How do you find the values of the other trigonometric functions of θ from the information given $cos θ = - \frac{5}{13}$ $cos theta = − 5/13$ , and theta is in quadrant III?

Question

5194 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-30T20:00:45

This comes from a $5$ $5$ , $12$ $12$ , $13$ $13$ triangle ( $5^{2} + 12^{2} = 13^{2}$ $5^2+12^2 = 13^2$ )

$sin θ = - \frac{12}{13}$ $sin theta = -12/13$ ( $sin < 0$ $sin < 0$ in QIII)

$tan θ = \frac{sin θ}{cos θ} = \frac{5}{12}$ $tan theta = sin theta / cos theta = 5/12$

$cot θ = \frac{cos θ}{sin θ} = \frac{12}{5}$ $cot theta = cos theta / sin theta = 12/5$

$sec θ = \frac{1}{cos θ} = - \frac{13}{5}$ $sec theta = 1 / cos theta = -13/5$

$csc θ = \frac{1}{sin θ} = - \frac{13}{12}$ $csc theta = 1 / sin theta = -13/12$

Incidentally, the $5 - 12 - 13$ $5-12-13$ triangle is the next right angled triangle in a sequence that begins with the $3 - 4 - 5$ $3-4-5$ triangle.

$a_{k} = 2 k + 3$ $a_k = 2k+3$

$b_{k} = \frac{a_{k}^{2} - 1}{2} = 2 k^{2} + 6 k + 4$ $b_k = (a_k^2-1)/2 = 2k^2+6k+4$

$c_{k} = \frac{a_{k}^{2} + 1}{2} = 2 k^{2} + 6 k + 5$ $c_k = (a_k^2+1)/2 = 2k^2+6k+5$

How do you find the values of the other trigonometric functions of θ from the information given cosθ=−513cos theta = − 5/13 , and theta is in quadrant III?