Trigonometry Examples

$\cos^{2} (θ) = \frac{1}{2}$

Step 1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$\cos (θ) = \pm \sqrt{\frac{1}{2}}$

Step 2

Simplify

$\pm \sqrt{\frac{1}{2}}$ .

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Step 2.1

Rewrite

$\sqrt{\frac{1}{2}}$ as

$\frac{\sqrt{1}}{\sqrt{2}}$ .

$\cos (θ) = \pm \frac{\sqrt{1}}{\sqrt{2}}$

Step 2.2

Any root of

$1$ is

$1$ .

$\cos (θ) = \pm \frac{1}{\sqrt{2}}$

Step 2.3

Multiply

$\frac{1}{\sqrt{2}}$ by

$\frac{\sqrt{2}}{\sqrt{2}}$ .

$\cos (θ) = \pm \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

Step 2.4

Combine and simplify the denominator.

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Step 2.4.1

Multiply

$\frac{1}{\sqrt{2}}$ by

$\frac{\sqrt{2}}{\sqrt{2}}$ .

$\cos (θ) = \pm \frac{\sqrt{2}}{\sqrt{2} \sqrt{2}}$

Step 2.4.2

Raise

$\sqrt{2}$ to the power of

$1$ .

$\cos (θ) = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{1} \sqrt{2}}$

Step 2.4.3

Raise

$\sqrt{2}$ to the power of

$1$ .

$\cos (θ) = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{1} {\sqrt{2}}^{1}}$

Step 2.4.4

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\cos (θ) = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{1 + 1}}$

Step 2.4.5

Add

$1$ and

$1$ .

$\cos (θ) = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{2}}$

Step 2.4.6

Rewrite

${\sqrt{2}}^{2}$ as

$2$ .

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Step 2.4.6.1

Use

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

$\sqrt{2}$ as

$2^{\frac{1}{2}}$ .

$\cos (θ) = \pm \frac{\sqrt{2}}{{(2^{\frac{1}{2}})}^{2}}$

Step 2.4.6.2

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$\cos (θ) = \pm \frac{\sqrt{2}}{2^{\frac{1}{2} \cdot 2}}$

Step 2.4.6.3

Combine

$\frac{1}{2}$ and

$2$ .

$\cos (θ) = \pm \frac{\sqrt{2}}{2^{\frac{2}{2}}}$

Step 2.4.6.4

Cancel the common factor of

$2$ .

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Step 2.4.6.4.1

Cancel the common factor.

$\cos (θ) = \pm \frac{\sqrt{2}}{2^{\frac{2}{2}}}$

Step 2.4.6.4.2

Rewrite the expression.

$\cos (θ) = \pm \frac{\sqrt{2}}{2^{1}}$

Step 2.4.6.5

Evaluate the exponent.

$\cos (θ) = \pm \frac{\sqrt{2}}{2}$

Step 3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.1

First, use the positive value of the

$\pm$ to find the first solution.

$\cos (θ) = \frac{\sqrt{2}}{2}$

Step 3.2

Next, use the negative value of the

$\pm$ to find the second solution.

$\cos (θ) = - \frac{\sqrt{2}}{2}$

Step 3.3

The complete solution is the result of both the positive and negative portions of the solution.

$\cos (θ) = \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}$

Step 4

Set up each of the solutions to solve for

$θ$ .

$\cos (θ) = \frac{\sqrt{2}}{2}$

$\cos (θ) = - \frac{\sqrt{2}}{2}$

Step 5

Solve for

$θ$ in

$\cos (θ) = \frac{\sqrt{2}}{2}$ .

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Step 5.1

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

$θ$ from inside the cosine.

$θ = \arccos (\frac{\sqrt{2}}{2})$

Step 5.2

Simplify the right side.

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Step 5.2.1

The exact value of

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

$45$ .

$θ = 45$

Step 5.3

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

$360$ to find the solution in the fourth quadrant.

$θ = 360 - 45$

Step 5.4

Subtract

$45$ from

$360$ .

$θ = 315$

Step 5.5

Find the period of

$\cos (θ)$ .

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Step 5.5.1

The period of the function can be calculated using

$\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Step 5.5.2

Replace

$b$ with

$1$ in the formula for period.

$\frac{360}{| 1 |}$

Step 5.5.3

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

$0$ and

$1$ is

$1$ .

$\frac{360}{1}$

Step 5.5.4

Divide

$360$ by

$1$ .

$360$

Step 5.6

The period of the

$\cos (θ)$ function is

$360$ so values will repeat every

$360$ degrees in both directions.

$θ = 45 + 360 n, 315 + 360 n$ , for any integer

$n$

$θ = 45 + 360 n, 315 + 360 n$ , for any integer

$n$

Step 6

Solve for

$θ$ in

$\cos (θ) = - \frac{\sqrt{2}}{2}$ .

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Step 6.1

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

$θ$ from inside the cosine.

$θ = \arccos (- \frac{\sqrt{2}}{2})$

Step 6.2

Simplify the right side.

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Step 6.2.1

The exact value of

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

$135$ .

$θ = 135$

Step 6.3

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from

$360$ to find the solution in the third quadrant.

$θ = 360 - 135$

Step 6.4

Subtract

$135$ from

$360$ .

$θ = 225$

Step 6.5

Find the period of

$\cos (θ)$ .

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Step 6.5.1

The period of the function can be calculated using

$\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Step 6.5.2

Replace

$b$ with

$1$ in the formula for period.

$\frac{360}{| 1 |}$

Step 6.5.3

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

$0$ and

$1$ is

$1$ .

$\frac{360}{1}$

Step 6.5.4

Divide

$360$ by

$1$ .

$360$

Step 6.6

The period of the

$\cos (θ)$ function is

$360$ so values will repeat every

$360$ degrees in both directions.

$θ = 135 + 360 n, 225 + 360 n$ , for any integer

$n$

$θ = 135 + 360 n, 225 + 360 n$ , for any integer

$n$

Step 7

List all of the solutions.

$θ = 45 + 360 n, 315 + 360 n, 135 + 360 n, 225 + 360 n$ , for any integer

$n$

Step 8

Consolidate the answers.

$θ = 45 + 90 n$ , for any integer

$n$