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Math Help - Need help on using trig functions to verify identities

  1. #1
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    Need help on using trig functions to verify identities

    I'm not sure if i'm doing this problem correctly. :

    cot z sin z + tan z cos z

    Heres one way i did it:

    cotz = (cos / sin) sin z

    = cos z

    = cot z + (sin/cos) cos z

    This is what i'm left with:

    = cot z + cos z
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  2. #2
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    Quote Originally Posted by MajorJohnson View Post
    I'm not sure if i'm doing this problem correctly. :

    cot z sin z + tan z cos z

    Heres one way i did it:

    cotz = (cos / sin) sin z

    = cos z

    = cot z + (sin/cos) cos z

    This is what i'm left with:

    = cot z + cos z
    Hi MajorJohnson,

    I don't see an identity here, but if you just want to simplify....

    \cot z \sin z + \tan z \cos z = \dfrac{\cos z}{\sin z}\sin z+\dfrac{\sin z}{\cos z}\cos z= \boxed{\cos z + sin z}

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    Quote Originally Posted by masters View Post
    Originally Posted by MajorJohnson
    I'm not sure if i'm doing this problem correctly. :

    cot z sin z + tan z cos z

    Heres one way i did it:

    cotz = (cos / sin) sin z

    = cos z

    = cot z + (sin/cos) cos z

    This is what i'm left with:

    = cot z + cos z


    Hi MajorJohnson,

    I don't see an identity here, but if you just want to simplify....

    \cot z \sin z + \tan z \cos z = \dfrac{\cos z}{\sin z}\sin z+\dfrac{\sin z}{\cos z}\cos z= \boxed{\cos z + sin z}
    Sorry that cot z should be a cos z


    And the method you did, was the other way i had tried to solve it and got the same answer. I just wasn't sure if it was right or not.
    Thanks.

    Code:
    Here's another one i'm stuck on:
    
    1 - sin^2 / csc^2 - 1 =
    
    Heres what i've gottten so far:
    
    
    1/csc ^ 2 - sin ^ 2 / 1 =
    I've i try converting them using fundamental I.D. they won't cancel out, it'll just be subtracted to equal 0. Would that be the answer then?

    Here's another one
    sec x * sin x/tan x =

    1 / cos x * sin x / sin x /cos x

    (sin x / cos x ) (sin x / cos x) =

    sin x /cos x ^ 2 * sin x =

    answer:
    sinx ^ 2 / cos x ^ 2
    Last edited by MajorJohnson; March 22nd 2011 at 08:02 AM.
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  4. #4
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    \cos z \sin z + \dfrac{\sin z}{\cos z} \cos z = \cos z \sin z + \sin z = \sin z(\cos z + 1)
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  5. #5
    A riddle wrapped in an enigma
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    Quote Originally Posted by MajorJohnson View Post
    Code:
    Here's another one i'm stuck on:
    
    1 - sin^2 / csc^2 - 1 =
    
    Heres what i've gottten so far:
    
    
    1/csc ^ 2 - sin ^ 2 / 1 =
    Hello again MajorJohnson,

    If I'm reading your expressions correctly, you have

    \dfrac{1-\sin^2 \theta}{\csc^2 \theta -1}

    Using your Pythagorean Identities, you can simplify this way:

    \dfrac{1-\sin^2 \theta}{\csc^2 \theta -1}=\dfrac{\cos^2 \theta}{\cot^2\theta}=\dfrac{\cos^2\theta}{\frac{\  cos^2\theta}{\sin^2\theta}}=\dfrac{\cos^2\theta}{1  } \times \dfrac{\sin^2\theta}{\cos^2\theta}=\sin^2\theta



    Quote Originally Posted by MajorJohnson View Post
    Here's another one
    sec x * sin x/tan x =

    1 / cos x * sin x / sin x /cos x

    (sin x / cos x ) (sin x / cos x) =

    sin x /cos x ^ 2 * sin x =

    answer:
    sinx ^ 2 / cos x ^ 2
    For this one...

    \dfrac{\sec x \sin x}{\tan x}=\dfrac{\frac{1}{\cos x} \sin x}{\frac{\sin x}{\cos x}}=\dfrac{\frac{\sin x}{\cos x}}{\frac{\sin x}{\cos x}}=1

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  6. #6
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    okay, thank you. One more question, how do you get your equations to look like that in your post?
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  7. #7
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    Quote Originally Posted by MajorJohnson View Post
    okay, thank you. One more question, how do you get your equations to look like that in your post?
    Latex Help Forum
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  8. #8
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    Thanks,

    I might just keep using this thread to post more problems, up just to make sure i understand this. After working through these for some time, i feel like i'm almost comfortable with proving trig identities.


    1) sin  x * tan x + cos  x

    sin  x  *  sin x  / cos x + cos x

    sin x ^ 2 / cos x + cos x =  Answer: sin x ^ 2


    2)       1\div 1 + cos x  + 1 \ div 1 - cos x =

       1 - cos x \div  1 + cos x + 1 + cos x \div   1 - cos x = Answer: 2

    For this problem, does it matter if the right side's answer is the same value on the left side but flipped?

    3)  sin t * tan t = 1 - cos ^ 2 t / cos t
    Doing the right side:

        sin ^ 2  t/ cos t


       sin t / cos t (sin t)

       tan t sec t
    Last edited by MajorJohnson; March 24th 2011 at 01:54 PM.
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  9. #9
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    Quote Originally Posted by MajorJohnson View Post
    Thanks,

    I might just keep using this thread to post more problems, up just to make sure i understand this. After working through these for some time, i feel like i'm almost comfortable with proving trig identities.


    1) sin  x * tan x + cos  x

    sin  x  *  sin x  / cos x + cos x

    sin x ^ 2 / cos x + cos x =  Answer: sin x ^ 2 no ... should be sec(x)

    2)       1\div 1 + cos x  + 1 \ div 1 - cos x =

       1 - cos x \div  1 + cos x + 1 + cos x \div   1 - cos x = Answer: 2    no ... 2csc^2(x)

    I'm confused with this one:

    3)  sin t * tan t = 1 - cos ^ 2 t / cos t
    Doing the right side:

    From this point i'm not sure about what to do next. Would i factor sin ^ 2 / cos t or convert it into tan ^ 2 t?
        sin ^ 2  t/ cos t   \sin^2(t)/cos(t) = sin(t) * sin(t)/cos(t) = sin(t) * tan(t)
    1. start a new thread ... this one is getting worn.

    2. use parentheses or learn how to form a latex fraction ... e.g. \dfrac{\sin{x}}{\cos{x}} inside of the math tags will yield \dfrac{\sin{x}}{\cos{x}}
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