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Math Help - trig help needed

  1. #1
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    trig help needed

    Hi
    I have a few questions in can't get the right answers:

    1)Simply the expression sin^4x-cos^4x

    2)Simply the expression tan^3x+tanx
    This is what i have done:
    = \frac{sinx^3}{cosx^3}+\frac{sinx}{cosx}

    = \frac{sinx^3}{cos^3}+\frac{sinxcosx^2}{cosx^3}

    what should i do next?
    3)if tanx=-4,x [ \frac{-\pi}{2},0] find: sec(x)

    P.S
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I have a few questions in can't get the right answers:

    1)Simply the expression sin^4x-cos^4x Note that this is the difference of squares.

    2)Simply the expression tan^3x+tanx Factor out a tanx, then note that (tan^2x+1)=sec^2x. What you have done is unneccesary.
    This is what i have done:
    = \frac{sinx^3}{cosx^3}+\frac{sinx}{cosx}

    = \frac{sinx^3}{cos^3}+\frac{sinxcosx^2}{cosx^3}

    what should i do next?
    3)if tanx=-4,x [ \frac{-\pi}{2},0] find: sec(x)

    P.S
    .
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  3. #3
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I have a few questions in can't get the right answers:

    1)Simply the expression sin^4x-cos^4x
    You can begin by saying \sin^4x-\cos^4x = \sin^2x\times \sin^2x - \cos^2x \times \cos^2x

    Now you can apply some identities for sin and cos all squared.
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  4. #4
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    Quote Originally Posted by VonNemo19 View Post
    1)Simply the expression Note that this is the difference of squares.
    How so?
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  5. #5
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I have a few questions in can't get the right answers:

    1)Simply the expression sin^4x-cos^4x

    2)Simply the expression tan^3x+tanx
    This is what i have done:
    = \frac{sinx^3}{cosx^3}+\frac{sinx}{cosx}

    = \frac{sinx^3}{cos^3}+\frac{sinxcosx^2}{cosx^3}

    what should i do next?
    3)if tanx=-4,x [ \frac{-\pi}{2},0] find: sec(x)

    P.S
    1) \sin^4{x} - \cos^4{x} = (\sin^2{x} + \cos^2{x})(\sin^2{x} - \cos^2{x})

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

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

     = \cos{2x}.


    2) \tan^3{x} + \tan{x} = \tan{x}(\tan^2{x} + 1)

     = \tan{x}\sec^2{x}


    3) \tan{x} = -4

    \tan^2{x} = 16

    \tan^2{x} + 1 = 17

    \sec^2{x} = 17

    \sec{x} = \pm \sqrt{17}.


    Since x is in the fourth quadrant, \cos{x} > 0 and so \sec{x} > 0.


    Therefore \sec{x} = \sqrt{17}.
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  6. #6
    Super Member bigwave's Avatar
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I have a few questions in can't get the right answers:

    1)Simply the expression sin^4x-cos^4x

    2)Simply the expression tan^3x+tanx
    This is what i have done:
    = \frac{sinx^3}{cosx^3}+\frac{sinx}{cosx}

    = \frac{sinx^3}{cos^3}+\frac{sinxcosx^2}{cosx^3}

    what should i do next?
    3)if tanx=-4,x [ \frac{-\pi}{2},0] find: sec(x)

    P.S
    1) sin^4x-cos^4x

    (sin^2{x} + cos^2{x})(sin^2{x}-cos^2{x})<br />
 \Rightarrow<br />
(1)(sin^2{x}-cos^2{x})<br />
 \Rightarrow<br />
-cos{2x}
    Last edited by bigwave; January 10th 2010 at 10:27 PM. Reason: wrong sign
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  7. #7
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by pickslides View Post
    How so?
    Let a=\sin^2{x} and b=\cos^2{x}

    We then have a^2-b^2. Your way of doing things was equally valid. It looks as though prove it has combined our efforts.
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  8. #8
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    Quote Originally Posted by Prove It View Post
    3) \tan{x} = -4

    \tan^2{x} = 16

    \tan^2{x} + 1 = 17

    \sec^2{x} = 17

    \sec{x} = \pm \sqrt{17}.


    Since x is in the fourth quadrant, \cos{x} > 0 and so \sec{x} > 0.


    Therefore \sec{x} = \sqrt{17}.
    just to clarify what your doing is working backwards to find out sec(x)?
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  9. #9
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    He is using the fact that

     <br />
\tan^2{x} + 1 = \sec^2{x}<br />
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  10. #10
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    can someone explained to me how do you get \frac{\sqrt17}{17} when you find cos(x) from tan(x)=-4.I understand that you get cos(x)=\frac{1}{sec(x)}then
    cos(x)=\frac{1}{\sqrt17} but where does the 17 come from?
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  11. #11
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    Quote Originally Posted by Paymemoney View Post
    can someone explained to me how do you get \frac{\sqrt17}{17} when you find cos(x) from tan(x)=-4.I understand that you get cos(x)=\frac{1}{sec(x)}then
    cos(x)=\frac{1}{\sqrt17} but where does the 17 come from?
    don't worry i undertsand how they got it, but why do you have to simply down to \frac{\sqrt17}{17}
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  12. #12
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    1)





    .
    Hey, shouldn't that be -\cos(2x)?
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  13. #13
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    Quote Originally Posted by bandedkrait View Post
    Hey, shouldn't that be -\cos(2x)?
    It seems so

    Table of Trigonometric Identities
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  14. #14
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    Quote Originally Posted by Paymemoney View Post
    don't worry i undertsand how they got it, but why do you have to simply down to \frac{\sqrt17}{17}
    \frac{1}{\sqrt{17}} = \frac{1}{\sqrt{17}}\times 1 = \frac{1}{\sqrt{17}}\times \frac{\sqrt{17}}{\sqrt{17}}= \frac{\sqrt17}{17}
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