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Anyone need help with any *EDEXCEL* C3 and C4 past paper questions? watch

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    Somebody please help me with a C3 TRIG question! :o:

    Mixed Exercise 7F. Q6. part e. + part f.

    I looked at the first line of the answers in hope that it'd click in my mind and i could do the rest of the proof. But i still can't

    Thank you
    + rep. :awesome:
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    ^ Can you post the question?
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    (Original post by Apple24)
    ^ Can you post the question?
    Sure

    Show that :

    1-cos2x
    --------- = (sec^2 x -1 )
    1+cos2x

    EDIT: I posted the wrong one! :facepalm: sorry!! :o:

    this is the one i'm stuck on:

    show that:

    sin(x+y)sin(x-y) = coz^2y - cos^2x
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    Use cos2x=(1-2sin^2x) on the top, and cos2x=(2cos^2x-1) on the bottom.
    These will cancel out the 1's and it'll look a bit nicer then!
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    (Original post by Apple24)
    ^ Can you post the question?
    Sorry, posted the wrong question. :o:
    I've edited the post, with the right one now.
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    (Original post by matty92)
    Use cos2x=(1-2sin^2x) on the top, and cos2x=(2cos^2x-1) on the bottom.
    These will cancel out the 1's and it'll look a bit nicer then!
    Thanks. Had a slightly blonde moment and posted the question i just did :o:
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    I think this question would be to do with the trig identity that cosA-cosB=-2sin(a+b/2)sin((a-b)/2)?

    I will re-edit later if I get it!
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    (Original post by time.to.dance)
    Thanks. Had a slightly blonde moment and posted the question i just did :o:
    Haha. Where have you got upto on this one? I started with multiplying out using the addition formula sin(a+b)=sina.cosb+cosa.sin.b It does actually cancel down a bit.
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    (Original post by matty92)
    Haha. Where have you got upto on this one? I started with multiplying out using the addition formula sin(a+b)=sina.cosb+cosa.sin.b It does actually cancel down a bit.
    Nowhere! i've been staring at it for 10 mins now :p: couldn't figure out the first step.
    Thanks for that! i'll attempt it again now
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    Expand
    sin(x+y).sin(x-y)

    = sin^2x.cos^2y - cos^2x.sin^2y

    = (1-cos^2x).cos^2y - cos^2x.(1-cos^2y)

    ...
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    (Original post by matty92)
    Expand
    sin(x+y).sin(x-y)

    = sin^2x.cos^2y - cos^2x.sin^2y

    = (1-cos^2x).cos^2y - cos^2x.(1-cos^2y)

    ...
    Thank you! i've done it now i'll rep as soon as i can :awesome:
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    (Original post by matty92)
    Haha. Where have you got upto on this one? I started with multiplying out using the addition formula sin(a+b)=sina.cosb+cosa.sin.b It does actually cancel down a bit.
    I'm stuck on the next question now :o: oh god that maths prediction does not look so promising :o:

    tan[(pie/4)+x] - tan[(pie/4) -- x] = 2tan2x (not to the power)

    sunday evening is not the time for last minute maths homework either :rolleyes:

    Any help would be really appreciated
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    how do I prove that: cot2x + cosec2x = cotx ????

    omg, this is annoying me soo much, been thinkin about this the whole day
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    ^Change cot and cosec into cos and sins and try and simplify.
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    (Original post by W.H.T)
    how do I prove that: cot2x + cosec2x = cotx ????

    omg, this is annoying me soo much, been thinkin about this the whole day

    1/tan2x + 1/sin2x

    = cos2x/sin2x + 1/sin2x

    = (1+cos2x)/sin2x

    = (2cosx^2 - 1 + 1)/2sinxcosx

    = 2cos^2x/2sinxcosx

    cancle by 2cosx

    = cosx/sinx

    = cotx
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    just a quick question as i cant seem to interpret what the MS has done..

    1b) Find the exact solutions to the equation:

    e^x+3e^-x = 4

    i do know that i had to turn it into an quadratic equations however how does the mark scheme manage to change the 4 and the 3 around so there isnt a e^-x?

    (e^x)^2 - 4e^x + 3 = 0

    thanks!
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    Because you're multiply both sides by e^x.

    (e^x)(e^x) + 3(e^-x)(e^x) = 4(e^x)
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    (Original post by time.to.dance)
    I'm stuck on the next question now :o: oh god that maths prediction does not look so promising :o:

    tan[(pie/4)+x] - tan[(pie/4)+x] = 2tan2x (not to the power)

    sunday evening is not the time for last minute maths homework either :rolleyes:

    Any help would be really appreciated
    Nope maybe not the best time...:rolleyes:
    Probably a bit late now but, its a bit of a strange question... doesn't the tan(\frac{\pi}{4}+x) - tan(\frac{\pi}{4}+x) cancel to make zero?

    So we're proving that  0=2tan2x ? or am I just completely misinterpreting it?
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    (Original post by matty92)
    Nope maybe not the best time...:rolleyes:
    Probably a bit late now but, its a bit of a strange question... doesn't the tan(\frac{\pi}{4}+x) - tan(\frac{\pi}{4} - x) cancel to make zero?

    So we're proving that  0=2tan2x ? or am I just completely misinterpreting it?
    Ahh it was rather late yesterday :o: I've edited it now. It was meant to be a minus before the x in the second bracket.
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    Good luck everyone
 
 
 
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