since cosθ > 0, we are in QIV.
cosθ = √3/2
even without figuring it, we have θ = -π/6, so tan -π/3 = -√3
or, we have
sin2θ = 2(-1/2)(√3/2) = -√3/2
cos2θ = 2(3/4)-1 = 1/2
tan2θ = -√3
or, since tanθ = -1/√3,
tan2θ = 2(-1/√3)/(1-1/3) = -√3
Given sin θ = −1/2 and sec θ > 0, find the exact value of tan 2θ.
Radical 3
Radical 3/2
Negative radical 3
Negative radical 3/2
1 answer