Solve for θ csc(theta) = square root of 1+(-12/5)^2

$\csc (θ) = \sqrt{1 + {(- \frac{12}{5})}^{2}}$

Step 1

Simplify $\sqrt{1 + {(- \frac{12}{5})}^{2}}$ .

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{12}{5}$ .

$\csc (θ) = \sqrt{1 + {(- 1)}^{2} {(\frac{12}{5})}^{2}}$

Apply the product rule to $\frac{12}{5}$ .

$\csc (θ) = \sqrt{1 + {(- 1)}^{2} \frac{12^{2}}{5^{2}}}$

Raise $- 1$ to the power of $2$ .

$\csc (θ) = \sqrt{1 + 1 \frac{12^{2}}{5^{2}}}$

Multiply $\frac{12^{2}}{5^{2}}$ by $1$ .

$\csc (θ) = \sqrt{1 + \frac{12^{2}}{5^{2}}}$

Raise $12$ to the power of $2$ .

$\csc (θ) = \sqrt{1 + \frac{144}{5^{2}}}$

Raise $5$ to the power of $2$ .

$\csc (θ) = \sqrt{1 + \frac{144}{25}}$

Write $1$ as a fraction with a common denominator.

$\csc (θ) = \sqrt{\frac{25}{25} + \frac{144}{25}}$

Combine the numerators over the common denominator.

$\csc (θ) = \sqrt{\frac{25 + 144}{25}}$

Add $25$ and $144$ .

$\csc (θ) = \sqrt{\frac{169}{25}}$

Rewrite $\sqrt{\frac{169}{25}}$ as $\frac{\sqrt{169}}{\sqrt{25}}$ .

$\csc (θ) = \frac{\sqrt{169}}{\sqrt{25}}$

Simplify the numerator.

Rewrite $169$ as $13^{2}$ .

$\csc (θ) = \frac{\sqrt{13^{2}}}{\sqrt{25}}$

Pull terms out from under the radical, assuming positive real numbers.

$\csc (θ) = \frac{13}{\sqrt{25}}$

Simplify the denominator.

Rewrite $25$ as $5^{2}$ .

$\csc (θ) = \frac{13}{\sqrt{5^{2}}}$

Pull terms out from under the radical, assuming positive real numbers.

$\csc (θ) = \frac{13}{5}$

Step 2

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

$θ = arccsc (\frac{13}{5})$

Step 3

Simplify the right side.

Evaluate $arccsc (\frac{13}{5})$ .

$θ = 0.39479111$

Step 4

The cosecant 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.

$θ = (3.14159265) - 0.39479111$

Step 5

Solve for $θ$ .

Remove parentheses.

$θ = 3.14159265 - 0.39479111$

Remove parentheses.

$θ = (3.14159265) - 0.39479111$

Subtract $0.39479111$ from $3.14159265$ .

$θ = 2.74680153$

Step 6

Find the period of $\csc (θ)$ .

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

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

$θ = 0.39479111 + 2 π n, 2.74680153 + 2 π n$ , for any integer $n$

Solve for θ csc(theta) = square root of 1+(-12/5)^2

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