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    I need a bit of help fully understanding this question.

    7. (a) Express 4 sin x + 3 cos x in the form R sin (x + alpha) where R > 0
    and 0 < alpha < pi/2 (4)

    (b) State the minimum value of 4 sin x + 3 cos x and the smallest positive
    value of x for which this minimum value occurs. (3)


    Part a I'm fine with. But on part b, I knew it was -5 since R from the previous question was 5 and I've just followed the pattern from previous exam papers. But I don't actually understand where it comes from.

    This is the answer from the mark scheme:

    minimum = -5 (B1)
    occurs when x + 0.6435 = 3Pi/2, x = 4.07 (3sf)
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    The reason is because when you write R\sin (x+\alpha) what you're doing is scaling and shifting the graph of y=\sin x. As you can tell from the graph, the maximum value \sin x can take is 1, and the minimum is -1, so when you scale it by R, the maximum and minimum values become \pm R.

    I can't work out the specifics of where you've gone wrong because your symbols seem to have come out weird, but anyway, when working out the smallest positive value, it's a good idea to make the substitution \theta = x + \alpha. Then you need to remember rules like \cos \theta = \cos (-\theta) and \sin \theta = \sin (\frac{\pi}{2}-\theta) and so on. Because you're shifting the graph, it might be that one of these gives you a smaller positive value of x despite the value of \theta being negative.
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    (Original post by nuodai)
    The reason is because when you write R\sin (x+\alpha) what you're doing is scaling and shifting the graph of y=\sin x. As you can tell from the graph, the maximum value \sin x can take is 1, and the minimum is -1, so when you scale it by R, the maximum and minimum values become \pm R.

    I can't work out the specifics of where you've gone wrong because your symbols seem to have come out weird, but anyway, when working out the smallest positive value, it's a good idea to make the substitution \theta = x + \alpha. Then you need to remember rules like \cos \theta = \cos (-\theta) and \sin \theta = \sin (\frac{\pi}{2}-\theta) and so on. Because you're shifting the graph, it might be that one of these gives you a smaller positive value of x despite the value of \theta being negative.

    Thank you for the help it made complete sense.
 
 
 
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