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C4 trig

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    I got :
    (1- tany) / (1+ tany) = k

    now i need to express k in terms of tany so i need tany= ?
    but how do i do it ?
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    done it.

    1-tany = k + k.tany
    1 - k = k.tany +tany
    1 - k = tany (k + 1)
    (1 - k) / (1+ k) = tany

    If anyones got a simpler way, i would still like to know :P
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    (Original post by uxa595)
    done it.

    1-tany = k + k.tany
    1 - k = k.tany +tany
    1 - k = tany (k + 1)
    (1 - k) / (1+ k) = tany

    If anyones got a simpler way, i would still like to know :P
    Nope that's the way to do it.

    I'm guessing you're on AQA cause I remember helping someone in my school do this question just a couple of days ago
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    (Original post by hassi94)
    Nope that's the way to do it.

    I'm guessing you're on AQA cause I remember helping someone in my school do this question just a couple of days ago

    Yep, AQA

    I've got another one for you.

    SR( 1+sin2A / 1-sin2A) = tan (A + pi/4) p262 ex6b q8

    i've got it to cosA + sinA / cosA - sinA
    but i'm lost from there on :confused:
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    (Original post by uxa595)
    Yep, AQA

    I've got another one for you.

    SR( 1+sin2A / 1-sin2A) = tan (A + pi/4) p262 ex6b q8

    i've got it to cosA + sinA / cosA - sinA
    but i'm lost from there on :confused:
    Yeah I had to do this one for my teacher!

    You used your identities wrong, but anyway don't do that.

    Start with the left hand side. Multiply both the top ans bottom by 1+ sin2a but don't multiply out the bracket I'm the numerator.

    Then figure out how you can get rid of the square root.
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    (Original post by hassi94)
    Yeah I had to do this one for my teacher!

    You used your identities wrong, but anyway don't do that.

    Start with the left hand side. Multiply both the top ans bottom by 1+ sin2a but don't multiply out the bracket I'm the numerator.

    Then figure out how you can get rid of the square root.
    Ahem he's done it right so far. If he multiplies top and bottom by cosA+sinA and simplifies it reduces to a well known identity.

    Spoiler:
    Show



    tan(A/2+ pi/4)= tan(A)+sec(A)

    Replace A with 2A and you get?

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    (Original post by f1mad)
    Ahem he's done it right so far. If he multiplies top and bottom by cosA+sinA and simplifies it reduces to a well known identity.

    Spoiler:
    Show



    tan(A/2+ pi/4)= tan(A)+sec(A)

    Replace A with 2A and you get?

    Oh sorry I didn't see it, I'll have a look later I did that while getting changed this morning. But I don't think you should use any identity except the main sin cos tang a + b and cos squared and sin squared. Otherwise its cheating a little
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    (Original post by hassi94)
    Oh sorry I didn't see it, I'll have a look later I did that while getting changed this morning. But I don't think you should use any identity except the main sin cos tang a + b and cos squared and sin squared. Otherwise its cheating a little
    Hmm maybe, I don't know.

    It's a well known result though.

    We're using identities (they hold true for any value of A), so I think it would be allowed .
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    I know what you mean but my solution assumes a lot less which I think is better.
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    (Original post by hassi94)
    I know what you mean but my solution assumes a lot less which I think is better.
    In which way are you doing it, i did it in a similar way to you by multiply the top and bottom in the LHS by 1+sin2a, getting rid of square root, but then also it wasn't simple to solve it.
    So i get LHS as:


     LHS = \dfrac{1+sin2a}{cos2a} = \dfrac{1+2sinacosa}{cos2a}

    Then i tried to show the expression i got for LHS by using RHS, and succeeded in it. So now by seeing the steps in reverse i was able to know how to do the question by just using the LHS and not making any changes to RHS. But how to do it directly, i don't have any clue of how could i have solved it if i wouldn't have tried using both sides.
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    (Original post by raheem94)
    In which way are you doing it, i did it in a similar way to you by multiply the top and bottom in the LHS by 1+sin2a, getting rid of square root, but then also it wasn't simple to solve it.
    So i get LHS as:


     LHS = \dfrac{1+sin2a}{cos2a} = \dfrac{1+2sinacosa}{cos2a}

    Then i tried to show the expression i got for LHS by using RHS, and succeeded in it. So now by seeing the steps in reverse i was able to know how to do the question by just using the LHS and not making any changes to RHS. But how to do it directly, i don't have any clue of how could i have solved it if i wouldn't have tried using both sides.
    After where you are, expand cos2a then divide by cos to make everything tan or sec. Then make equations in tan and take common factor (1 + tan a I think) then its clear. And I only manipulated the LHS to work it out
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    (Original post by hassi94)
    I know what you mean but my solution assumes a lot less which I think is better.
    But we're not making assumptions. Identities serve useful purposes, for example in this case .

    Anyway fair enough.
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    (Original post by f1mad)
    But we're not making assumptions. Identities serve useful purposes, for example in this case .

    Anyway fair enough.
    It is an assumption if you don't prove it. In a level maths you can only assume what's in the formula booklet and slight variations (like the ones derived from cos squared + sin squared. It's not right to show an identity is true by using some other identity that isn't given. You may or may not get the marks but it's definitely not proper - I think it's worth the extra effort to do it my way

    Sorry for poor formatting throughout. I don't like typing on phones.: p
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    (Original post by raheem94)
    In which way are you doing it, i did it in a similar way to you by multiply the top and bottom in the LHS by 1+sin2a, getting rid of square root, but then also it wasn't simple to solve it.
    So i get LHS as:


     LHS = \dfrac{1+sin2a}{cos2a} = \dfrac{1+2sinacosa}{cos2a}

    Then i tried to show the expression i got for LHS by using RHS, and succeeded in it. So now by seeing the steps in reverse i was able to know how to do the question by just using the LHS and not making any changes to RHS. But how to do it directly, i don't have any clue of how could i have solved it if i wouldn't have tried using both sides.
    Now I'm on my PC I can type it out fully.
    Spoiler:
    Show

    \dfrac{1+sin2a}{cos2a} = \dfrac{1+2sinacosa}{cos^2a - sin^2a}

    Diving all terms by cos^2a to get \dfrac{sec^2a + 2tana}{1-tan^2a} = \dfrac{tan^2a + 2tana + 1}{(1+tana)(1-tana)} = \dfrac{1+tana}{1-tana}

    Then since tan45 = 1, \dfrac{1+tana}{1-tana} = \dfrac{tana + tan45}{1-(tana)(tan45)} = tan(a+\frac{\pi}{4})

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    (Original post by hassi94)
    It is an assumption if you don't prove it. In a level maths you can only assume what's in the formula booklet and slight variations (like the ones derived from cos squared + sin squared. It's not right to show an identity is true by using some other identity that isn't given. You may or may not get the marks but it's definitely not proper - I think it's worth the extra effort to do it my way

    Sorry for poor formatting throughout. I don't like typing on phones.: p
    Again that's like saying you're assuming sin(A+B)= sinAcosB+cosAsinB (simply because you haven't proved it).

    Also it is an identity; so you wouldn't be penalised at all. My way is just a shortcut.
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    http://en.wikipedia.org/wiki/List_of...ric_identities

    Scroll down you'll see where I got it from. It's not an assumption at all.
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    I got it finally :P
    All i did was divide everything by cos and from there on, it was easy.
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    (Original post by f1mad)
    http://en.wikipedia.org/wiki/List_of...ric_identities

    Scroll down you'll see where I got it from. It's not an assumption at all.
    No no that's not what I'm saying. It is an assumption, anything is an assumption if you don't prove it. What makes the difference is what you're ALLOWED to assume, and you're allowed to quote the formula booklet without proof.
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    One thing i found confusing though is, how just before you get to this step:
    http://www.wolframalpha.com/input/?i...28pi%2F4%29%29

    You get this:
    http://www.wolframalpha.com/input/?i...in%5E2+x%29%29

    and the denominator can be factorised to either:
    (sinx - cosx)^2
    or
    (cosx - sinx)^2

    but only the second one in right. How would you know which one works in the exam without having to do them both fully?
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    (Original post by uxa595)
    I got it finally :P
    All i did was divide everything by cos and from there on, it was easy.
    Good That's a good direction to take from there too well done (sorry I told you that you were wrong at the start, I didn't look properly - was in the middle of getting changed for 6th form! )

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Updated: March 16, 2012
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