What is #cottheta-sectheta+costheta# in terms of #sintheta#?

1 Answer
mason m
Jan 15, 2016

#(2(1-sin^2theta-sin^3theta))/(sin2theta)#

Explanation:

Express everything in terms of #sintheta# and #costheta#.

#=costheta/sintheta-1/costheta+costheta#

Get a common denominator.

#=cos^2theta/(costhetasintheta)-sintheta/(costhetasintheta)+(cos^2thetasintheta)/(costhetasintheta)#

Combine.

#=(cos^2theta-sintheta+cos^2thetasintheta)/(costhetasintheta)#

Here, use the identities:

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

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

#=(1-sin^2theta-sintheta+sin-sin^3theta)/((sin2theta)/2)#

#=(2(1-sin^2theta-sin^3theta))/(sin2theta)#

Related questions
See all questions in Fundamental Identities
Impact of this question
1220 views around the world
You can reuse this answer
Creative Commons License