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Math Help - More Trig/Pre-Cal Identity Proofs

  1. #1
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    More Trig/Pre-Cal Identity Proofs

    Sorry to ask again, but I need help with these problems:

    (secx+1)/tanx = sinx/(1-cosx)
    I got (1+cosx)/sinx which is the opposite of what I need.

    tan^4x + tan^2x = sec^4 - sec^2x

    tan^2x - sin^2x = tan^2x * sin^2x

    sinx/(1+cosx) + (1+cosx)/sinx = 2cscx

    I could only figure out about half of my homework and I would ask the teacher, but she isn't at school in the morning for tutoring.
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  2. #2
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    Quote Originally Posted by Purpledog100 View Post
    Sorry to ask again, but I need help with these problems:

    (secx+1)/tanx = sinx/(1-cosx)
    I got (1+cosx)/sinx which is the opposite of what I need. (This is good)

    tan^4x + tan^2x = sec^4 - sec^2x

    tan^2x - sin^2x = tan^2x * sin^2x

    sinx/(1+cosx) + (1+cosx)/sinx = 2cscx

    I could only figure out about half of my homework and I would ask the teacher, but she isn't at school in the morning for tutoring.
    Now multiply top and bottom by  1- \cos x

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

    expand the numerator, use some identites and simplify.
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  3. #3
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    (secx+1)/tanx = sinx/(1-cosx)
    I got (1+cosx)/sinx which is the opposite of what I need.
    go from there ...

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

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

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

    \frac{\sin{x}}{1 - \cos{x}}
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  4. #4
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    For the third one:

     \tan^2 x -\sin^2 x

     = \frac{\sin^2 x}{\cos^2 x} - \tan^2 x \cos^2 x
     = \frac{\sin^2 x}{\cos^2 x} - \frac{\sin^2 x}{\cos^2 x} \cos^2x
     = \frac{\sin^2 x - \sin^2 x \cos^2 x} {\cos^2 x}
     = \frac {\sin^2 x (1- \cos^2 x)}{\cos^2 x}
     = \frac {\sin^2 x}{\cos^2 x} (1- \cos^2 x)
     = \tan^2 x \sin^2 x
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  5. #5
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    Quote Originally Posted by danny arrigo View Post
    Now multiply top and bottom by  1- \cos x

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

    expand the numerator, use some identites and simplify.
    For the second, factor a \tan^2 x from the LHS, i.e.

    \tan^2x \left( 1 + \tan^2x \right)

    then use the identity

    1 + \tan^2x = \sec^2x to eliminate the tan

    For the forth one, multiply the first term by

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

    then follow the idea what skeeter did above.
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  6. #6
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    Quote Originally Posted by Amanda H View Post
    For the third one:

     \tan^2 x -\sin^2 x

     = \frac{\sin^2 x}{\cos^2 x} - \tan^2 x \cos^2 x
     = \frac{\sin^2 x}{\cos^2 x} - \frac{\sin^2 x}{\cos^2 x} \cos^2x
     = \frac{\sin^2 x - \sin^2 x \cos^2 x} {\cos^2 x}
     = \frac {\sin^2 x (1- \cos^2 x)}{\cos^2 x}
     = \frac {\sin^2 x}{\cos^2 x} (1- \cos^2 x)
     = \tan^2 x \sin^2 x
    Why did you change -sin^2x to -tan^2x*cos^2x

    For the forth one, multiply the first term by



    then follow the idea what skeeter did above.
    I'm sorry, but I still don't understand how to do it. I have:
    {sinx(1-cosx)}/sin^2x + (1+cosx)/sinx

    Do I multiply the second fraction by sinx to have equal denominators?
    Last edited by Purpledog100; February 3rd 2009 at 03:59 PM.
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  7. #7
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    It comes from
     \frac {\sin x}{\cos x} = \tan x
    rearranged it becomes
     \sin x = \tan x \cos x
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  8. #8
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    Quote Originally Posted by Amanda H View Post
    It comes from
     \frac {\sin x}{\cos x} = \tan x
    rearranged it becomes
     \sin x = \tan x \cos x
    this works, but it's overkill. you could go directly from \frac {\sin^2 x}{\cos^2 x} - \sin^2 x \to \frac {\sin^2 x - \sin^2 x \cos^2 x}{\cos^2 x}. just combining fractions, as we know \frac ab \pm \frac cd = \frac {ad \pm bc}{bd}. from there, it's nice

    we could also do

    \frac {\sin^2 x}{\cos^2 x} - \sin^2 x

    = \sin^2 x \left( \frac 1{\cos^2 x} - 1 \right)

    = \sin^2 x \cdot \frac {1 - \cos^2 x}{\cos^2 x}

    = \sin^2 x \tan^2 x

    @ OP: the nice thing about trig identities is that there are usually several ways to solve them. it is good practice to use your imagination to come up with alternative methods of proving the identities than the ones posted here. just take note of the techniques used, like changing everything to sines and cosines. usually a good way to go.
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  9. #9
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    Quote Originally Posted by Amanda H View Post
    It comes from
     \frac {\sin x}{\cos x} = \tan x
    rearranged it becomes
     \sin x = \tan x \cos x
    Oh, right. Duh. I had that in my notes and still couldn't figure it out.

    @ OP: the nice thing about trig identities is that there are usually several ways to solve them. it is good practice to use your imagination to come up with alternative methods of proving the identities than the ones posted here. just take note of the techniques used, like changing everything to sines and cosines. usually a good way to go.
    Yeah, but I usually pick the wrong ways to solve them.

    Thanks to everyone for all of their help!!
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Purpledog100 View Post
    Yeah, but I usually pick the wrong ways to solve them.
    that's the point. as long as your moves are mathematically sound, and you can get from one side to the other, there is no wrong way. a problem here can be done several valid ways. the difference is, one way might be easier or quicker than another. if you do enough practice, this won't matter though, as you will either be able to move so quickly that the extra steps hardly slow you down, or you will naturally be able to spot the easier ways first
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