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    Hi there,

    I am trying to solve the following two (separate) questions.

    i) SinA= Cos2A

    ii) Sin2A -1 = Cos2A

    Where A must be presented in radians

    Find A.

    I'm afraid I'm a bit rusty on trigonometry as I have not done this in the 3 years I have been at Uni, so couldn't find many helpful resources on how to get started on this problem.
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    (Original post by VincentCheung)
    Hi there,

    I am trying to solve the following two (separate) questions.

    i) SinA= Cos2A

    ii) Sin2A -1 = Cos2A

    Where A must be presented in radians

    Find A.

    I'm afraid I'm a bit rusty on trigonometry as I have not done this in the 3 years I have been at Uni, so couldn't find many helpful resources on how to get started on this problem.
    Think of how you can use double angle formulae here.
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    For the first one use the double angle formula for cos and form a quadratic in sinA.
    For the second one it will probably be helpful to cancel the -1 on the LHS so you may want to choose the appropriate double angle formula for cos2A.
    (What course aware you doing at uni?)
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    Oh I see, I completely forgot about focusing on a quadratic in Sine! Perfect part i is all good.

    For part 2 I am pretty sure I have done it right. So I use Cos2A = (CosA)^2 - 1 and remove the -1 as you have said. Then I rearrange to get tanA =0 and finally A = 0.

    I did Maths and Economics, although throughout the course I hardly touched trigonometric identities.
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    (Original post by VincentCheung)
    Oh I see, I completely forgot about focusing on a quadratic in Sine! Perfect part i is all good.

    For part 2 I am pretty sure I have done it right. So I use Cos2A = (CosA)^2 - 1 and remove the -1 as you have said. Then I rearrange to get tanA =0 and finally A = 0.

    I did Maths and Economics, although throughout the course I hardly touched trigonometric identities.
    Your penultimate step isn't quite right. You have \sin A = \cos^2 A, which doesn't get you \tan A = 0.

    Instead, write everything in terms of \sin A again so you have \sin A - 1 = 1 - 2\sin^2 A, now re-arrange to get a quadratic in \sin A.
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    Sorry it's so I used the double angle on both sides of the equation. So the and then after cancelling the -1 I was left with which I then divided by CosA and 2 on both sides to get .... But now I do see I've gone wrong as it should be 45 since dividing by CosA I put as 0 on the RHS, whereas it should be 1.
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    (Original post by VincentCheung)
    Sorry it's so I used the double angle on both sides of the equation. So the and then after cancelling the -1 I was left with which I then divided by CosA and 2 on both sides to get .... But now I do see I've gone wrong as it should be 45 since dividing by CosA I put as 0 on the RHS, whereas it should be 1.
    You cannot divide by cosA since cosA=0 is a solution to the equation - in effect you are losing infinitely many solutions by dividing by cosA. You need to factorise.
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    Oh yes you're right.. So if I factorise then I get 0 is a solution along with the one I have found already? (A=45 degrees)
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    (Original post by VincentCheung)
    Oh yes you're right.. So if I factorise then I get 0 is a solution along with the one I have found already? (A=45 degrees)
    No, arccos(0) is 90 degrees.
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    Ah, another silly mistake.. But yes, thank you very much for your help!
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    (Original post by VincentCheung)
    Ah, another silly mistake.. But yes, thank you very much for your help!
    Also remember that there are infinitely many solutions unless the range of x values are restricted in the question.
    If  \sin x= 1/2 for example, there will be infinitely many solutions of the form  x=\pi /6 +2n\pi and also  x= 5\pi /6 +2n\pi, n \in \mathbb{Z} .
 
 
 
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