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    Hi, I am having difficulties with this question in a paper I have just done.

    Using the identity (1-cos2x)/sin2x = tanx

    Solve, giving your answer in terms of π, 2(1-cos2x) = tanx for 0<x<π

    Really appreciate your advice. I currently cannot see how to solve the equation by rearrangement.
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    (Original post by gabby07)
    Hi, I am having difficulties with this question in a paper I have just done.

    Using the identity (1-cos2x)/sin2x = tanx

    Solve, giving your answer in terms of π, 2(1-cos2x) = tanx for 0<x<π

    Really appreciate your advice. I currently cannot see how to solve the equation by rearrangement.
    Well, if \frac{1-\cos 2x}{\sin 2x} = \tan x then (1-\cos 2x) = \tan x\sin 2x so: 2(1-\cos 2x) = 2\sin 2x\tan x

    Which makes your equation 2\sin 2x\tan x = \tan x \Rightarrow \tan x(2\sin 2x-1) =0 which you can then solve.
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    I would substitute the tanx for (1-cos2x)/sin2x and then bring the 2(1-cos2x) from the LHS to the RHS were it becomes the denominator, and you bring the sin2x up on the LHS so you get sin2x=(1-cos2x)/2(1-cos2x) so the brackets cancel and you get sin2x=1/2 and then you solve for 2x where 0<2x<2pi and divide your answers by 2 to get x where 0<x<pi

    I don't know if this is right I have done it quickly in my head, but I think it is.
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    (Original post by Zacken)
    Well, if \frac{1-\cos 2x}{\sin 2x} = \tan x then (1-\cos 2x) = \tan x\sin 2x so: 2(1-\cos 2x) = 2\sin 2x\tan x

    Which makes your equation 2\sin 2x\tan x = \tan x \Rightarrow \tan x(2\sin 2x-1) =0 which you can then solve.
    Thanks! Letting tanx = 0 though isn't giving me a solution in the range - 0 is the primary answer on my calculator ! I will try with the 2sin2x-1 = 0 now
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    (Original post by PLM98)
    I don't know if this is right I have done it quickly in my head, but I think it is.
    :no: You know something is wrong when you talk about "cancelling" when dealing with trig functions. It's lucky that in this case, because of the range, it works out - but you'd need to specify the validity of your cancelling.
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    (Original post by gabby07)
    Thanks! Letting tanx = 0 though isn't giving me a solution in the range - 0 is the primary answer on my calculator ! I will try with the 2sin2x-1 = 0 now
    Yep, \tan x = 0 \Rightarrow x = 0, \pi so you can discard both of these solutions, but you'll be expected to find them and then specify that you're discarding them on your exam paper so they can away you the marks.
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    (Original post by Zacken)
    :no: You know something is wrong when you talk about "cancelling" when dealing with trig functions. It's lucky that in this case, because of the range, it works out - but you'd need to specify the validity of your cancelling.
    Hahahaha it probably is, just had a quick try for fun. You better trust Zacken then @gabby07 hahahaha
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    Oh no im not wrong, I got the same thing as 2sin2x - 1=0 which is what Zacken got in the bracket
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    (Original post by Zacken)
    Yep, \tan x = 0 \Rightarrow x = 0, \pi so you can discard both of these solutions, but you'll be expected to find them and then specify that you're discarding them on your exam paper so they can away you the marks.
    Ah ok - will do that.
    And with sin2x = 1/2 how do you go about solving? Do you rewrite sin2x as 2 sin x cos x? I would have thought not as that introduces another trig function, but I'm not sure how to go about dealing with the sin2x otherwise.
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    (Original post by gabby07)
    Ah ok - will do that.
    And with sin2x = 1/2 how do you go about solving? Do you rewrite sin2x as 2 sin x cos x? I would have thought not as that introduces another trig function, but I'm not sure how to go about dealing with the sin2x otherwise.
    Nopes, just write: u = 2x solve \sin u = \frac{1}{2} normally and then find all your answer.

    Then, for example: u = \text{answer 1} so 2x = \text{answer 1} so x = \frac{\text{answer 1}}{2}.
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    (Original post by PLM98)
    Oh no im not wrong, I got the same thing as 2sin2x - 1=0 which is what Zacken got in the bracket
    Yes, like I said - in this case, you were lucky because of the range. But other solutions are generated from \tan x= 0 which you cancelled out.
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    No you do the inverse sin of 1/2 which gives you the answers for 2x (and since it is 2x and not x the answers are 0<2x<2pi), then once youve got your answers in that interval divide by 2, and you get x in the interval 0<x<pi

    EDIT: Zacken was faster than me again hahahahaha
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    (Original post by Zacken)
    Yes, like I said - in this case, you were lucky because of the range. But other solutions are generated from \tan x= 0 which you cancelled out.
    True
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    Another thread I missed to show off my trigonometric skills ...
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    (Original post by TeeEm)
    Another thread I missed to show off my trigonometric skills ...
    Want me to create a thread so you can show off?
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    (Original post by drandy76)
    Want me to create a thread so you can show off?
    save it ... I am not really that good... I was only saying that because the problem was already solved.
    Name:  f8.jpg
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    (Original post by TeeEm)
    save it ... I am not really that good... I was only saying that because the problem was already solved.
    Name:  f8.jpg
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    That's a shame, was about to pull out an off syllabus fp3 question that OCR just throw in for the masochists out there
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    (Original post by Zacken)
    Well, if \frac{1-\cos 2x}{\sin 2x} = \tan x then (1-\cos 2x) = \tan x\sin 2x so: 2(1-\cos 2x) = 2\sin 2x\tan x

    Which makes your equation 2\sin 2x\tan x = \tan x \Rightarrow \tan x(2\sin 2x-1) =0 which you can then solve.
    Hey Zacken, just a question. How do you jump from 2(1-cos2x) = 2sin2x tanx to 2sin2x tanx=tanx ? And if you end up with that then then 2sin2x=1 so sin2x=1/2 and you dont get any bracket with a tanx outside or anything.
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    (Original post by drandy76)
    That's a shame, was about to pull out an off syllabus fp3 question that OCR just throw in for the masochists out there
    I wish I had the time to do it ...
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    (Original post by TeeEm)
    I wish I had the time to do it ...
    i could post it if you like,then you could get around to doing it whenever you're free?
 
 
 
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