Question

watch the video and then solve the problem given below. click here to watch the video. a point p(x,y) is shown on the unit circle corresponding to a real number t. find the values of the trigonometric functions at t.

Explanation:

Step1: Recall trigonometric function definitions

For a point $P(x,y)$ on the unit - circle $x^{2}+y^{2}=1$, $\sin t=y$, $\cos t = x$, $\tan t=\frac{y}{x}(x
eq0)$, $\csc t=\frac{1}{y}(y
eq0)$, $\sec t=\frac{1}{x}(x
eq0)$, $\cot t=\frac{x}{y}(y
eq0)$.

Step2: Identify $x$ and $y$ values

Given $P(x,y)=P(-\frac{1}{2},-\frac{\sqrt{3}}{2})$, so $x = -\frac{1}{2}$ and $y=-\frac{\sqrt{3}}{2}$.

Step3: Calculate sine function

$\sin t=y=-\frac{\sqrt{3}}{2}$.

Step4: Calculate cosine function

$\cos t=x = -\frac{1}{2}$.

Step5: Calculate tangent function

$\tan t=\frac{y}{x}=\frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=\sqrt{3}$.

Step6: Calculate cosecant function

$\csc t=\frac{1}{y}=\frac{1}{-\frac{\sqrt{3}}{2}}=-\frac{2\sqrt{3}}{3}$.

Step7: Calculate secant function

$\sec t=\frac{1}{x}=\frac{1}{-\frac{1}{2}}=-2$.

Step8: Calculate cotangent function

$\cot t=\frac{x}{y}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=\frac{\sqrt{3}}{3}$.

Answer:

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