How do you express #cos theta - cos^2 theta + cot^2 theta # in terms of #sin theta #?

1 Answer
mason m
Jan 15, 2016

#(2sin^4theta-4sin^2theta+sinthetasin2theta+2)/(2sin^2theta)#

Explanation:

Write in terms of #sintheta# and #costheta#.

#=costheta-cos^2theta+cos^2theta/sin^2theta#

Find a common denominator.

#=(costhetasin^2theta)/sin^2theta-(cos^2thetasin^2theta)/sin^2theta+cos^2theta/sin^2theta#

Combine.

#=(costhetasin^2theta-cos^2thetasin^2theta+cos^2theta)/sin^2theta#

The following simplification may seem unecessary, but is actually relevant. Its purpose will become clear in the following step.

#=(sintheta(color(blue)(costhetasintheta))-color(green)(cos^2theta)sin^2theta+color(green)(cos^2theta))/sin^2theta#

Use the following identities:

  • #color(green)(cos^2theta=1-sin^2theta#
  • #2costhetasintheta=sin2theta=>color(blue)(costhetasintheta=(sin2theta)/2#

#=(sintheta((sin2theta)/2)-(1-sin^2theta)sin^2theta+(1-sin^2theta))/sin^2theta#

#=((sinthetasin2theta)/2-sin^2theta+sin^4theta+1-sin^2theta)/sin^2theta#

#=((sinthetasin2theta)/2-2sin^2theta+sin^4theta+1)/sin^2theta#

#=(2sin^4theta-4sin^2theta+sinthetasin2theta+2)/(2sin^2theta)#

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