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Math Help - Trigo Sums

  1. #1
    Junior Member
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    Trigo Sums

    Dear all,

    Could you kindly help me with these 2 sums - Q 18 and Q21 - many, many thanks
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Hello Drdj
    Quote Originally Posted by Drdj View Post
    Dear all,

    Could you kindly help me with these 2 sums - Q 18 and Q21 - many, many thanks
    I think one of the reasons no-one has attempted to answer question 18 is that the diagram is not clear. We can't see the graph of y = 4\cos3x at all. This means that it's not possible to be sure what the value of k is. However, here is some working and a diagram showing two possible solutions. You will have to take it from here.

    First y = 4\cos3x crosses the x-axis where:
    \cos3x = 0
    So, working in degrees:
    \Rightarrow 3x = 90, 270, 450, 630, ...

    \Rightarrow x = 30, 90, 150, 210, ...
    See the black graph in the attached diagram.

    So let's suppose first that this graph meets y= 2\sin x + k where x = 30^o. Then:
    2\sin30+k=0

    \Rightarrow k = -1
    So that's one possible value of k. See the yellow coloured graph.

    But perhaps the diagram shows a different position of the graph, and they actually meet at a different point at which the first graph crosses the x-axis; for instance, where x = 210^o. Then we get:
    2\sin210+k = 0

    \Rightarrow k = 1
    This is the pink graph.

    Over to you. Can you sort it out now?

    For question 21: multiply top-and-bottom by (1-\sin x):
    \sqrt{\frac{1-\sin x}{1+\sin x}}= \sqrt{\frac{(1-\sin x)^2}{1-\sin^2 x}}
    = \sqrt{\frac{(1-\sin x)^2}{\cos^2 x}}
    =\frac{1-\sin x}{\cos x}, if we are given that x is acute, because \cos x is then positive.

    =\sec x - \tan x
    But if x is obtuse, \sin x > 0 and \cos x < 0. So \cos x = -\sqrt{\cos^2x}. So we would have to say:
    \sqrt{\frac{(1-\sin x)^2}{\cos^2 x}}=-\frac{1-\sin x}{\cos x}
    =\tan x - \sec x
    Can you see what happens for other values of x? If you're not sure, look at the graphs of \sec x and \tan x, and see whereabouts \sec x >\tan x and whereabouts \tan x > \sec x. And then bear in mind that \sqrt{\frac{1-\sin x}{1+\sin x}} has to be positive.

    Grandad
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  3. #3
    Junior Member
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    Thanks

    Dear Grandad,

    You're simply wonderful !!

    Apologies for the not-so-clear scan of Qn 18 - causing the confusion ....
    But thank you so much for your detailed and clear explanation, not to mention to the colorful graphs plotted !! )
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