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?
回答 (3)
2016-11-28 10:22 pm
(-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
cosx/2=-15/(17(2)=-15/34
tanx/2==-8/15(2)=-4/15
2016-11-19 7:54 pm
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
2016-11-19 7:49 pm
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
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
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