What is #(sintheta)/4# in terms of #sectheta#?

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Answer 1 · 2016-02-09T03:18:34

#sqrt(sec^2theta-1)/(4sectheta)#

First, we should write #sintheta/4# in terms of #costheta#. Then, since #sectheta=1/costheta#, it's a simple fix to switch from one to the other.

We should make use of the Pythagorean identity:

#sin^2theta+cos^2theta=1" "=>" "sin^2theta=1-cos^2theta" "=>" "sintheta=sqrt(1-cos^2theta)#

Thus,

#sintheta/4=sqrt(1-cos^2theta)/4=sqrt(1-1/sec^2theta)/4#

Although this does meet the parameters of the question, we can algebra our way to a prettier looking answer.

#sintheta/4=sqrt((sec^2theta-1)/sec^2theta)/4=sqrt(sec^2theta-1)/(4sectheta)#