(1)
− 1 + i = 2 ( − 1 2 + i 1 2 ) = 2 ( cos 3 π 4 + i sin 3 π 4 ) -1+i =\sqrt 2\big(\frac{-1}{\sqrt 2}+i \frac{1}{\sqrt 2}\big)=\sqrt 2(\cos \frac{3\pi}{4}+i\sin \frac{3\pi}{4}) − 1 + i = 2 ( 2 − 1 + i 2 1 ) = 2 ( cos 4 3 π + i sin 4 3 π )
Using de Moivre's formula, we have:
( − 1 + i ) 16 = ( 2 ) 16 ( cos ( 16 × 3 π 4 ) + i sin ( 16 × 3 π 4 ) ) = 256 ( cos 12 π + i sin 12 π ) = 256 (-1+i)^{16}=(\sqrt 2)^{16} (\cos (16\times \frac{3\pi}{4})+i \sin(16\times \frac{3\pi}{4}))=256(\cos 12\pi+i\sin 12\pi)=256 ( − 1 + i ) 16 = ( 2 ) 16 ( cos ( 16 × 4 3 π ) + i sin ( 16 × 4 3 π )) = 256 ( cos 12 π + i sin 12 π ) = 256
( − 1 + i ) 16 = 256 (-1+i)^{16}=256 ( − 1 + i ) 16 = 256
1 ( − 1 + i ) 6 = ( − 1 + i ) − 6 = ( 2 ) − 6 ( cos ( − 6 × 3 π 4 ) + i sin ( − 6 × 3 π 4 ) ) = 1 / 8 ( cos − 4 , 5 π + i sin − 4 , 5 π ) = − i / 8 \frac{1}{(-1+i)^6}=(-1+i)^{-6}=(\sqrt 2)^{-6} (\cos (-6\times \frac{3\pi}{4})+i \sin(-6\times \frac{3\pi}{4}))=1/8(\cos -4,5\pi+i\sin -4,5\pi)=-i /8 ( − 1 + i ) 6 1 = ( − 1 + i ) − 6 = ( 2 ) − 6 ( cos ( − 6 × 4 3 π ) + i sin ( − 6 × 4 3 π )) = 1/8 ( cos − 4 , 5 π + i sin − 4 , 5 π ) = − i /8
1 ( − 1 + i ) 6 = − i / 8 \frac{1}{(-1+i)^6}=-i /8 ( − 1 + i ) 6 1 = − i /8
(2)
Let z = cos α + i sin α z=\cos \alpha +i \sin \alpha z = cos α + i sin α
Using de Moivre's formula, we have that z 5 = cos 5 α + i sin 5 α z^5=\cos 5 \alpha +i \sin 5 \alpha z 5 = cos 5 α + i sin 5 α
z 5 = ( cos α + i sin α ) 5 = cos 5 α + 5 cos 4 α ( i sin α ) + 10 cos 3 α ( i sin α ) 2 + 10 cos 2 α ( i sin α ) 3 + 5 cos α ( i sin α ) 4 + ( i sin α ) 5 = ( cos 5 α − 10 cos 3 α sin 2 α + 5 cos α sin 4 α ) + i ( 5 cos 4 α sin α − 10 cos 2 α sin 3 α + sin 5 α ) z^5=(\cos \alpha +i \sin \alpha )^5=\cos^5 \alpha +5\cos ^4 \alpha (i \sin \alpha )+10\cos ^3 \alpha (i \sin \alpha )^2+10\cos ^2 \alpha (i \sin \alpha )^3+5\cos \alpha (i \sin \alpha )^4+(i \sin \alpha )^5=(\cos ^5 \alpha -10\cos ^3 \alpha \sin^2 \alpha +5\cos \alpha \sin^4 \alpha )+i (5\cos^4 \alpha \sin \alpha -10\cos ^2 \alpha \sin^3 \alpha +\sin^5 \alpha ) z 5 = ( cos α + i sin α ) 5 = cos 5 α + 5 cos 4 α ( i sin α ) + 10 cos 3 α ( i sin α ) 2 + 10 cos 2 α ( i sin α ) 3 + 5 cos α ( i sin α ) 4 + ( i sin α ) 5 = ( cos 5 α − 10 cos 3 α sin 2 α + 5 cos α sin 4 α ) + i ( 5 cos 4 α sin α − 10 cos 2 α sin 3 α + sin 5 α )
cos 5 α = cos 5 α − 10 cos 3 α sin 2 α + 5 cos α sin 4 α = cos α ( cos 4 α − 10 cos 2 α sin 2 α + 5 sin 4 α ) = cos α ( ( 1 − sin 2 α ) 2 − 10 ( 1 − sin 2 α ) sin 2 α + 5 sin 4 α ) = cos α ( 16 sin 4 α − 12 sin 2 α + 1 ) \cos 5 \alpha = \cos ^5 \alpha -10\cos ^3 \alpha \sin^2 \alpha +5\cos \alpha \sin^4 \alpha =\cos \alpha (\cos^4 \alpha -10\cos^2 \alpha \sin^2 \alpha +5\sin^4 \alpha )=\cos \alpha ((1-\sin^2 \alpha )^2-10(1-\sin^2 \alpha )\sin^2 \alpha +5\sin^4 \alpha )=\cos \alpha (16\sin^4 \alpha -12 \sin^2 \alpha +1) cos 5 α = cos 5 α − 10 cos 3 α sin 2 α + 5 cos α sin 4 α = cos α ( cos 4 α − 10 cos 2 α sin 2 α + 5 sin 4 α ) = cos α (( 1 − sin 2 α ) 2 − 10 ( 1 − sin 2 α ) sin 2 α + 5 sin 4 α ) = cos α ( 16 sin 4 α − 12 sin 2 α + 1 )
cos 5 α cos α = 16 sin 4 α − 12 sin 2 α + 1 \frac{\cos 5 \alpha }{\cos \alpha }=16\sin^4 \alpha -12\sin^2 \alpha +1 c o s α c o s 5 α = 16 sin 4 α − 12 sin 2 α + 1
sin 5 α = 5 cos 4 α sin α − 10 cos 2 α sin 3 α + sin 5 α = 5 ( 1 − sin 2 α ) 2 sin α − 10 ( 1 − sin 2 α ) sin 3 α + sin 5 α = 16 sin 5 α − 20 sin 3 α + 5 sin α \sin 5 \alpha =5\cos^4\alpha \sin \alpha -10\cos^2 \alpha \sin^3 \alpha +\sin ^5 \alpha =5(1-\sin^2
\alpha )^2\sin \alpha -10(1-\sin^2 \alpha )\sin^3 \alpha +\sin ^5 \alpha =16\sin ^5 \alpha -20\sin^3
\alpha +5\sin\alpha sin 5 α = 5 cos 4 α sin α − 10 cos 2 α sin 3 α + sin 5 α = 5 ( 1 − sin 2 α ) 2 sin α − 10 ( 1 − sin 2 α ) sin 3 α + sin 5 α = 16 sin 5 α − 20 sin 3 α + 5 sin α
sin 5 α = 16 sin 5 α − 20 sin 3 α + 5 sin α \sin 5\alpha= 16\sin ^5 \alpha -20\sin^3
\alpha +5\sin \alpha sin 5 α = 16 sin 5 α − 20 sin 3 α + 5 sin α
#333702 on Dec 2022
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