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

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Answer 1 · 2016-11-10T16:02:07

With #theta in [0.pi], f(theta)= -sqrt((1-sqrt((1+cos theta)/2))/2)-sqrt((1+cos theta)/2)#

#+6sqrt((1-cos theta)/2)#

I assume that #theta in [0, pi]# so that

#theta/2 and theta/4 in [0, pi/2]

Use #sin A =+-sqrt((1-cos 2A)/2) and cos A = +-sqrt((1+cos 2A)/2)#

Under the assumption, prefixing signs are both positive, for our

application

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

#+6sqrt((1-cos theta)/2)#

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

#+6sqrt((1-cos theta)/2)#