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

Question

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

in progress 0
Madeline 4 days 2021-10-11T01:49:23+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-10-11T01:50:25+00:00

    \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}

    .

Leave an answer

Browse

14:4+1-6*5-7*14:3+5 = ? ( )