Solve for θ sin(theta)=-1/( square root of 2)

$\sin (θ) = - \frac{1}{\sqrt{2}}$

Step 1

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

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

$\sin (θ) = - (\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}})$

Combine and simplify the denominator.

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

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

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

$\sin (θ) = - \frac{\sqrt{2}}{{\sqrt{2}}^{1} \sqrt{2}}$

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

$\sin (θ) = - \frac{\sqrt{2}}{{\sqrt{2}}^{1} {\sqrt{2}}^{1}}$

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

$\sin (θ) = - \frac{\sqrt{2}}{{\sqrt{2}}^{1 + 1}}$

Add $1$ and $1$ .

$\sin (θ) = - \frac{\sqrt{2}}{{\sqrt{2}}^{2}}$

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

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

$\sin (θ) = - \frac{\sqrt{2}}{{(2^{\frac{1}{2}})}^{2}}$

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

$\sin (θ) = - \frac{\sqrt{2}}{2^{\frac{1}{2} \cdot 2}}$

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

$\sin (θ) = - \frac{\sqrt{2}}{2^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\sin (θ) = - \frac{\sqrt{2}}{2^{\frac{2}{2}}}$

Rewrite the expression.

$\sin (θ) = - \frac{\sqrt{2}}{2^{1}}$

Evaluate the exponent.

$\sin (θ) = - \frac{\sqrt{2}}{2}$

Step 2

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

$θ = \arcsin (- \frac{\sqrt{2}}{2})$

Step 3

Simplify the right side.

The exact value of $\arcsin (- \frac{\sqrt{2}}{2})$ is $- \frac{π}{4}$ .

$θ = - \frac{π}{4}$

Step 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.

$θ = 2 π + \frac{π}{4} + π$

Step 5

Simplify the expression to find the second solution.

Subtract $2 π$ from $2 π + \frac{π}{4} + π$ .

$θ = 2 π + \frac{π}{4} + π - 2 π$

The resulting angle of $\frac{5 π}{4}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{4} + π$ .

$θ = \frac{5 π}{4}$

Step 6

Find the period of $\sin (θ)$ .

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

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

Step 7

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

Add $2 π$ to $- \frac{π}{4}$ to find the positive angle.

$- \frac{π}{4} + 2 π$

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$2 π \cdot \frac{4}{4} - \frac{π}{4}$

Combine fractions.

Combine $2 π$ and $\frac{4}{4}$ .

$\frac{2 π \cdot 4}{4} - \frac{π}{4}$

Combine the numerators over the common denominator.

$\frac{2 π \cdot 4 - π}{4}$

Simplify the numerator.

Multiply $4$ by $2$ .

$\frac{8 π - π}{4}$

Subtract $π$ from $8 π$ .

$\frac{7 π}{4}$

List the new angles.

$θ = \frac{7 π}{4}$

Step 8

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

$θ = \frac{5 π}{4} + 2 π n, \frac{7 π}{4} + 2 π n$ , for any integer $n$

Solve for θ sin(theta)=-1/( square root of 2)

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