Solve for θ in Degrees cos(theta)^2=1/2

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Trigonometry

$\cos^{2} (θ) = \frac{1}{2}$

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

$\cos (θ) = \pm \sqrt{\frac{1}{2}}$

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

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

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

Any root of $1$ is $1$ .

$\cos (θ) = \pm \frac{1}{\sqrt{2}}$

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

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

Combine and simplify the denominator.

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

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

Raise $\sqrt{2}$ to the power of $1$ .

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

Raise $\sqrt{2}$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

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

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

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

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Combine $\frac{1}{2}$ and $2$ .

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

$\cos (θ) = \pm \frac{\sqrt{2}}{2}$

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

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First, use the positive value of the $\pm$ to find the first solution.

$\cos (θ) = \frac{\sqrt{2}}{2}$

Next, use the negative value of the $\pm$ to find the second solution.

$\cos (θ) = - \frac{\sqrt{2}}{2}$

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

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

Set up each of the solutions to solve for $θ$ .

$\cos (θ) = \frac{\sqrt{2}}{2}$

$\cos (θ) = - \frac{\sqrt{2}}{2}$

Solve for $θ$ in $\cos (θ) = \frac{\sqrt{2}}{2}$ .

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Take the inverse cosine of both sides of the equation to extract $θ$ from inside the cosine.

$θ = \arccos (\frac{\sqrt{2}}{2})$

Simplify the right side.

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The exact value of $\arccos (\frac{\sqrt{2}}{2})$ is $45$ .

$θ = 45$

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.

$θ = 360 - 45$

Subtract $45$ from $360$ .

$θ = 315$

Find the period of $\cos (θ)$ .

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The period of the function can be calculated using $\frac{360}{| b |}$ .

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

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

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

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

$\frac{360}{1}$

Divide $360$ by $1$ .

$360$

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 $n$

Solve for $θ$ in $\cos (θ) = - \frac{\sqrt{2}}{2}$ .

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Take the inverse cosine of both sides of the equation to extract $θ$ from inside the cosine.

$θ = \arccos (- \frac{\sqrt{2}}{2})$

Simplify the right side.

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The exact value of $\arccos (- \frac{\sqrt{2}}{2})$ is $135$ .

$θ = 135$

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.

$θ = 360 - 135$

Subtract $135$ from $360$ .

$θ = 225$

Find the period of $\cos (θ)$ .

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The period of the function can be calculated using $\frac{360}{| b |}$ .

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

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

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

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

$\frac{360}{1}$

Divide $360$ by $1$ .

$360$

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 $n$

List all of the solutions.

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

Consolidate the answers.

$θ = 45 + 90 n$ , for any integer $n$

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