tanθ= -2sqrt10/3 and π/2<θ< π find
a. cos(-θ)
b. cos(11π/6 - θ)
c.sin2θ
d.tanθ/2
can some please explain to me how to solve these questions? thanks
The angle is in the 2nd quadrant, so x is negative, and y is positive. The hypotenuse has length 7.
So, since tanθ = -2√10/3,
y = 2√10
x = -3
sinθ = y/h = 2√10/7
cosθ = x/h = -3/7
Use these values in your double-angle, half-angle and sum/difference formulas