Question

Let $\cos A+\cos B=x$; $\cos 2A+\cos 2B=y$; $\cos 3A+\cos 3B=z$, then which of the following is true?

• A. ${\cos }^{2}A+{\cos }^{2}B=1+\frac{y}{2}$
• B. $\cos A.\cos B=\frac{1}{4}(2x^2-y-2)$
• C. $2x^3+z=3x(1+y)$
• D. $xyz=0,\forall A,B\in R$

Answer 1

Answer: (A), (B), (C)

$2x^3+z=3x(1+y)$

$y=2({\cos }^{2}A+{\cos }^{2}B)-2$
$\Rightarrow {\cos }^{2}A+{\cos }^{2}B=1+\frac{y}{2}$

$x^2={\cos }^{2}A+{\cos }^{2}B+2\cos A\cos B$
$\cos A\cos B=\frac{1}{4}(2x^2-y-2)$

$z=4({\cos }^{3}A+{\cos }^{3}B)-3(\cos A+\cos B)$
$\Rightarrow z=4\left[\right(\cos A+\cos B\left)\right({\cos }^{2}A+{\cos }^{2}B-\cos A\cos B\left)\right]-3x$
$\Rightarrow z=x(6+3y-2x^2)-3x$
$\Rightarrow z=-2x^3+3xy+3x$
$\Rightarrow 2x^3+z=3x(y+1)$

$xyz=0\forall A,B\in R$ not true.