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Math Help - Trig identities

  1. #1
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    Trig identities

    Hey guys, i am doing a problem and one of the terms is "sec(2pi + x)", so I made that "1/cos(2pi + x)", but i'm stuck. I don't know what cos(2pi + x) is equal to.
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  2. #2
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    sec(2{\pi}+x)=sec(x)
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  3. #3
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    thanks. so does that also mean:
    sin(2pi + x) = sin(x)
    cos(2pi + x) = cos(x)
    tan(2pi + x) = tan(x)
    csc(2pi + x) = csc(x)
    cot(2pi + x) = cot(x)
    ?

    we weren't taught the (2x + x) identities, only the (2x - 1)
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  4. #4
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    thanks. so does that also mean:
    sin(2pi + x) = sin(x)
    cos(2pi + x) = cos(x)
    tan(2pi + x) = tan(x)
    csc(2pi + x) = csc(x)
    cot(2pi + x) = cot(x)
    Yep.
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  5. #5
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    ok another one.

    i've gotten the left side down to

    cos(x/2) - sin(x/2)
    _________________

    cos(x/2) + sin(x/2)

    and the final right side equals:

    1 - tan(x/2)
    __________

    1 + tan(x/2)

    please help...


    ps. i'm sorry for making it look so noobish. I am not sure how you guys do that fancy way of writing the equation down. The way Galactus over there answered me.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CenturionMonkey View Post
    ok another one.

    i've gotten the left side down to

    cos(x/2) - sin(x/2)
    _________________

    cos(x/2) + sin(x/2)

    and the final right side equals:

    1 - tan(x/2)
    __________

    1 + tan(x/2)

    please help...


    ps. i'm sorry for making it look so noobish. I am not sure how you guys do that fancy way of writing the equation down. The way Galactus over there answered me.
    start with the right side:

    \frac {1 - \tan (x/2)}{1 + \tan (x/2)} = \frac {1 - \frac {\sin (x/2)}{\cos (x/2)}}{1 + \frac {\sin (x/2)}{\cos (x/2)}}

    now combine the fractions in the numerator and denominator and simplify
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  7. #7
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    yeah i would, but the left side i gave there, is like 8 steps down from the original left side. and it's too difficult to to go from right side, all the way to the original left side. :/

    original LS: sec(x) - tan(x)

    orignial RS: tan((pi/4) - (x/2))
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CenturionMonkey View Post
    yeah i would, but the left side i gave there, is like 8 steps down from the original left side. and it's too difficult to to go from right side, all the way to the original left side. :/

    original LS: sec(x) - tan(x)

    orignial RS: tan((pi/4) - (x/2))
    there is no need to go to the original left side. you will prove they are equivalent if you can bring them both to the same thing. you take advantage of the law of transitivity here.
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  9. #9
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    oh i see what you're saying. alright, thanks a lot.
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