{2}$

Step 1

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

θ

$θ$ from inside the cosine.

θ = \arccos (\frac{\sqrt{3}}{2})

$θ = \arccos (\frac{\sqrt{3}}{2})$

Step 2

Simplify the right side.

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Step 2.1

The exact value of

\arccos (\frac{\sqrt{3}}{2})

$\arccos (\frac{\sqrt{3}}{2})$ is

\frac{π}{6}

$\frac{π}{6}$ .

θ = \frac{π}{6}

$θ = \frac{π}{6}$

θ = \frac{π}{6}

$θ = \frac{π}{6}$

Step 3

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from

2 π

$2 π$ to find the solution in the fourth quadrant.

θ = 2 π - \frac{π}{6}

$θ = 2 π - \frac{π}{6}$

Step 4

Simplify

2 π - \frac{π}{6}

$2 π - \frac{π}{6}$ .

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Step 4.1

To write

2 π

$2 π$ as a fraction with a common denominator, multiply by

\frac{6}{6}

$\frac{6}{6}$ .

θ = 2 π \cdot \frac{6}{6} - \frac{π}{6}

$θ = 2 π \cdot \frac{6}{6} - \frac{π}{6}$

Step 4.2

Combine fractions.

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Step 4.2.1

Combine

2 π

$2 π$ and

\frac{6}{6}

$\frac{6}{6}$ .

θ = \frac{2 π \cdot 6}{6} - \frac{π}{6}

$θ = \frac{2 π \cdot 6}{6} - \frac{π}{6}$

Step 4.2.2

Combine the numerators over the common denominator.

θ = \frac{2 π \cdot 6 - π}{6}

$θ = \frac{2 π \cdot 6 - π}{6}$

θ = \frac{2 π \cdot 6 - π}{6}

$θ = \frac{2 π \cdot 6 - π}{6}$

Step 4.3

Simplify the numerator.

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Step 4.3.1

Multiply

6

$6$ by

2

$2$ .

θ = \frac{12 π - π}{6}

$θ = \frac{12 π - π}{6}$

Step 4.3.2

Subtract

π

$π$ from

12 π

$12 π$ .

θ = \frac{11 π}{6}

$θ = \frac{11 π}{6}$

θ = \frac{11 π}{6}

$θ = \frac{11 π}{6}$

θ = \frac{11 π}{6}

$θ = \frac{11 π}{6}$

Step 5

Find the period of

\cos (θ)

$\cos (θ)$ .

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Step 5.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 5.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{2 π}{1}

$\frac{2 π}{1}$

Step 5.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 6

The period of the

\cos (θ)

$\cos (θ)$ function is

2 π

$2 π$ so values will repeat every

2 π

$2 π$ radians in both directions.

θ = \frac{π}{6} + 2 π n, \frac{11 π}{6} + 2 π n

$θ = \frac{π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer

n

$n$