Question

If $\cos x+\cos y=a,\cos 2x+\cos 2y=b,\cos 3x+\cos 3y=c$, then

• A. ${\cos }^{2}x+{\cos }^{2}y=1+\frac{b}{2}$
• B. $\cos x\cos y=\frac{a^2}{2}-\left(\frac{b+2}{4}\right)$
• C. $2a^3+c=3a(1+b)$
• D. $a+b+c=3abc$

Answer 1

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

$2a^3+c=3a(1+b)$
${\cos }^{2}x+{\cos }^{2}y=1+\frac{b}{2}$
$\cos x\cos y=\frac{a^2}{2}-\left(\frac{b+2}{4}\right)$

$\cos x+\cos y=a\cos 2x+\cos 2y=b⟹2{\cos }^{2}x-1+2{\cos }^{2}y-1=b⟹2({\cos }^{2}x+{\cos }^{2}y)=b+2⟹{\cos }^{2}x+{\cos }^{2}y=\frac{b+2}{2}=1+\frac{b}{2}-----\left(A\right){(\cos x+\cos y)}^2-2\cos x\cos y=\frac{b+2}{2}⟹a^2-2\cos x\cos y=\frac{b+2}{2}⟹a^2-\left(\frac{b+2}{2}\right)=2\cos x\cos y⟹\cos x\cos y=\frac{a^2}{2}-\frac{b+2}{4}-----\left(B\right)$

Also,

$\cos 3x+\cos 3y=c⟹4{\cos }^{3}x-3\cos x+4{\cos }^{3}y-3\cos y=c⟹4({\cos }^{3}x+{\cos }^{3}y)-3(\cos x+\cos y)=c⟹4\left[\right(\cos x+\cos y\left)\right({\cos }^{2}x-\cos x\cos y+{\cos }^{2}y\left)\right]-3(\cos x+\cos y)=c⟹4\left(a(\frac{b+2}{2}-\frac{a^2}{2}+\frac{b+2}{4})\right)-3a=c⟹4\left(a\left(\frac{3}{4}\right(b+2)-\frac{a^2}{2})\right)-3a=c⟹3a(b+2)-2a^3-3a=c⟹3a(b+2-1)=2a^3+c⟹3a(1+b)=2a^3+c-----\left(C\right)$