How do you simplify #cot4theta# to trigonometric functions of a unit #theta#?

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Answer 1 · 2017-01-28T14:32:55

As #f(cottheta)#,
#cot4theta= (cottheta(cot^4theta-6cot^2theta+1))/(4(cot^2theta-1)#

Thanks to Abraham de Moivre (1667-1754),

#(costheta+isintheta)^4=(cos4theta+isin4theta)#,

the real part of the RHS

#cos4theta#

=the real part in the expansion of LHS

#=cos^4theta-4C_2cos^2thetasin^2theta+4C_4sin^4theta#

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

Likewise, the imaginary parts give

#sin4theta=4(cos^3thetasintheta-costhetasin^3theta)#

The ratio

#cos(4theta)/sin(4theta)#

#=cot4theta#

#=(cos^4theta-6cos^2thetasin^2theta+sin^4theta)/(4(cos^3thetasintheta-costhetasin^3theta))#

#=(cottheta(cot^4theta-6cot^2theta+1))/(4(cot^2theta-1)#