Question

If $\sin A+\cos A=m$ and ${\sin }^{3}A+{\cos }^{3}A=n$, then

• A. $m^3-3m+n=0$
• B. $n^3-3n+2m=0$
• C. $m^3-3m+2n=0$
• D. $m^3+3m+2n=0$

Answer 1

Answer: (C)

$m^3-3m+2n=0$

${\sin }^{3}A+{\cos }^{3}A=n\sin A+\cos A=m{(\sin A+\cos A)}^3=m^3{\sin }^{3}A+{\cos }^{3}A+3\sin A\cos A(\sin A+\cos A)=m^{3}n+3m\sin A\cos A=m^3.......\left(i\right){(\sin A+\cos A)}^2=m^2{\sin }^{2}A+{\cos }^{2}A+2\sin A\cos A=m^{2}1+2\sin A\cos A=m^2\sin A\cos A=\frac{m^2-1}{2}$

Substituting in $\left(i\right)$

$\Rightarrow n+3m\left(\frac{m^2-1}{2}\right)=m^3\Rightarrow 2n+3m^3-3m=2m^3{\Rightarrow m}^3-3m+2n=0$

So option $C$ is correct