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Math Help - trig identities

  1. #1
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    trig identities

    (sinx)/(1+cosx)=(1-cosx)/(sinx)
    simplify so RS=LS
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  2. #2
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    Re: trig identities

    Quote Originally Posted by ahmedb View Post
    (sinx)/(1+cosx)=(1-cosx)/(sinx)
    simplify so RS=LS
    What does \frac{\sin(x)}{1+\cos(x)}\frac{1-\cos(x)}{1-\cos(x}}=~?
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  3. #3
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    Re: trig identities

    thats not what i am talking about, i am talking about proofs.
    like proof that RS=LS by simplifying it
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  4. #4
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    Re: trig identities

    hi ahmedb

    Spoiler:

    <br />
\frac{\sin(x)}{1+\cos(x)}=\frac{1-\cos(x)}{\sin(x)} \\<br />
\\<br />
\frac{\sin(x)\cdot(1-\cos(x))}{(1+\cos(x))(1-\cos(x))}=\frac{1-\cos(x)}{\sin(x)} \\<br />
\\<br />
\frac{\sin(x)\cdot(1-\cos(x))}{1-\cos^2(x)}=\frac{1-\cos(x)}{\sin(x)} \\<br />
\\<br />
\frac{\sin(x)\cdot(1-\cos(x))}{\sin^2(x)}=\frac{1-\cos(x)}{\sin(x)} \\<br />
\\<br />
\frac{1-\cos(x)}{\sin(x)}=\frac{1-\cos(x)}{\sin(x)}<br />
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  5. #5
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    Re: trig identities

    what did you multiply by 1-cosx?
    y can't u use 1-sinx or anything else???
    thats what confused me :S
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  6. #6
    Member anonimnystefy's Avatar
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    Re: trig identities

    when you multiply by 1-sin(x) you don't get anything pretty.by multiplying by 1-cos(x) you get 1-cos^2(x) which you should now is sin^2(x).another way to do this is to multiply the fraction on the RHS by 1+cos(x).
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  7. #7
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    Re: trig identities

    for the RHS why is it 1+cosx??
    is there a method like a rule in which you have to follow?
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  8. #8
    MHF Contributor Siron's Avatar
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    Re: trig identities

    You can also use \frac{a}{b}=\frac{c}{d} \Leftrightarrow a\cdot d=c\cdot b
    That means for this exercice:
    \frac{\sin(x)}{1+\cos(x)}=\frac{1-\cos(x)}{\sin(x)} \Leftrightarrow \sin^2(x)=[1-\cos(x)]\cdot[1+\cos(x)] \Leftrightarrow \sin^2(x)=1-\cos^2(x) \Leftrightarrow \sin^2(x)=\sin^2(x)
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