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    state the max. possible value of 6sin(2x) - 4cos(2x) and determine the smallest value of x for which the maximum occurs.

    i know that 6sinx-4cosx= (52^1/2)sin(x-33.7)
    and that between 0 and 360, the values of x for the equation 6sin(2x) - 4cos(2x)=3 are 58.3 and 189.1
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    (Original post by Princess_perfect786)
    state the max. possible value of 6sin(2x) - 4sin(2x) and determine the smallest walue of x for which the maximum occurs.

    i know that 6sinx-4sinx= (52^1/2)sin(x-33.7)
    and that between 0 and 360, the values of x for the equation 6sin(2x) - 4sin(2x)=3 are 58.3 and 189.1
    6sin(2x) - 4sin(2x) = 2sin(2x) ...

    Consider the graph of y=sin(x) and what transformations are applied to it to get it to y=2sin(2x).
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    (Original post by dknt)
    6sin(2x) - 4sin(2x) = 2sin(2x) ...

    Consider the graph of y=sin(x) and what trnaformations are applied to it to get it to y=2sin(2x).
    oops i wrote the question wrong. it should say 6sin2x- 4cos2x
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    Generally (AQA) these questions require you to write the entire formula as either

    RSin(x\pm \ a)\ or\ RCos(x\pm \ a)

    From there you can then work out the coefficient (R) and then using what's inside the bracket work out the maximum value (cos and sin both maximum of 1, unless coefficient >1 in which case it is the value of the coefficient).

    I hope this helps!
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    (Original post by Princess_perfect786)
    oops i wrote the question wrong. it should say 6sin2x- 4cos2x
    Assuming you've done the R-alpha method correctly (too lazy to check )

    6sin2x- 4cos2x = (52^1/2)sin(x-33.7)

    First consider the graph of y=sinx. What is it's maximum value?

    Now consider y=(52^1/2)sin(x-33.7) . How do these transformations affect this maximum value?

    Once you've done this, set (52^1/2)sin(x-33.7) = to its maximum value.

    You won't need to worry about the "principle angle" and things like that, as you're trying to find the smallest value for which this occurs.
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    (Original post by dknt)

    6sin2x- 4cos2x = (52^1/2)sin(x-33.7)
    does this mean that whatever the coefficient of x is, the Rsin(x-a) formula will stay the same (as long as the coefficients of x for both sin x and cos x are the same)?
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    (Original post by Princess_perfect786)
    does this mean that whatever the coefficient of x is, the Rsin(x-a) formula will stay the same (as long as the coefficients of x for both sin x and cos x are the same)?
    Well, consider this with transformation of functions.

    First consider y=sinx, tranformed into y=sin(x-a). This is a translation of "a" in the x-direction. In otherwords it gets moved left and right. This has no effect on the maximum value. Now consider y=sinx to y=Rsin(x-a). This gives a translation of a, as already said, and also a stretch of scale factor R, in the y direction, so now the min. and max. points become -R and R. So you nedmax and mins for y=(52^1/2)sin(x-33.7) are 52^1/2 and -52^1/2
 
 
 
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