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Math Help - Trig Identity

  1. #1
    Senior Member sakonpure6's Avatar
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    Trig Identity

    (1+cosx)/sinx = cot(x/2)
    Very tough... I am thinking now of working with the right side and doing :

    cot(x/2) * cot(2x)/cot(2x)
    = cot^2x/cot2x // Is this true?

    because I have no idea what to do when I am given some thing like (x/2)

    Any hints are really appreciated
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  2. #2
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    Re: Trig Identity

    Hello, sakonpure6!

    You are expected to know these double-angle identities:

    . . \cos^2\theta \:=\:\frac{1+\cos2\theta}{2} \quad\Rightarrow\quad 1 + \cos x \:=\:2\cos^2\tfrac{x}{2}

    . . \sin2\theta \:=\:2\sin\theta\cos\theta \quad\Rightarrow\quad \sin x \:=\:\2\sin\tfrac{x}{2}\cos\tfrac{x}{2}



    \text{Prove: }\:\frac{1+\cos x}{\sin x} \:=\:\cot\tfrac{x}{2}

    We have: . \frac{1+\cos x}{\sin x} \;=\;\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2} \cos\frac{x}{2}} \;=\;\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}} \;=\;\cot\tfrac{x}{2}
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  3. #3
    MHF Contributor

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    Re: Trig Identity

    Do you know the "sum" formulas: sin(A+ B)= sin(A)cos(B)+ cos(A)sin(B) and cos(A+ B)= cos(A)cos(B)- sin(A)sin(B)?
    Setting A= B= X gives the "double angle" formulas: sin(2X)= 2sin(X)cos(X) and cos(2X)= cos^2(X)- sin^2(X)

    Letting Y= 2X in that last lets us derive the "half angle" formulas:
    cos(Y)= cos^2(Y/2)- sin^2(Y/2)= cos^2(Y/2)- (1- cos^2(Y/2))= 2cos^2(Y/2)- 1
    2cos^2(Y/2)= cos(Y)+ 1
    cos(Y/2)= \dfrac{\sqrt{(cos(Y)+ 1)}}{2}

    cos(Y)= cos^(Y/2)- sin^2(Y/2)= 1- sin^2(Y/2)- sin^2(Y/2)= 1- 2sin^2(Y/2)
    2sin^2(Y/2)= 1- cos(Y)
    sin(Y/2)= \sqrt{(1- cos(Y))/2}

    And, of course, tan(x/2)= \frac{sin(x/2)}{cos(x/2)} and cos(x/2)= \frac{cos(x/2)}{sin(x/2)}
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