Question

Let A, B, C are 3 angles such that $cosA+cosB+cosC=0$ and if $cosAcosBcosC=\lambda (cos3A+cos3B+cos3C)$ then $\lambda$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{1}{6}$
• C. $\frac{1}{9}$
• D. $\frac{1}{12}$

Answer 1

Answer: (D)

$\frac{1}{12}$

$cos3A=4co{s}^{3}A-3cosA$ similarly other two.

$cosAcosBcosC=\lambda (4co{s}^{3}A-3cosA+4co{s}^{3}B-3cosB+4co{s}^{3}C-3cosC$

$cosAcosBcosC=\lambda (4co{s}^{3}A+4co{s}^{3}B+4co{s}^{3}C-(cosA+cosB+cosC\left)\right)$

$cosAcosBcosC=\lambda (4co{s}^{3}A+4co{s}^{3}B+4co{s}^{3}C)$

$cosA+cosB+cosC=0=>cosA+cosB=-cosC$

cubing on both sides

$co{s}^{3}A+co{s}^{3}B+3cosAacosB(cosA+cosB)=-co{s}^{3}C$

$co{s}^{3}A+co{s}^{3}B+co{s}^{3}C=-3cosAacosB(cosA+cosB)$

$co{s}^{3}A+co{s}^{3}B+co{s}^{3}C=3cosAacosB\left(cosC\right)$

$cosAcosBcosC=\lambda 4(co{s}^{3}A+co{s}^{3}B+co{s}^{3}C)$

$cosAcosBcosC=\lambda 4\left(3cosAcosBcosC\right)$

Therefore $\lambda =\frac{1}{12}$