Use demoiveres theoreum to show that ;? #tan 4theta= (4tan theta — 4tan³theta)/ (1 — 6tan²theta+tan^4theta)#

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Answer 1 · 2018-01-21T10:35:37

By de Moiveres theorem we can write

#cos4theta+isin4theta=(costheta+isin theta)^4#

Expanding RHS by binomial theorem we get

#cos4theta +isin4theta=cos^4theta+i4cos^3thetasintheta-6cos^2thetasin^2theta-i4costhetasin^3theta+sin^4theta#

Equating imaginary parts from both sides we get

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

Equating real parts from both sides we get

#cos4theta=cos^4theta+sin^4theta-6cos^2thetasin^2theta#

Now #tan4theta=(sin4theta)/(cos4theta)#

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

#=(4tantheta-4tan^3theta)/(1-6tan^3theta+tan^4theta)#

#color(red)(["Dividing both numerator and denominator by "cos^4theta])#