How do you find the exact value of the six trigonometric functions of θ when your given a point (-4,-6)?

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Answer 1 · 2016-07-15T17:01:45

$sintheta = -3sqrt13/13$

$costheta = -2sqrt13/13$

$tantheta = 3/2$

$sectheta = -sqrt13/2$

$csctheta = -sqrt13/3$

$ctntheta = 2/3$

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For the point $(-4,-6)$ we can create a right triangle with legs of $x = -4$ and $y =-6$

Using the pythagorean theorem we get the hypotenuse.

$a^2 + b^2 = c^2$

$(-4)^2 + (-6)^2 = hyp^2$

$16+36=hyp^2$

$52 = hyp^2$

$sqrt52 = hyp$

$2sqrt13 = hyp$

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The six trig functions are

$sintheta = (opp)/(hyp)$

$costheta = (adj)/(hyp)$

$tantheta = (opp)/(adj)$

$sectheta = (hyp)/(adj)$

$csctheta = (hyp)/(opp)$

$ctntheta = (adj)/(opp)$

$sintheta = (-6)/(2sqrt13) = -3/sqrt13 * sqrt13/sqrt13 = -3sqrt13/13$

$costheta = (-4)/(2sqrt13) = -2/sqrt13 * sqrt13/sqrt13 = -2sqrt13/13$

$tantheta = (-6)/-4 = 3/2$

$sectheta = (2sqrt13)/-4 = -sqrt13/2$

$csctheta = (2sqrt13)/-6 = -sqrt13/3$

$ctntheta = (-4)/-6 = 2/3$