Answer:
The imaginary part is 0
Step-by-step explanation:
The number given is:
[tex]x=(\cos(12)+i\sin(12)+ \cos(48)+ i\sin(48))^6[/tex]
First, we can expand this power using the binomial theorem:
[tex](a+b)^k=\sum_{j=0}^{k}\binom{k}{j}a^{k-j}b^{j}[/tex]
After that, we can apply De Moivre's theorem to expand each summand:[tex](\cos(a)+i\sin(a))^k=\cos(ka)+i\sin(ka)[/tex]
The final step is to find the common factor of i in the last expansion. Now:
[tex]x^6=((\cos(12)+i\sin(12))+(\cos(48)+ i\sin(48)))^6[/tex]
[tex]=\binom{6}{0}(\cos(12)+i\sin(12))^6(\cos(48)+ i\sin(48))^0+\binom{6}{1}(\cos(12)+i\sin(12))^5(\cos(48)+ i\sin(48))^1+\binom{6}{2}(\cos(12)+i\sin(12))^4(\cos(48)+ i\sin(48))^2+\binom{6}{3}(\cos(12)+i\sin(12))^3(\cos(48)+ i\sin(48))^3+\binom{6}{4}(\cos(12)+i\sin(12))^2(\cos(48)+ i\sin(48))^4+\binom{6}{5}(\cos(12)+i\sin(12))^1(\cos(48)+ i\sin(48))^5+\binom{6}{6}(\cos(12)+i\sin(12))^0(\cos(48)+ i\sin(48))^6[/tex]
[tex]=(\cos(72)+i\sin(72))+6(\cos(60)+i\sin(60))(\cos(48)+ i\sin(48))+15(\cos(48)+i\sin(48))(\cos(96)+ i\sin(96))+20(\cos(36)+i\sin(36))(\cos(144)+ i\sin(144))+15(\cos(24)+i\sin(24))(\cos(192)+ i\sin(192))+6(\cos(12)+i\sin(12))(\cos(240)+ i\sin(240))+(\cos(288)+ i\sin(288))[/tex]
The last part is to multiply these factors and extract the imaginary part. This computation gives:
[tex]Re x^6=\cos 72+6cos 60\cos 48-6\sin 60\sin 48+15\cos 96\cos 48-15\sin 96\sin 48+20\cos 36\cos 144-20\sin 36\sin 144+15\cos 24\cos 192-15\sin 24\sin 192+6\cos 12\cos 240-6\sin 12\sin 240+\cos 288[/tex]
[tex]Im x^6=\sin 72+6cos 60\sin 48+6\sin 60\cos 48+15\cos 96\sin 48+15\sin 96\cos 48+20\cos 36\sin 144+20\sin 36\cos 144+15\cos 24\sin 192+15\sin 24\cos 192+6\cos 12\sin 240+6\sin 12\cos 240+\sin 288[/tex]
(It is not necessary to do a lengthy computation: the summands of the imaginary part are the products sin(a)cos(b) and cos(a)sin(b) as they involve exactly one i factor)
A calculator simplifies the imaginary part Im(x⁶) to 0
