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    Currently C3 trig is making me want to die because I've spent about an hour on about 3 questions. Here are the 3 I'm struggling with (hint, they all have sin2x in)
    1) cos^2(x) - sin2(x) = sin^2(x)
    2) 2sin2(x) = 3tan(x)
    3) 5sin2(x) + 4sin(x) = 0

    If anyone could help on any of them then I'd be really grateful.
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    I'll go one by one. For the first, assuming you mean \cos^2(x) - \sin^2(x) = \sin^2(x), remember that \displaystyle \tan x = \frac{\sin x}{\cos x} and so \displaystyle \tan^2 x = \frac{\sin^2 x}{\cos^2 x}. How does this help here?
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    (Original post by AxSirlotl)
    Currently C3 trig is making me want to die because I've spent about an hour on about 3 questions. Here are the 3 I'm struggling with (hint, they all have sin2x in)
    1) cos^2(x) - sin2(x) = sin^2(x)
    2) 2sin2(x) = 3tan(x)
    3) 5sin2(x) + 4sin(x) = 0

    If anyone could help on any of them then I'd be really grateful.
    for 1.)
    cos^2(x)-sin^2(x)=sin(2x)
    cos^2(x)-sin^2(x)=cos(2x) so

    cos(2x)=sin(2x)
    divide by cos(2x) but remember you get cos(2x)=0 as a lost solution.
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    (Original post by _gcx)
    I'll go one by one. For the first, assuming you mean \cos^2(x) - \sin^2(x) = \sin^2(x), remember that \displaystyle \tan x = \frac{\sin x}{\cos x} and so \displaystyle \tan^2 x = \frac{\sin^2 x}{\cos^2 x}. How does this help here?
    Is there a guide on how to type maths equations like that because I've been looking but I can't find one.
    What I mean is cos squared x - sin 2x = sin squared x (I hope that makes more sense.
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    (Original post by AxSirlotl)
    Is there a guide on how to type maths equations like that because I've been looking but I can't find one.
    What I mean is cos squared x - sin 2x = sin squared x (I hope that makes more sense.
    There's a guide here. From \cos^2(x) - \sin(2x) = \sin^2(x) you have \cos^2(x) - \sin^2(x) = \sin(2x). Is there any identity that you've learnt that you can apply to the LHS here?
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    (Original post by _gcx)
    I'll go one by one. For the first, assuming you mean \cos^2(x) - \sin^2(x) = \sin^2(x)
    But... but... that's not an identity otherwise \cot(x) \equiv 2 :/
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    (Original post by RDKGames)
    But... but... that's not an identity otherwise \cot(x) \equiv 2 :/
    Did I say it was an identity? :rofl:

    Spoiler:
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    I meant the double angle formula for the LHS
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    (Original post by RDKGames)
    But... but... that's not an identity otherwise \cot(x) \equiv 2 :/
    It's not clear from the question, but I think that these are equations to be solved, rather than identities. OP - can you clarify?
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    (Original post by Pangol)
    It's not clear from the question, but I think that these are equations to be solved, rather than identities. OP - can you clarify?
    I think it's pretty safe to assume that they're equations to be solved.
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    (Original post by _gcx)
    There's a guide here. From \cos^2(x) - \sin(2x) = \sin^2(x) you have \cos^2(x) - \sin^2(x) = \sin(2x). Is there any identity that you've learnt that you can apply to the LHS here?
    Thanks, should I get  Sin2x = 1 ?
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    (Original post by Pangol)
    It's not clear from the question, but I think that these are equations to be solved, rather than identities. OP - can you clarify?
    They need to be solved
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    (Original post by _gcx)
    Did I say it was an identity? :rofl:
    (Original post by Pangol)
    It's not clear from the question, but I think that these are equations to be solved, rather than identities. OP - can you clarify?
    Eh yeah, they're equations to solve, I thought OP meant identities to prove without knowing he can use the equivalence relation in his post, then gcx just skipping the \equiv part by being lazy and use = instead

    Most of these threads are identities so I just skipped to assume without reading the thread properly lol
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    (Original post by AxSirlotl)
    Thanks, should I get  Sin2x = 1 ?
    Easy mistake to make, \cos^2(x) + \sin^2(x) \equiv 1, but \cos^2(x) - \sin^2(x) \not\equiv 1 . Think double angle formulas.
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    (Original post by _gcx)
    Easy mistake to make, \cos^2(x) + \sin^2(x) \equiv 1, but \cos^2(x) - \sin^2(x) \not\equiv 1 . Think double angle formulas.
    Oh yeah I've just realised that it's the  Cos2A double angle formula.
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    (Original post by AxSirlotl)
    Oh yeah I've just realised that it's the  Cos2A double angle formula.
    So are you fine with saying we have \cos 2x = \sin 2x? Where can we go from there?

    Btw, you can click on LaTeX here to see what code was used to produce it
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    (Original post by _gcx)
    So are you fine with saying we have \cos 2x = \sin 2x? Where can we go from there?

    Btw, you can click on LaTeX here to see what code was used to produce it
    I got to  cos2x = sin2x and then from there got to  tan^2(x) + 2tan(x) -1 = 0 which is a quadratic but I don't know if I've made a mistake somewhere or if I've done the right thing.
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    (Original post by AxSirlotl)
    I got to  cos2x = sin2x and then from there got to  tan^2(x) + 2tan(x) -1 = 0 which is a quadratic but I don't know if I've made a mistake somewhere or if I've done the right thing.
    Not quite sure what you did there. You can just divide both sides by \cos 2x to get \tan 2x = 1
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    (Original post by _gcx)
    Not quite sure what you did there. You can just divide both sides by \cos 2x to get \tan 2x = 1
    I only managed to get one of the correct solutions from what I did, thanks for the help, I have a habit of over-complicating things.
 
 
 
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