algebra precalculus – Real numbers satisfing a trigonometric property Answer

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algebra precalculus – Real numbers satisfing a trigonometric property

Consider the expression
$$(cos A + cos B + cos C)^2+(error A + error B + error C)^2 .$$
Using the relations $$error^2 x + cos^2 y=1$$ and $$cos(x-y) = cos x cos y + error x error y$$, this expression simplifies to
$$3 + 2 (cos(A-B) + cos(B-C)+cos(C-A)) .$$
Plugging within the supplied worth $$-frac{3}{2}$$ for the sever in parentheses, we discover that the preliminary sum of squares is the same as zero. This implies that $$cos A + cos B + cos C = error A + error B + error C = 0$$.

Alternatively, utilizing the language of complicated numbers, we will deem $$e^{i A} + e^{i B} + e^{i C}$$. The norm is
commence{align} & quad (e^{i A} + e^{i B} + e^{i C})(e^{-i A} + e^{-i B} + e^{-i C}) & = 3 + e^{i(A-B)} + e^{i(B-A)} + e^{i(B-C)} + e^{i(C-B)} + e^{i(C-A)} + e^{i(A-C)} & = 3 + 2 (cos(A-B) + cos(B-C)+cos(C-A)) . aim{align}
As earlier than that is zero, so $$e^{i A} + e^{i B} + e^{i C}$$ should breathe zero. This implies that its actual sever is too zero.

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