Suppose that tanθ =-8 over 15 and 90 degrees < θ < 180 degrees Find the exact values of .. A) cos θ over 2 B) tan θ over 2?

(-8^2)+(15)^2=64+225=289 then hypotenuse=sqr289=17
cosx/2=-15/(17(2)=-15/34
tanx/2==-8/15(2)=-4/15

　
You really should use math symbols ----> use slash / instead of over

Since 90° < θ < 180°, then 45° < θ < 90°
Therefore, sinθ > 0, cosθ < 0, cos(θ/2) > 0, tan(θ/2) > 0

tanθ = −8/15
sinθ = 8/17
cosθ = −15/17

A)

cos²(θ/2) = 1/2 (1 + cosθ) = 1/2 (1 − 15/17) = 1/17
cos(θ/2) = 1/√17

B)

tan(θ/2) = (1 − cosθ) / sinθ = (1 + 15/17) / (8/17)
tan(θ/2) = 4

tanθ = -8/15
θ in quadrant II, so cosθ < 0 and sinθ > 0.

√((-8)²+15²) = √289 = 17
cosθ = -15/17
sinθ = 8/17

cos²(θ/2) = (1+cosθ)/2 = (1-15/17)/2 = (2/17)/2 = 1/17
cos(θ/2) = 1/√17