Given tanΘ=-2 and π/2 < Θ < π; How would you find cos2Θ?

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Answer 1 · 2018-04-21T10:35:59

#:.cos2theta=-3/5#

Here,

#tantheta=-2#

Where, #pi/2< theta < pi..toII^(nd) Quadrant#

#=>2xxpi/2 < 2xxtheta < 2xxpi#

#=>pi < 2theta <2pi=>III^(rd) or IV^(th) Quadrant#

Now,

#cos2theta=(1-tan^2theta)/(1+tan^2theta)...to,where,tantheta=-2#

#=(1-(-2)^2)/(1+(-2)^2#

#=(1-4)/(1+4)#

#=-3/5 < 0toIII^(rd) Quadrant#

#:.cos2theta=-3/5#