Question

1. (14 pts.) let cosθ = - 3/11 and θ terminates in qiii. find all six trigonometric functions of θ. give exact answers. a. draw angle in standard position cosθ=-3/11

Explanation:

Step1: Recall cosine definition

We know that $\cos\theta=\frac{x}{r}$, where $x$ is the $x -$coordinate of a point on the terminal - side of the angle $\theta$ in standard position and $r$ is the distance from the origin to the point $(x,y)$ with $r=\sqrt{x^{2}+y^{2}}$ and $r>0$. Given $\cos\theta =-\frac{3}{11}$, we can let $x=- 3$ and $r = 11$.

Step2: Find $y$ using $r=\sqrt{x^{2}+y^{2}}$

Since $r=\sqrt{x^{2}+y^{2}}$, then $11=\sqrt{(-3)^{2}+y^{2}}$. Squaring both sides gives $121 = 9 + y^{2}$, so $y^{2}=121 - 9=112$. Because the angle $\theta$ terminates in QIII, $y<0$, so $y =-\sqrt{112}=-4\sqrt{7}$.

Step3: Calculate sine function

$\sin\theta=\frac{y}{r}=\frac{-4\sqrt{7}}{11}$.

Step4: Calculate tangent function

$\tan\theta=\frac{y}{x}=\frac{-4\sqrt{7}}{-3}=\frac{4\sqrt{7}}{3}$.

Step5: Calculate cosecant function

$\csc\theta=\frac{r}{y}=\frac{11}{-4\sqrt{7}}=-\frac{11\sqrt{7}}{28}$.

Step6: Calculate secant function

$\sec\theta=\frac{r}{x}=\frac{11}{-3}=-\frac{11}{3}$.

Step7: Calculate cotangent function

$\cot\theta=\frac{x}{y}=\frac{-3}{-4\sqrt{7}}=\frac{3\sqrt{7}}{28}$.

Answer:

$\sin\theta=-\frac{4\sqrt{7}}{11}$, $\cos\theta =-\frac{3}{11}$, $\tan\theta=\frac{4\sqrt{7}}{3}$, $\csc\theta=-\frac{11\sqrt{7}}{28}$, $\sec\theta=-\frac{11}{3}$, $\cot\theta=\frac{3\sqrt{7}}{28}$