Question

If sin x, cos x is the roots of r^2-pt+q=0 then show that sin^3x + cos^3x is equal to p^3-3pq

Answer 1

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Answer 2

$\huge\bf{ \underline{ \underline{Question}}}$

• If Sinx , Cosx is the roots of $\sf{\ r^{2}-pt+q\:=\:0}$ then show that $\sf{\ Sin^{3}x+\ Cos^{3}x}$ is equal to $\sf{(\ p^{3}-3pq)}$

$\huge\bf{ \underline{ \underline{Solution}}}$

• $\sf{Sinx\:+\:Cosx\:=\:p}$

• $\sf{SinxCosx\:=\:q}$

$\sf{\ Sin^{3}x+\ Cos^{3}x\:can\:be\: written\:as}$

$\tt{\ (Sinx+Cosx)^{3}-3SinxCosx(Sinx+Cosx)}$

$\bf{ Put\:value\:in\:formula}$

${}$

$\mapsto\sf{\ p^{3}-3p(q)}$

${}$

$\mapsto\sf{\ p^{3}-3pq}$

$\bf{ \underline{ \underline{Hence\:Proved}}}$

$\sf{ }$