How do you express #f(theta)=-sin(theta/4)-cos(theta/2)+6cos(theta/2)# in terms of trigonometric functions of a whole theta?

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Answer 1 · 2017-11-19T22:54:34

#f(theta)=+-sqrt((1+-sqrt((1+costheta)/2))/2)+-5sqrt((1+costheta)/2)#

We know:

#Sin(x/2)=+-sqrt((1-cosx)/2)#

#cos(x/2)=+-sqrt((1+cosx)/2)#

Therefore:

#sin(theta/4)=+-sqrt((1-cos(theta/2))/2#

#sin(theta/4)=+-sqrt((1+-sqrt((1+costheta)/2))/2#

Therefore:

#f(theta)=-sin(theta/4)-cos(theta/2)+6cos(theta/2)=-sin(theta/4)+5cos(theta/2)#

#f(theta)=+-sqrt((1+-sqrt((1+costheta)/2))/2)+-5sqrt((1+costheta)/2)#