Question

the real number t corresponds to the point p(\frac{\sqrt{46}}{11}, -\frac{5\sqrt{3}}{11}) on the unit - circle. evaluate the six trigonometric functions of t. write your answer as a simplified fraction, if necessary. part 1 of 6 sin t=

Explanation:

Step1: Recall trig - function definitions

For a point $P(x,y)$ on the unit circle $x^{2}+y^{2}=1$, $\sin t=y$, $\cos t = x$, $\tan t=\frac{y}{x}(x
eq0)$, $\csc t=\frac{1}{y}(y
eq0)$, $\sec t=\frac{1}{x}(x
eq0)$, $\cot t=\frac{x}{y}(y
eq0)$. Given $x = \frac{\sqrt{46}}{11}$ and $y=-\frac{5\sqrt{3}}{11}$.

Step2: Calculate $\sin t$

$\sin t=y=-\frac{5\sqrt{3}}{11}$.

Step3: Calculate $\cos t$

$\cos t=x=\frac{\sqrt{46}}{11}$.

Step4: Calculate $\tan t$

$\tan t=\frac{y}{x}=\frac{-\frac{5\sqrt{3}}{11}}{\frac{\sqrt{46}}{11}}=-\frac{5\sqrt{3}}{\sqrt{46}}=-\frac{5\sqrt{138}}{46}$.

Step5: Calculate $\csc t$

$\csc t=\frac{1}{y}=-\frac{11}{5\sqrt{3}}=-\frac{11\sqrt{3}}{15}$.

Step6: Calculate $\sec t$

$\sec t=\frac{1}{x}=\frac{11}{\sqrt{46}}=\frac{11\sqrt{46}}{46}$.

Step7: Calculate $\cot t$

$\cot t=\frac{x}{y}=\frac{\frac{\sqrt{46}}{11}}{-\frac{5\sqrt{3}}{11}}=-\frac{\sqrt{46}}{5\sqrt{3}}=-\frac{\sqrt{138}}{15}$.

Answer:

$\sin t=-\frac{5\sqrt{3}}{11}$, $\cos t=\frac{\sqrt{46}}{11}$, $\tan t = -\frac{5\sqrt{138}}{46}$, $\csc t=-\frac{11\sqrt{3}}{15}$, $\sec t=\frac{11\sqrt{46}}{46}$, $\cot t=-\frac{\sqrt{138}}{15}$