Question

use the information contained in the figure to determine the values of the six trigonometric functions of $\theta$. write the exact answers and simplify. do not round.

Explanation:

Step1: Find the hypotenuse

By the Pythagorean theorem $c=\sqrt{a^{2}+b^{2}}$, where $a = 9$ and $b=9\sqrt{3}$. So $c=\sqrt{9^{2}+(9\sqrt{3})^{2}}=\sqrt{81 + 243}=\sqrt{324}=18$.

Step2: Calculate sine

$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{9}{18}=\frac{1}{2}$.

Step3: Calculate cosine

$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}$.

Step4: Calculate tangent

$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}=\frac{9}{9\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$.

Step5: Calculate cosecant

$\csc\theta=\frac{1}{\sin\theta}=2$.

Step6: Calculate secant

$\sec\theta=\frac{1}{\cos\theta}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$.

Step7: Calculate cotangent

$\cot\theta=\frac{1}{\tan\theta}=\sqrt{3}$.

Answer:

$\sin\theta=\frac{1}{2},\cos\theta=\frac{\sqrt{3}}{2},\tan\theta=\frac{\sqrt{3}}{3},\csc\theta = 2,\sec\theta=\frac{2\sqrt{3}}{3},\cot\theta=\sqrt{3}$