Question

determine the exact values of the six trigonometric functions of each angle θ. (a) ( - √2, - 1) sin θ = cos θ = tan θ = csc θ = sec θ = cot θ = (b)

Explanation:

Step1: Calculate the radius $r$

For a point $(x,y)=(-\sqrt{2}, - 1)$ in the coordinate - plane, use the formula $r=\sqrt{x^{2}+y^{2}}$. So, $r=\sqrt{(-\sqrt{2})^{2}+(-1)^{2}}=\sqrt{2 + 1}=\sqrt{3}$.

Step2: Calculate $\sin\theta$

By the definition $\sin\theta=\frac{y}{r}$, substituting $y=-1$ and $r = \sqrt{3}$, we get $\sin\theta=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}$.

Step3: Calculate $\cos\theta$

Using the definition $\cos\theta=\frac{x}{r}$, substituting $x = -\sqrt{2}$ and $r=\sqrt{3}$, we have $\cos\theta=-\frac{\sqrt{2}}{\sqrt{3}}=-\frac{\sqrt{6}}{3}$.

Step4: Calculate $\tan\theta$

By the definition $\tan\theta=\frac{y}{x}$, substituting $x=-\sqrt{2}$ and $y = - 1$, we obtain $\tan\theta=\frac{-1}{-\sqrt{2}}=\frac{\sqrt{2}}{2}$.

Step5: Calculate $\csc\theta$

Since $\csc\theta=\frac{1}{\sin\theta}$, substituting $\sin\theta=-\frac{\sqrt{3}}{3}$, we get $\csc\theta=-\sqrt{3}$.

Step6: Calculate $\sec\theta$

Since $\sec\theta=\frac{1}{\cos\theta}$, substituting $\cos\theta=-\frac{\sqrt{6}}{3}$, we have $\sec\theta=-\frac{\sqrt{3}}{\sqrt{2}}=-\frac{\sqrt{6}}{2}$.

Step7: Calculate $\cot\theta$

Since $\cot\theta=\frac{1}{\tan\theta}$, substituting $\tan\theta=\frac{\sqrt{2}}{2}$, we obtain $\cot\theta=\sqrt{2}$.

Answer:

$\sin\theta=-\frac{\sqrt{3}}{3}$
$\cos\theta=-\frac{\sqrt{6}}{3}$
$\tan\theta=\frac{\sqrt{2}}{2}$
$\csc\theta=-\sqrt{3}$
$\sec\theta=-\frac{\sqrt{6}}{2}$
$\cot\theta=\sqrt{2}$