How do you express #cos theta - cos^2 theta + cot^2 theta # in terms of #sin theta #?
1 Answer
Explanation:
Write in terms of
#=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)#