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Thread: Trigo Sums

  1. #1
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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
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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 $\displaystyle y = 4\cos3x$ at all. This means that it's not possible to be sure what the value of $\displaystyle k$ is. However, here is some working and a diagram showing two possible solutions. You will have to take it from here.

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

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

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

    $\displaystyle \Rightarrow k = -1$
    So that's one possible value of $\displaystyle k$. 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 $\displaystyle x$-axis; for instance, where $\displaystyle x = 210^o$. Then we get:
    $\displaystyle 2\sin210+k = 0$

    $\displaystyle \Rightarrow k = 1$
    This is the pink graph.

    Over to you. Can you sort it out now?

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

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

    Grandad
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  3. #3
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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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