Find the area of this shape

Find the imaginary part of\[(\cos12^\circ+i\sin12^\circ+\cos48^\circ+i\sin48^\circ)^6.\]

iren [92.7K]

Answer:

The imaginary part is 0

Step-by-step explanation:

The number given is:

$x=(\cos(12)+i\sin(12)+ \cos(48)+ i\sin(48))^6$

First, we can expand this power using the binomial theorem:

$(a+b)^k=\sum_{j=0}^{k}\binom{k}{j}a^{k-j}b^{j}$

After that, we can apply De Moivre's theorem to expand each summand: $(\cos(a)+i\sin(a))^k=\cos(ka)+i\sin(ka)$

The final step is to find the common factor of i in the last expansion. Now:

$x^6=((\cos(12)+i\sin(12))+(\cos(48)+ i\sin(48)))^6$

$=\binom{6}{0}(\cos(12)+i\sin(12))^6(\cos(48)+ i\sin(48))^0+\binom{6}{1}(\cos(12)+i\sin(12))^5(\cos(48)+ i\sin(48))^1+\binom{6}{2}(\cos(12)+i\sin(12))^4(\cos(48)+ i\sin(48))^2+\binom{6}{3}(\cos(12)+i\sin(12))^3(\cos(48)+ i\sin(48))^3+\binom{6}{4}(\cos(12)+i\sin(12))^2(\cos(48)+ i\sin(48))^4+\binom{6}{5}(\cos(12)+i\sin(12))^1(\cos(48)+ i\sin(48))^5+\binom{6}{6}(\cos(12)+i\sin(12))^0(\cos(48)+ i\sin(48))^6$

$=(\cos(72)+i\sin(72))+6(\cos(60)+i\sin(60))(\cos(48)+ i\sin(48))+15(\cos(48)+i\sin(48))(\cos(96)+ i\sin(96))+20(\cos(36)+i\sin(36))(\cos(144)+ i\sin(144))+15(\cos(24)+i\sin(24))(\cos(192)+ i\sin(192))+6(\cos(12)+i\sin(12))(\cos(240)+ i\sin(240))+(\cos(288)+ i\sin(288))$

The last part is to multiply these factors and extract the imaginary part. This computation gives:

$Re x^6=\cos 72+6cos 60\cos 48-6\sin 60\sin 48+15\cos 96\cos 48-15\sin 96\sin 48+20\cos 36\cos 144-20\sin 36\sin 144+15\cos 24\cos 192-15\sin 24\sin 192+6\cos 12\cos 240-6\sin 12\sin 240+\cos 288$

$Im x^6=\sin 72+6cos 60\sin 48+6\sin 60\cos 48+15\cos 96\sin 48+15\sin 96\cos 48+20\cos 36\sin 144+20\sin 36\cos 144+15\cos 24\sin 192+15\sin 24\cos 192+6\cos 12\sin 240+6\sin 12\cos 240+\sin 288$

(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)

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