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Math Help - Identities in Terms of Trigonometry

  1. #1
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    Identities in Terms of Trigonometry

    I have two questions below.

    Establish each identity.

    (1) cos(x)/[1 - tan(x)] + sin(x)/[1 - cot(x)] = sin(x) + cos(x)

    This is my set up for question 1:

    Let tan(x) = sin(x)/cos(x)

    Let cot(x) = cos(x)/sin(x)

    After plugging and simplifying the complex fractions, I get this on the left side:

    cos^2 (x)/[cos(x) - sin(x)] + sin^2 (x)/[sin(x) - cos(x)]

    Where do I go from there?

    Question 2:

    3 sin^2 (x) + 4 cos^2 (x) = 3 + cos^2 (x)

    I know that sin^2 (x) = 1 - cos^2 (x)

    This is my set up:

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

    I then use the distributive rule on the left side and get this:

    3 - 3cos^2 (x) + 4 cos^2 (x) = 3 + cos^2 (x)

    Where do I go from there?

    Thanks
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  2. #2
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    Quote Originally Posted by magentarita View Post
    I have two questions below.

    Establish each identity.

    (1) cos(x)/[1 - tan(x)] + sin(x)/[1 - cot(x)] = sin(x) + cos(x)

    This is my set up for question 1:

    Let tan(x) = sin(x)/cos(x)

    Let cot(x) = cos(x)/sin(x)

    After plugging and simplifying the complex fractions, I get this on the left side:

    cos^2 (x)/[cos(x) - sin(x)] + sin^2 (x)/[sin(x) - cos(x)]

    Where do I go from there?
    Get it all over the common denominator:

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

    Factorise the numerator (difference of two squares). Cancel the common factor. Viola. You're left with the RHS. Q.E.D.
    Last edited by mr fantastic; August 26th 2008 at 10:45 PM.
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  3. #3
    Super Member 11rdc11's Avatar
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    1st problem

    \frac{cosx}{1-tanx} +\frac{sinx}{1-cotx} = sinx + cosx

    \frac{cosx}{1-\frac{sinx}{cosx}} + \frac{sinx}{1-\frac{cosx}{sinx}}

    \frac {cos^2x}{cosx- sinx} + \frac{-sin^2x}{cosx - sinx}

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

    \frac {(cosx - sinx)(cosx + sinx)}{cosx - sinx}

    cosx + sinx

    oops I was to slow again
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  4. #4
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    Quote Originally Posted by magentarita View Post
    I have two questions below.

    Establish each identity.



    [snip]

    Question 2:

    3 sin^2 (x) + 4 cos^2 (x) = 3 + cos^2 (x)

    I know that sin^2 (x) = 1 - cos^2 (x)

    This is my set up:

    Mr F edits:


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


    I then use the distributive rule on the left side and get this:

    = 3 - 3cos^2 (x) + 4 cos^2 (x) = 3 + cos^2 (x)

    = RHS.


    Where do I go from there? Mr F says: You write Q.E.D. and move on to the next question.

    Thanks
    ..
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  5. #5
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    I thank you both

    Thanks a lot. I am learning so much math here.

    Math has now become my passion thanks to this site.
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