Question

If cos2x-cosx= sin4x - sinx(
where tan doesnot equal to 1) find value of cos3x - sin3x

Answer 1

$\textbf{Given:}$

$cos2x-cosx=sin4x-sinx$

$\textbf{To find:}$

$\text{The value of cos3x-sin3x }$

$\textbf{Solution:}$

$\text{Consider,}$

$cos3x-sin3x$

$=(4\,cos^3x-3\,cosx)-(3\,sinx-4\,sin^3x)$

$=4\,cos^3x-3\,cosx-3\,sinx+4\,sin^3x$

$=4(cos^3x+sin^3x)-3(cosx+sinx)$

$=4(cosx+sinx)(cos^2x-cosx\,sinx+sin^2x)-3(cosx+sinx)$

$=4(cosx+sinx)(1-cosx\,sinx)-3(cosx+sinx)$

$=(cosx+sinx)(4(1-cosx\,sinx)-3)$

$=(cosx+sinx)(4-4\,cosx\,sinx-3)$

$=(cosx+sinx)(1-4\,cosx\,sinx)$

$=(cosx+sinx)(1-2(2\,sinx\,cosx))$

$\implies\bf\,cos3x-sin3x=(cosx+sinx)(1-2\,sin\,2x))$.........(1)

$\text{Now,}$

$cos2x-cosx=sin4x-sinx$

$\text{Rearranging terms we get}$

$sinx-cosx=sin4x-cos2x$

$sinx-cosx=2\,sin2x\,cos2x-cos2x$

$sinx-cosx=cos2x(2\,sin2x-1)$

$\dfrac{sinx-cosx}{cos2x}=2\,sin2x-1$

$\implies\,1-2\,sin2x=\dfrac{cosx-sinx}{cos2x}$.........(2)

$\text{Using (2) in (1), we get}$

$cos3x-sin3x=(cosx+sinx)(\dfrac{cosx-sinx}{cos2x})$

$cos3x-sin3x=\dfrac{cos^2x-sin^2x}{cos2x})$

$cos3x-sin3x=\dfrac{cos^2x-sin^2x}{cos2x}$

$cos3x-sin3x=\dfrac{cos2x}{cos2x}$

$\implies\bf\,cos3x-sin3x=1$

$\textbf{Answer:}$

$\textbf{The value of \bf\,cos3x-sin3x is 1}$

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