How do you find the five remaining trigonometric function satisfying $costheta=-2/5$ , $sintheta>0$ ?

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Answer 1 · 2018-08-04T12:29:17

$sin theta= sqrt 21/5 , csc theta=5/sqrt 21 ,tan theta= - sqrt 21/2$
$cot theta= -2/sqrt 21 , sec theta=-5/2$

In right triangle $cos theta = a/h =-2 /5 :.a=-2 , h=5$

$:.o=sqrt(5^2-2^2)= sqrt 21 ; a,,o, h$ are adjacent side, opposite

side and hypotenuse. $sin theta= o/h= sqrt 21/5$ ,

$tan theta= o/a= - sqrt 21/2, csc theta= h/o= 5/sqrt 21$ ,

$sec theta= h/a= - 5/2, cot theta= a/o= -2/sqrt 21$ [Ans]

How do you find the five remaining trigonometric function satisfying costheta=-2/5costheta=-2/5, sintheta>0sintheta>0?