How do you simplify #f(theta)=csc2theta+sin4theta+cos4theta# to trigonometric functions of a unit #theta#?

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Answer 1 · 2018-08-03T10:50:19

#= 1/( 2 sin theta cos theta ) #

#+ 2 sin theta cos theta ( 2 ( cos^2theta - sin^2theta )- 3 sin theta cos theta ) + 1#

#f theta ) =csc 2theta + sin 4theta + cos 4theta#

#= 1/sin (2theta) + 2 sin 2theta cos 2theta#

#+ (cos^2 2theta - sin^2 2theta)#

#= 1/( 2 sin theta cos theta ) #

#+ 4 sin theta cos theta ( cos^2theta - sin^2theta)#

#+ (cos^2theta - sin^2theta )^2 - 4 sin^2theta cos^2theta#

#= 1/( 2 sin theta cos theta ) #

#+ 4 sin theta cos theta ( cos^2theta - sin^2theta)#

#+ (cos^2theta + sin^2theta )^2 - 6 sin^2theta cos^2theta#

#= 1/( 2 sin theta cos theta ) #

#+ 4 sin theta cos theta ( cos^2theta - sin^2theta)#

#+ 1 - 6 sin^2theta cos^2theta#

#= 1/( 2 sin theta cos theta ) #

#+ 2 sin theta cos theta ( 2 ( cos^2theta - sin^2theta )- 3 sin theta cos theta ) + 1#