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    Can anyone teach me how to do this question fully?
    The mark scheme is missing so many steps
    I cannot find R and alpha

    Thanks in advance
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    (Original post by mystreet091234)
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    Can anyone teach me how to do this question fully?
    The mark scheme is missing so many steps
    I cannot find R and alpha

    Thanks in advance

    Ok, so I'm assuming you know how to use the product rule to get to e^-x(2cos2x-sin2x)?


    Ok from there normally to find R, you square root a^2 and b^2so cosx=a and sinx=b

    we have the form RCOS(X+alpha) this can also be written as acosx-bsinx

    Using the equation we found before if we compare whats in the bracket to the version that we are given we can see that:

    2cos2x-sinx2x
    acos2x-bsin2x

    so a=2 and b=1 (you don't need to worry about the signs here)

    so to find R= square root 2^2 + 1^2
    so we get square root of 5

    Using what we know before we can see that cos2x=a which is 2
    and sin2x=b which is 1

    normally tanalpha= sin2x/cos2x (as tanx is sinx/cosx)

    so we can see that tanalpha= 1/2

    so inverse tan (1/2) we get 0.46

    Does that help? It was kinda difficult to explain it on here,
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    (Original post by cookiemonster15)
    Ok, so I'm assuming you know how to use the product rule to get to e^-x(2cos2x-sin2x)?


    Ok from there normally to find R, you square root a^2 and b^2so cosx=a and sinx=b

    we have the form RCOS(X+alpha) this can also be written as acosx-bsinx

    Using the equation we found before if we compare whats in the bracket to the version that we are given we can see that:

    2cos2x-sinx2x
    acos2x-bsin2x

    so a=2 and b=1 (you don't need to worry about the signs here)

    so to find R= square root 2^2 + 1^2
    so we get square root of 5

    Using what we know before we can see that cos2x=a which is 2
    and sin2x=b which is 1

    normally tanalpha= sin2x/cos2x (as tanx is sinx/cosx)

    so we can see that tanalpha= 1/2

    so inverse tan (1/2) we get 0.46

    Does that help? It was kinda difficult to explain it on here,
    Yes!! it helps a lot thank you so much!
 
 
 
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