Sqrt(3)(-9/2-9isqrt(3)/2)

is the answer, but I came up with a postive result, why is it negative?
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2 Answers

????????????????? both ansers ??????????

yu hav sumthun raesed tu power=4...

that meen yu hav 1 anser

Inside the parens, yu hav kosine av wot look like a KOMPLEX NUMBER

That meen yer anser kan be put in serkel form (e^iangel)

That meen theer be infinut angels that giv the same value for kosine
by

cos(A+B)=cosAcosB-sinAsinB (trig identity).

cos(5(pi)/6)=-sqrt(3)/2; sin(5(pi)/6)=1/2. Putting A=5(pi)/6 and B=isin(5(pi)/6)=i/2:

cos(A+B)=(-sqrt(3)/2)(cos(i/2))-(1/2)sin(i/2).

Also, by de Moivre: e^(ix)=cosx+isinx; e^-(ix)=cosx-isinx,

so e^(ix)+e^-(ix)=2cosx and e^(ix)-e^-(ix)=2isinx.

2cos(ix)=e^-x+e^x; 2isin(ix)=e^-x-e^x; -2sin(ix)=i(e^-x-e^x).

cos(i/2)=(e^-(1/2)+e^(1/2))/2; sin(i/2)=(-i(e^-(1/2)-e^(1/2))/2=i(e^(1/2)-e^-(1/2))/2.

[sqrt(3)(cos(A+B)]^4=9[(-sqrt(3)/2)(e^-(1/2)+e^(1/2))/2-(1/2)i(e^(1/2)-e^-(1/2))/2]^4.

Now, this answer is clearly very complicated and we still have to calculate the fourth power. In order to assess whether this is leading to the expected answer provided, we need to work out the expression in square brackets as far as possible. I believe that the original question may have been wrongly entered, because of the complexity, and should have been [sqrt(3)(cos(5pi/6)+isin(5pi/6)]^4, which would result in an easier application of de Moivre. Proceeding briefly on this assumption we get:

9e^(i5(pi)/6)^4=9(cos(20(pi)/6)+isin(20(pi)/6) applying e^(iX)=cosX+isinX, so that e^(niX)=cos(nX)+isin(nX), where n=4 and X=5(pi)/6.

20(pi)/6=10(pi)/3=2(pi)+4(pi)/3, in the third quadrant. cos(4(pi)/3)=-1/2 and sin(4(pi)/3)=-sqrt(3)/2.

The expression becomes: 9(-1/2-isqrt(3)/2) or -9/2-i9sqrt(3)/2 in standard form (a+ib), which is the provided answer, proving that the question was misstated. Both the real and imaginary parts are negative, caused by multiplication of the angle taking it into the third quadrant where sine and cosine are negative.

 

by Top Rated User (1.1m points)

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