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Math Help - Analytic Trig- Sin, Cos, Tan

  1. #1
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    Analytic Trig- Sin, Cos, Tan

    Hi, I was wondering how to solve this problem....

    "Write each of the other trigonometric functins of x in tems of sin x."

    I know that sin x= (1/csc x), but that's about it.

    Thanks!
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  2. #2
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    [IMG]file:///C:/Users/ADVENT%7E1/AppData/Local/Temp/moz-screenshot.jpg[/IMG]http://en.wikipedia.org/wiki/Trigonometric_identies
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  3. #3
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    Quote Originally Posted by checkthiskid View Post
    [IMG]file:///C:/Users/ADVENT%7E1/AppData/Local/Temp/moz-screenshot.jpg[/IMG]

    List of trigonometric identities - Wikipedia, the free encyclopedia
    I appreciate your help.... I think it's supposed to be a picture? I'm sorry for some reason it's not working!
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  4. #4
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    Quote Originally Posted by falloutatthedisco View Post
    I appreciate your help.... I think it's supposed to be a picture? I'm sorry for some reason it's not working!
    yeah im rubbish with computers

    List of trigonometric identities - Wikipedia, the free encyclopedia

    it was supposed to be a link but i still had the image copied and pasted
    scroll down and there will be a table of all the identities
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  5. #5
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    For example,

    \tan x=\frac{\sin x}{\cos x}=\frac{\sin x}{\sqrt{1-{{\sin }^{2}}x}}\implies {{\tan }^{2}}x=\frac{{{\sin }^{2}}x}{1-{{\sin }^{2}}x}, so \sin x=\frac{\tan x}{\sqrt{1+{{\tan }^{2}}x}}.

    Obviously, this requires restrictions for x. For example, \cos x=\sqrt{1-\sin^2x} works when 0\le x\le\frac\pi2 or \frac{3\pi}2\le x\le2\pi.
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  6. #6
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    Checkthiskid-

    it's fine.... I'm rubbish at pre-calc!

    thanks so much for the help!


    Krisalid-

    thanks so much!
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  7. #7
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    Quote Originally Posted by Krizalid View Post
    For example,

    \tan x=\frac{\sin x}{\cos x}=\frac{\sin x}{\sqrt{1-{{\sin }^{2}}x}}\implies {{\tan }^{2}}x=\frac{{{\sin }^{2}}x}{1-{{\sin }^{2}}x}, so \sin x=\frac{\tan x}{\sqrt{1+{{\tan }^{2}}x}}.

    Obviously, this requires restrictions for x. For example, \cos x=\sqrt{1-\sin^2x} works when 0\le x\le\frac\pi2 or \frac{3\pi}2\le x\le2\pi.
    you may need to show your working out so i would suggest doing it step by step using what was shown in krizalid's post and using cosx=\sqrt{1-sin^{2}x}
    derived from cos^{2}x +sin^{2}x=1
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