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Math Help - Verifying Identities

  1. #1
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    Verifying Identities

    So I've been looking at this homework that's due tommorrow for a grade for the past hour and I can't seem to figure these out.

    1. sin^3x - cos^3x = (1 + sinx cosx)(sinx - cosx)

    2. (1/(siny - 1)) - (1/(siny + 1)) = -2sec^2y

    3. 1 - 2sin^2r + sin^4r = cos^4r

    4. tanu + (cosu/(1 + sinu)) = secu

    5. (tanx + secx)/(secx - cosx + tanx) = cscx

    If I could get a hint or something on how to do these I would appreciate it, I don't want to ask anyone to do my homework for me (although I won't complain if you show me the whole problem) but I don't even know where to start on these problems. Thanks!
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  2. #2
    Senior Member I-Think's Avatar
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    For number one, use the identity: a^3-b^3=(a-b)(a^2+ab+b^2)

    Number 2
    Place the L.H.S. under a common denominator and solve from there

    Number 3
    Use (1-sin^2r)=cos^2r on the R.H.S.

    Number 4
    Do the same thing as number 2

    Number 5
    A bit more tricky

    \frac{tanx+secx}{secx-cosx+tanx}
    becomes
    (\frac{sinx+1}{cosx})/(\frac{1-cos^2x+sinx}{cosx})
    becomes
    \frac{sinx+1}{sin^2x+sinx}

    Solve from there
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  3. #3
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    Quote Originally Posted by Skizye View Post
    So I've been looking at this homework that's due tommorrow for a grade for the past hour and I can't seem to figure these out.

    1. sin^3x - cos^3x = (1 + sinx cosx)(sinx - cosx)

    2. (1/(siny - 1)) - (1/(siny + 1)) = -2sec^2y

    3. 1 - 2sin^2r + sin^4r = cos^4r

    4. tanu + (cosu/(1 + sinu)) = secu

    5. (tanx + secx)/(secx - cosx + tanx) = cscx

    If I could get a hint or something on how to do these I would appreciate it, I don't want to ask anyone to do my homework for me (although I won't complain if you show me the whole problem) but I don't even know where to start on these problems. Thanks!
    Hi Skizye,

    Here's a start for #2

    \frac{1}{\sin y-1}-\frac{1}{\sin y+1}=-2\sec^2 y

    Work with the left side.

    The common denominator will be \sin^2-1

    \frac{\sin y+1-\sin y+1}{\sin^2 y-1}

    Simplify the numerator.

    \frac{2}{\sin^2 y-1}

    \frac{2}{-(1-\sin^2 y)}

    Can you take it from here?
    Last edited by masters; November 19th 2009 at 11:16 AM.
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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by Skizye View Post
    So I've been looking at this homework that's due tommorrow for a grade for the past hour and I can't seem to figure these out.

    1. sin^3x - cos^3x = (1 + sinx cosx)(sinx - cosx)

    2. (1/(siny - 1)) - (1/(siny + 1)) = -2sec^2y

    3. 1 - 2sin^2r + sin^4r = cos^4r

    4. tanu + (cosu/(1 + sinu)) = secu

    5. (tanx + secx)/(secx - cosx + tanx) = cscx

    If I could get a hint or something on how to do these I would appreciate it, I don't want to ask anyone to do my homework for me (although I won't complain if you show me the whole problem) but I don't even know where to start on these problems. Thanks!
    Hello again,

    And for #5, work with the left side.

    \dfrac{\tan x+ \sec x}{\sec x-\cos x+\tan x}=\csc x

    \dfrac{\dfrac{\sin x}{\cos x}+\dfrac{1}{\cos x}}{\dfrac{1}{\cos x}-\cos x+ \dfrac{\sin x}{\cos x}}=

    Do you think you can handle it from here?
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  5. #5
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    Thanks so much for the help!

    On number 2 I can't figure out I would get the secant to become a sine or the other way around so they will be equal, how can I do this (or should I even be trying to do this)? Thanks!
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  6. #6
    A riddle wrapped in an enigma
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    Quote Originally Posted by masters View Post
    Hi Skizye,

    Here's a start for #2

    \frac{1}{\sin y-1}-\frac{1}{\sin y+1}=-2\sec^2 y

    Work with the left side.

    The common denominator will be \sin^2-1

    \frac{\sin y+1-\sin y+1}{\sin^2 y-1}=

    Simplify the numerator.

    \frac{2}{\sin^2 y-1}=

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

    Can you take it from here?
    Quote Originally Posted by Skizye View Post
    Thanks so much for the help!

    On number 2 I can't figure out I would get the secant to become a sine or the other way around so they will be equal, how can I do this (or should I even be trying to do this)? Thanks!
    I had a typo in my first post. I've fixed it now.

    Notice the next two steps...

    \dfrac{2}{-\cos^2 y}=

    -2 \sec^2 y
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  7. #7
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    Alright. I finally got all of them except for 4! I found a common denominator but I'm stuck at this step

    \frac{(1 + sinu)(tanu) + cosu}{1 + sinu} = secu

    Have I done it right so far? What can I do from here?

    Sorry for all the questions and thanks again for the help!
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  8. #8
    A riddle wrapped in an enigma
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    Quote Originally Posted by Skizye View Post
    Alright. I finally got all of them except for 4! I found a common denominator but I'm stuck at this step

    \frac{(1 + sinu)(tanu) + cosu}{1 + sinu} = secu

    Have I done it right so far? What can I do from here?

    Sorry for all the questions and thanks again for the help!
    This is how I proceeded.

    \tan u+\dfrac{\cos u}{1+ \sin u}=\sec u

    \frac{\sin u}{\cos u}+\frac{\cos u}{1+\sin u}=

    \frac{\sin u(1+\sin u)+\cos^2 u}{\cos u(1+\sin u)}=

    \frac{\sin u+\sin^2u+\cos^2 u}{\cos u(1+\sin u)}=

    \frac{\sin u+1}{\cos u(1+\sin u)}=

    \frac{1}{\cos u}=

    \sec u
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