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Math Help - Trig proofs need help ASAP!

  1. #1
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    Trig proofs need help ASAP!

    Need to prove these 2
    1.(sin(x) - cos(x) -1)/(sin(x)+cos(x)-1) = - ((cos(x)+1)/sin(x))

    2. 1-cos(2x) + cos(4x) - cos(6x) = 4sin(x)cos(2x)sin(3x)


    Thanks!
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  2. #2
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    I just want you to know that I have the solutions and explanations for these two trig proofs. However, as a result of recent elbow and nerve surgery, it takes me seemingly forever to type up something (esp math) accurately. So, if you have other problems and can afford waiting another few minutes, I can certainly help. Just letting you know this.
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  3. #3
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    Maybe this will help. I can't get it to verify...Maybe someone can fix my mistake. I can't figure out what I did wrong, but maybe you can see it. Or someone can just fix my mistake so they don't have to rewrite.

    \frac{\sin{x}-(\cos{x}-1)}{\sin{x}+(\cos{x}-1)}=\frac{-\cos{x}+1}{\sin{x}}

    I work the left side
    \frac{\sin{x}-(\cos{x}-1)}{\sin{x}+(\cos{x}-1)}

    Multiply by \frac{\cos{x}+1}{\cos{x}+1}

    \frac{\sin{x}-(\cos{x}-1)}{\sin{x}+(\cos{x}-1)} ({\frac{\cos{x}+1}{\cos{x}+1}})

    This leaves you with:
    \frac{-\cos^2{x}+\cos{x}-\cos{x}+1}{\cos^2{x}-\cos{x}+\cos{x}-1}

    Combine like terms
    \frac{-\cos^2{x}+1}{\cos^2{x}-1}

    Simplify
    \frac{-\cos^2{x}+1}{\sin^2{x}} or \frac{\sin^2{x}}{\sin^2{x}}=1


    Oh dear I see my mistake...I feel stupid...I will fix it as fast as possible. Nevermind...I didn't find it. I noticed I just randomly dropped off the sinx but even when I try to do it it cancels out. Sorry....I tried. *goes back to hole*
    Last edited by Ericonda; April 7th 2008 at 11:17 AM. Reason: Im dumb
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  4. #4
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    you changed the paranthesis around so they are not correct it is sin(x) -cos(x) -1 no parenthesis and the same for the bottom
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  5. #5
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    Hello,

    I'll try ^^

    We have :

    \frac{\sin{x}-\cos{x}-1}{\sin{x}+\cos{x}-1}

    Consider the denominator : (sin(x)+cos(x))-1.

    Use the formula (a-b)(a+b)=aē-bē
    So here, you're going to multiply up and down by (sin(x)+cos(x))+1

    This will make :

    \frac{(\sin{x}-\cos{x}-1)(\sin x+\cos x+1)}{\underbrace{(\sin{x}+\cos{x}-1)(\sin x+\cos x+1)}_{(\sin x+\cos x)^2-1^2}}

    The denominator :

    (\sin x+\cos x)^2-1^2=\underbrace{\sin^2 x+ \cos^2 x}_{1} +2 \cos x \sin x -1 =2 \cos x \sin x

    The numerator :

    (\sin{x}-\cos{x}-1)(\sin x+\cos x+1)=(\sin x - (\cos x +1))(\sin x + (\cos x +1))

    Which is (a-b)(a+b) form again
    Hence the numerator is equal to :

    \sin^2 x - (\cos x+1)^2=\underbrace{\sin^2x}_{1-\cos^2 x}-\cos^2x-2 \cos x -1

    =1-2\cos^2x-2\cos x-1=-2 \cos x (\cos x +1)


    And here is the grand final (music please xD)

    \frac{\sin{x}-\cos{x}-1}{\sin{x}+\cos{x}-1}=\frac{-2 \cos x (\cos x +1)}{2 \cos x \sin x}=- \frac{\cos x+1}{\sin x}

    Is it clear enough ?
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  6. #6
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    I see what I did. For some reason I grouped it wrong. Well I thought I would try, sorry.
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  7. #7
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    I apologize, my college's internet connection went down. Nonetheless, Mr/Mrs/Dr. Moo got you what ya needed: - ( cos(x) + 1 ) / sin(x)

    This can alternatively be written using our "co"-functions, namely a cotangent term and a cosecant term.

    - ( cos(x) + 1 ) / sin(x) = - ( (cos(x)/sin(x)) + (1/sin(x)) )
    = - ( cot(x) + csc(x) )
    = - cot(x) - csc(x)

    This is the same answer as - ( cos(x) + 1 ) / sin(x)
    Both work.

    For number 2,

    2. 1-cos(2x) + cos(4x) - cos(6x) = 4sin(x)cos(2x)sin(3x)

    1 - [(cos^2 x)-(sin^2 x)] + cos(4x) - cos(6x) = 4sin(x)cos(2x)sin(3x)

    Simplification:

    We know (cos^2 x) = 1-(sin^2 x) from the
    trig identity sin^2 x +cos^2 x=1.
    So, 1 - [(cos^2 x)-(sin^2 x)] = 1 - {[1-(sin^2 x)]-(sin^2 x)} = (line below)
    1 -{[1-(sin^2 x)]-(sin^2 x)} = 1 - [1-(2sin^2 x)] = 2sin^2 x .

    Or, you could have simply looked that one up.

    plugging and chugging, we get: 2(sin^2 x) + cos(4x) - cos(6x) = 4sin(x)cos(2x)sin(3x)

    We now deal with the cos(4x) term:

    cos4x = (cos^4 x) - [(6cos^2 x)(sin^2 x)] + (sin^4 x)

    Now, we have:
    (2sin^2 x) + {(cos^4 x) - [(6cos^2 x)(sin^2 x)] + (sin^4 x)} - cos(6x) = 4sin(x)cos(2x)sin(3x)

    Note that I am doing this without reference, therefore, I am positive that there is a much quicker way to do this. There are one or two theorems/identities/axioms that would make this go much faster. However, I usually just like to get rid of the double,triple,quadruple,..... angles before I move on. This is how I would go about this, unless I had a test, then I would memorize shortcuts. You kind of see "the big picture" this way.

    Note also that I used { and } inappropriately. I used them so you could see better. However, mathematically, { and } are usually used to denote collections (sets, sequences, streams, etc...)

    Sorry that I could not get you this earlier. I am glad Moo got too, though. Good luck!
    -Andy
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  8. #8
    Moo
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    I hope your elbow will soon recover, if it's for such a purpose... This is well explained !

    Ow, last thing, if you want to remember, i think that students are learnt the following formulaes :

    cos(2x)=1-2sinē(x)=2cosē(x)-1=cosē(x)-sinē(x)

    This reduces the demonstration of yours about 2-3 lines
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