How do you simplify #sin8theta-tan2theta# to trigonometric functions of a unit #theta#?

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Answer 1 · 2018-07-30T03:51:32

# 8 sin theta cos theta ( cos^2theta- sin^2theta )#
#( 1 - 8 cos^2theta sin^2theta )#
# - ( 2 tan theta )/( 1 - tan^2theta )#

Using De Moivre and Binomial theorems,

equate imaginary parts in

#( cos 8theta + i sin 8theta ) = ( cos theta + i sin theta )^8#

#= ( real ) C_8 + i ( real ) S_8 # and get

#sin 8theta = S_8 #

#= 8C_1 ( cos^7 theta sin theta - cos theta sin^7theta )#

#- 8C_3 ( cos^5theta sin^3theta - cos^3theta sin^5theta )#

#= 8 sin theta cos theta ( cos^6theta - sin^6theta#

# -7cos^2theta sin^2theta ( cos^2theta - sin^2theta ))# .

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

#( 1 - 8 cos^2theta sin^2theta )#

I have used #nC_r = nC_(n-r), cos^2theta + sin^2theta = 1# and

#a^3 - b^3 = ( a - b )(a^2 +ab + b^2 ) #, for this simplification.

Also #tan 2theta = ( 2 tan theta )/( 1 - tan^2theta )#.

And so,

#sin 8theta - tan 2theta#

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

#( 1 - 8 cos^2theta sin^2theta )#

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

Verification is done, with #theta = pi/8#.