If $x = cos A + i sin A$ and $y = cos B + i sin B$ then the value of $cos (A + B)$ in terms $x$ and $y$ is :

x y + \frac{1}{x y}

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\frac{1}{2} (x y + \frac{1}{x y})

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\frac{1}{4} (x + y)

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\frac{1}{4} (x y + \frac{1}{x y})

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Solution

The correct option is B $\frac{1}{2} (x y + \frac{1}{x y})$
$x = cos A + i sin A = c i s A = e^{i A}$
$y = cos B + i sin B = c i s B = e^{i B}$
$∴ x y = e^{i A} . e^{i B} = e^{i (A + B)} = c i s (A + B)$
$∴ x y = c i s (A + B) = cos (A + B) + i sin (A + B)$ ......... $(1)$
$∴ \frac{1}{x y} = \frac{1}{c i s (A + B)} = e^{- i (A + B)} = cos (A + B) - i sin (A + B)$ .......... $(2)$
Adding $(1)$ & $(2)$
$x y + \frac{1}{x y} = 2 cos (A + B)$
$∴ cos (A + B) = \frac{1}{2} (x y + \frac{1}{x y})$

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If x=cosA+isinA and y=cosB+isinB then the value of cos(A+B) in terms x and y is :

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If $x = cos A + i sin A$ and $y = cos B + i sin B$ then the value of $cos (A + B)$ in terms $x$ and $y$ is :