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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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