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3 years ago

What are two methods for proving the two triangles

Mathematics

1 answer:

Feliz [49]3 years ago

5 0

Congruence method: AAA, ASA, SSS,SAS,HL(For right triangle).

For coordinates proof, if you are proficient with vector, then it will be easier to calculate the angles, otherwise, you should know the distance formula and calculate the length of the vertices.

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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)

A calculator simplifies the imaginary part Im(x⁶) to 0

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3 years ago