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

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Answer 1 · 2018-07-10T13:06:00

See the long answer, in the explanation

For #theta > 0, theta/4 in Q_1# and #theta/2 in Q_1 or Q_2#.

So, the sine and cosine values are #> 0#.

#f( theta)#

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

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

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

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

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

If #theta < 0#, the prefixing signs would become # ( + - - )#.