Question #113437
Using de Moivres theorem, express

Cos5A and sin4A as polynomials in sin A and cos A.
1
Expert's answer
2020-05-04T06:27:09-0400

De Moivre's Theorem:

For any complex number z=r(cosθ+isinθ)z=r(\cos \theta +i \sin \theta) and any integer nn,

zn=rn(cos(nθ)+isin(nθ))z^n=r^n(\cos (n\theta)+i \sin(n\theta))


We will consider z=cosA+isinAz=\cos A+i \sin A

z5=cos5A+isin5Az^5=\cos 5A+i \sin 5A

z5=(cosA+isinA)5=cos5A+5cos4A(isinA)+10cos3A(isinA)2+10cos2A(isinA)3+5cosA(isinA)4+(isinA)5=(cos5A10cos3Asin2A+5cosAsin4A)+i(5cos4AsinA10cos2Asin3A+sin5A)z^5=(\cos A+i \sin A)^5=\cos^5A+5\cos ^4 A(i \sin A)+10\cos ^3 A(i \sin A)^2+10\cos ^2A(i \sin A)^3+5\cos A(i\sin A)^4+(i \sin A)^5=(\cos ^5A-10\cos ^3 A\sin^2 A+5\cos A\sin^4A)+i(5\cos^4 A\sin A-10\cos^2 A\sin ^3 A+\sin^5A)

Therefore, cos5A=cos5A10cos3Asin2A+5cosAsin4A\cos 5A= \cos ^5A-10\cos ^3 A\sin^2 A+5\cos A\sin^4A


z4=cos4A+isin4Az^4=\cos 4A+i \sin 4A

z4=(cosA+isinA)4=cos4A+4cos3A(isinA)+6cos2A(isinA)2+4cosA(isinA)3+(isinA)4=(cos4A6cos2Asin2A+sin4A)+i(4cos3AsinA4cosAsin3A)z^4=(\cos A+i \sin A)^4=\cos ^4A+4\cos ^3A(i\sin A)+6\cos^2 A(i\sin A)^2+4\cos A(i \sin A)^3+(i\sin A)^4=(\cos^4A-6\cos^2 A\sin^2A+\sin^4A)+i(4\cos ^3A\sin A-4\cos A\sin^3A)

Therefore, sin4A=4cos3AsinA4cosAsin3A\sin 4A= 4\cos ^3A\sin A-4\cos A\sin^3A


Answer: cos5A=cos5A10cos3Asin2A+5cosAsin4A\cos 5A= \cos ^5A-10\cos ^3 A\sin^2 A+5\cos A\sin^4A

sin4A=4cos3AsinA4cosAsin3A\sin 4A= 4\cos ^3A\sin A-4\cos A\sin^3A


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