How do you simplify #cos4theta-sin2theta+tan2theta# to trigonometric functions of a unit #theta#?

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Answer 1 · 2018-08-09T04:53:37

#= 1 - 8 sin^2theta cos^2theta + (4 sin^3theta cos theta )/( cos^2 theta - sin^2theta )#

Equating real parts in

#e^(i4theta) = (e^(itheta))^4#,

#cos 4theta = cos^4theta - 6sin^2theta cos^2theta + sin^4theta#

#= (sin^2theta +cos^2theta )^2 - 8 sin^2theta cos^2theta#

#= 1 - 8 sin^2theta cos^2theta#. So,

#cos 4theta - sin2theta + tan 2theta#

#= 1 - 8 sin^2theta cos^2theta - 2sin theta cos theta #

#+ (2tan theta)/( 1 - tan^2theta )#

#= .1 - 8 sin^2theta cos^2theta - 2sin theta cos theta #

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

#= .1 - 8 sin^2theta cos^2theta#

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

#= 1 - 8 sin^2theta cos^2theta + (4 sin^3theta cos theta )/( cos^2 theta - sin^2theta )#