tanØ= -√5/2 , so Ø must be in II
make a sketch of a right-angled triangle with sides √5 and 2, the hypotenuse is √9 or 3
since tanØ = -√5/2 = y/x
then
in quadrant II, x = -2, y = √5 and r = 3
cosØ = -2/3
Another: theta is a second quadrant angle in standard position and tan theta = - (sqrt5)/(2) .
Find the exact value of cos theta.
I assume I use the equation 1 / cos(theta)^2 = 1 + tan (theta)^2
However how does this work out algebraically?
2 answers
thanks!