Question

cos^2x-sin^2x=. ? want to know the proof​

Answer 1

$\huge{ANSWER}$

$\boxed{\rm{1}}$ $\rm{Cos^2x-Sin^2x=Cos2x}$

$\boxed{\rm{2}}$ $\rm{Cos^2x-Sin^2x=2Cos^2x-1}$

$\boxed{\rm{3}}$ $\rm{Cos^2x-Sin^2x=1-2Sin^2x}$

$\mathbb{EXPLANATION}$

$\boxed{\rm{1}}$

$\rm{Cos^2x-Sin^2x}$

$\rm{Cosx\:\times\:Cosx-Sinx\:\times\:Sinx}$

$\rm{Cos\left(2x\right)}$

$\boxed{Becoz\:CosA\:CosB\:-SinA\:SinB\:=Cos\left(A+B\right)}$

$\therefore$ $\rm{Cos^2x-Sin^2x\:=Cos2x}$

$\boxed{\rm{2}}$

$\rm{Cos^2x-Sin^2x}$

$\rm{Cos^2x-1+Cos^2x}$

$\boxed{Becoz\:\:Sin^2x=1-Cos^2x}$

$\rm{2Cos^2x-1}$

$\therefore$ $\rm{Cos^2x-Sin^2x\:=2Cos^2x-1}$

$\boxed{\rm{3}}$

$\rm{Cos^2x-Sin^2x}$

$\rm{1-Sin^2x-Sin^2x}$

$\boxed{Becoz\:\:Cos^2x=1-Sin^2x}$

$\rm{1-2Sin^2x}$

$\therefore$ $\rm{Cos^2x-Sin^2x\:=1-2Sin^2x}$