The point $(- 4, 10)$ $(-4,10)$ is on the terminal side of an angle in standard position, how do you determine the exact values of the six trigonometric functions of the angle?

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Answer 1 · 2016-12-14T18:33:35

Please see the explanation.

Let $x = - 4$ $x = -4$

Let $y = 10$ $y = 10$

Let $r =$ $r =$ the length of a line segment drawn from the origin to the point:

$r = \sqrt{x^{2} + y^{2}}$ $r = sqrt(x^2 + y^2)$

$r = \sqrt{{(- 4)}^{2} + 10^{2}}$ $r = sqrt((-4)^2 + 10^2)$

$r = \sqrt{116} = 2 \sqrt{29}$ $r = sqrt(116) = 2sqrt(29)$

$sin (θ) = \frac{y}{r}$ $sin(theta) = y/r$

$sin (θ) = \frac{10}{2 \sqrt{29}}$ $sin(theta) = 10/(2sqrt(29))$

$sin (θ) = \frac{5 \sqrt{29}}{29}$ $sin(theta) = (5sqrt(29))/29$

$csc (θ) = \frac{1}{sin (θ)} = \frac{\sqrt{29}}{5}$ $csc(theta) = 1/sin(theta) = sqrt(29)/5$

$cos (θ) = \frac{x}{r}$ $cos(theta) = x/r$

$cos (θ) = - \frac{4}{2 \sqrt{29}}$ $cos(theta) = -4/(2sqrt(29))$

$cos (θ) = - \frac{2 \sqrt{29}}{29}$ $cos(theta) = -(2sqrt(29))/29$

$sec (θ) = \frac{1}{cos (θ)} = - \frac{\sqrt{29}}{2}$ $sec(theta) = 1/cos(theta) = -sqrt(29)/2$

$tan (θ) = \frac{y}{x}$ $tan(theta) = y/x$

$tan (θ) = \frac{10}{- 4}$ $tan(theta) = 10/-4$

$tan (θ) = - 2.5$ $tan(theta) = -2.5$

$cot (θ) = \frac{1}{tan (θ)} = - \frac{2}{5}$ $cot(theta) = 1/tan(theta) = -2/5$

The point (−4,10)(-4,10) is on the terminal side of an angle in standard position, how do you determine the exact values of the six trigonometric functions of the angle?