De Moivre's Theorem:
For any complex number z=r(cosθ+isinθ) and any integer n,
zn=rn(cos(nθ)+isin(nθ))
We will consider z=cosA+isinA
z5=cos5A+isin5A
z5=(cosA+isinA)5=cos5A+5cos4A(isinA)+10cos3A(isinA)2+10cos2A(isinA)3+5cosA(isinA)4+(isinA)5=(cos5A−10cos3Asin2A+5cosAsin4A)+i(5cos4AsinA−10cos2Asin3A+sin5A)
Therefore, cos5A=cos5A−10cos3Asin2A+5cosAsin4A
z4=cos4A+isin4A
z4=(cosA+isinA)4=cos4A+4cos3A(isinA)+6cos2A(isinA)2+4cosA(isinA)3+(isinA)4=(cos4A−6cos2Asin2A+sin4A)+i(4cos3AsinA−4cosAsin3A)
Therefore, sin4A=4cos3AsinA−4cosAsin3A
Answer: cos5A=cos5A−10cos3Asin2A+5cosAsin4A
sin4A=4cos3AsinA−4cosAsin3A
Comments