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Maths C3 - Trigonometry... Help?? Watch

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    (Original post by RDKGames)
    Multiply top and bottom by cosine.
    Thank you!!! How did you know to multiply top and bottom by cos(x)?


    (Original post by notnek)
     \displaystyle \frac{1-tan x}{1+tan x} = \frac{1-\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}}

    The next step is to multiply top and bottom of the fraction by \cos x.
    Thank you. I never would have known to multiply the top and bottom by cos(x)
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    (Original post by Philip-flop)
    Thank you!!! How did you know to multiply top and bottom by cos(x)?



    Thank you. I never would have known to multiply the top and bottom by cos(x)
    Whenever you see a fraction on the top/bottom of another fraction then it's often useful to get rid of it as soon as possible by multiplying top and bottom by the denominator.

    E.g.

    \displaystyle \frac{y-x}{\frac{y}{x}-1}

    If you were trying to simplify this then you could get rid of \frac{y}{x} by multiplying top and bottom of the fraction by x. This gives

    \displaystyle \frac{y-x}{\frac{y}{x}-1} = \frac{yx-x^2}{y-x}
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    (Original post by notnek)
    Whenever you see a fraction on the top/bottom of another fraction then it's often useful to get rid of it as soon as possible by multiplying top and bottom by the denominator.

    E.g.

    \displaystyle \frac{y-x}{\frac{y}{x}-1}

    If you were trying to simplify this then you could get rid of \frac{y}{x} by multiplying top and bottom of the fraction by x. This gives

    \displaystyle \frac{y-x}{\frac{y}{x}-1} = \frac{yx-x^2}{y-x}
    Thanks! That's definitely helped fill a gap in my knowledge I really appreciate it!
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    notnek but for the next bit after I get...

     \frac{cos x - sinx}{cos x + sinx}

    how am I meant to know to do \frac{cos x - sinx}{cos x + sinx} \times \frac{cos x - sinx}{cos x - sinx}??
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    (Original post by Philip-flop)
    notnek but for the next bit after I get...

     \frac{cos x - sinx}{cos x + sinx}

    how am I meant to know to do \frac{cos x - sinx}{cos x + sinx} \times \frac{cos x - sinx}{cos x - sinx}??
    This comes a bit from practice/experience.

    It's similar to why you multiply top and bottom of e.g. \displaystyle \frac{3}{2+\sqrt{3}} by 2-\sqrt{3} to rationalise it.

    The difference of two squares method can sometimes simplify the denominator or put it in another form.

    In this case I noticed that (\cos x + \sin x)(\cos x - \sin x) = \cos^2 - \sin^2x = \cos 2x

    Using difference of two squares like this in trig can be a bit of a last resort method if you can't get anywhere from standard manipulation.
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    (Exe7C Question 6)

    For part (b) how do I know which solutions from...
     \theta = \frac{\pi}{24}, \frac{5 \pi}{24}, \frac{13\pi}{24}, \frac{17\pi}{24}

    ...satisfy the equation...
     cos2 \theta - sin2 \theta = \frac{1}{\sqrt 2} ?? <<I'm guessing it has something to do with the square root on the RHS of this equation.

    Do I just have to test each "solution" of theta by subbing them into the equation above using my calculator or is there a quicker way?
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    (Original post by Philip-flop)
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    (Exe7C Question 6)

    For part (b) how do I know which solutions from...
     \theta = \frac{\pi}{24}, \frac{5 \pi}{24}, \frac{13\pi}{24}, \frac{17\pi}{24}

    ...satisfy the equation...
     cos2 \theta - sin2 \theta = \frac{1}{\sqrt 2} ??

    Do I just have to test each "solution" of theta by subbing them into the equation above using my calculator or is there a quicker way?
    They all satisfy it, it even says "answers" (plural) in the question so you just list them.
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    (Original post by RDKGames)
    They all satisfy it, it even says "answers" (plural) in the question so you just list them.
    But the book says its only...
     \theta = \frac{\pi}{24}, \frac{17\pi}{24}

    Which is why I'm confused
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    Where the heck do I even start with part (b)i) ?? I'm not sure I even understand what the question is asking
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    (Original post by Philip-flop)
    Where the heck do I even start with part (b)i) ??
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    For the first one if  \tan \frac{\theta }{2}=t we have  \displaystyle \frac{2t}{1+t^2}+\frac{2(1-t^2)}{1+t^2}=1 .
    SO just solve the quadratic equation as you normally would.
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    (Original post by Philip-flop)
    Where the heck do I even start with part (b)i) ?? I'm not sure I even understand what the question is asking
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    get  tan\theta from the previous part, using \frac{sin\theta}{cos\theta}, in terms of  tan\frac{\theta}{2}
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    (Original post by B_9710)
    For the first one if  \tan \frac{\theta }{2} we have  \displaystyle \frac{2t}{1+t^2}+\frac{2(1-t^2)}{1+t^2}=1 .
    SO just solve the quadratic equation as you normally would.
    But how do you get...  \tan \frac{\theta }{2} >>>  \displaystyle \frac{2t}{1+t^2}+\frac{2(1-t^2)}{1+t^2}=1

    from... (b)i)  sin \theta + 2cos \theta = 1 ?? Sorry if I'm being silly
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    (Original post by NotNotBatman)
    get  tan\theta from the previous part, using \frac{sin\theta}{cos\theta}, in terms of  tan\frac{\theta}{2}
    Oh right! It's starting to make sense now. But where have those 't's come from?
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    (Original post by Philip-flop)
    Oh right! It's starting to make sense now. But where have those 't's come from?
    The above user has shortened it, so  t = tan\frac{\theta}{2} and has done  sin\theta +2cos\theta = 1 in terms of t.
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    (Original post by NotNotBatman)
    The above user has shortened it, so  t = tan\frac{\theta}{2} and has done  sin\theta +2cos\theta = 1 in terms of t.
    Thank you so much
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    For part (a) can someone please explain how this goes from....
     2cos^2 \frac{\theta}{2}

    to this...
     = \frac{1+cos \theta}{2} ??

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    (Original post by Philip-flop)
    But the book says its only...
     \theta = \frac{\pi}{24}, \frac{17\pi}{24}

    Which is why I'm confused
    Ah yes, looked at the question and it makes sense. The two extra solutions come from the fact that when you square \frac{1}{\sqrt{2}} you create an answer which can be achieved from squaring -\frac{1}{\sqrt{2}} so two of your solutions satisfy \sin(2\theta)-\cos(2\theta)=-\frac{1}{\sqrt{2}} which is not what we want.

    The easiest way around is to plug the values back through the equation and collect the values you need. You can also work out for which angles \cos2 \theta - \sin2 \theta is positive and consider only the answers which fall into those domains.
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    (Original post by Philip-flop)
    For part (a) can someone please explain how this goes from....
     2cos^2 \frac{\theta}{2}

    to this...
     = \frac{1+cos \theta}{2} ??

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    It doesn't.

    \cos(\theta)=\cos^2(\frac{\theta  }{2})-\sin^2(\frac{\theta}{2})=2\cos^2  (\frac{\theta}{2})-1 \implies \cos(\theta)+1=2\cos^2(\frac{ \theta }{2} )
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    (Original post by RDKGames)
    Ah yes, looked at the question and it makes sense. The two extra solutions come from the fact that when you square \frac{1}{\sqrt{2}} you create an answer which can be achieved from squaring -\frac{1}{\sqrt{2}} so two of your solutions satisfy \sin(2\theta)-\cos(2\theta)=-\frac{1}{\sqrt{2}} which is not what we want.

    The easiest way around is to plug the values back through the equation and collect the values you need. You can also work out for which angles \cos2 \theta - \sin2 \theta is positive and consider only the answers which fall into those domains.
    Thank you. That makes perfect sense to me now!

    (Original post by RDKGames)
    It doesn't.

    \cos(\theta)=\cos^2(\frac{\theta  }{2})-\sin^2(\frac{\theta}{2})=2\cos^2  (\frac{\theta}{2})-1 \implies \cos(\theta)+1=2\cos^2(\frac{ \theta }{2} )
    Thaaaank you!! I never thought I could re-arrange the double angle formulae in that way!! Seriously appreciate your help
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    Can someone help explain from where it says "Occurs when cos( \theta - 22.6) = 1 ......" <<Where has the number 1 come from?..
    I've worked out the rest of the question and can see that the maximum is 13 but I can't seem to understand the last bit. Why is the smallest value of  \theta 22.6 ??

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