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Math Help - help please cos x

  1. #1
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    help please cos x

    By writing cos x in terms of 1/2x show that

    1 - cosx / 1 + cosx = tan^2 1/2x

    thankyou :0
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  2. #2
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    Quote Originally Posted by gracey View Post
    By writing cos x in terms of 1/2x show that

    1 - cosx / 1 + cosx = tan^2 1/2x

    thankyou :0
    \frac{1 - \cos\left[ 2 \left (\frac{x}{2} \right] \right)}{1 + \cos \left (2 \left[ \frac{x}{2} \right] \right)} = \frac{1 - \cos^2 \left (\frac{x}{2}\right) + \sin^2 \left (\frac{x}{2}\right)}{1 + \cos^2 \left (\frac{x}{2}\right) - \sin^2 \left (\frac{x}{2}\right)} = \frac{2\sin^2 \left (\frac{x}{2}\right)}{2 \cos^2 \left (\frac{x}{2}\right)} = \tan^2 \left (\frac{x}{2}\right)

    where the Pythagorean Identity \sin^2 \left ( \frac{x}{2} \right) + \cos^2 \left ( \frac{x}{2} \right) = 1 has been used in the numerator and the denominator.
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  3. #3
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    Hello, gracey!

    A slightly different approach . . .

    We have these identities:

    . . \begin{array}{ccccccc}\sin^2\!\frac{x}{2} &=& \dfrac{1-\cos x}{2} & \Rightarrow & 1-\cos x &=&2\sin^2\!\frac{x}{2} \\ \\[-4mm]<br />
\cos^2\!\frac{x}{2} &=&\dfrac{1+\cos x}{2} & \Rightarrow & 1 + \cos x &=& 2\cos^2\!\frac{x}{2} \end{array}


    By writing \cos x in terms of \frac{x}{2} show that: . \frac{1 - \cos x}{1 + \cos x} \:= \:\tan^2\!\frac{x}{2}

    From the identities, we have: . \frac{1-\cos x}{1 + \cos x} \;=\;\frac{2\sin^2\!\frac{x}{2}}{2\cos^2\!\frac{x}  {2}} \;=\;\left(\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}  \right)^2 \;=\;\tan^2\!\frac{x}{2}

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  4. #4
    Senior Member nikhil's Avatar
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    Here is what you need!

    Hi!gracey,hope ur having a great time!
    cosx=(cos(x/2))^2-(sin(x/2))^2
    since (cos(x/2))^2+(sin(x/2))^2=1
    and A=A/1
    so cosx=[(cos(x/2))^2-(sin(x/2))^2]/1
    =[(cos(x/2))^2-(sin(x/2))^2]/
    [(cos(x/2))^2+(sin(x/2))]
    dividing numerator and denominator by (cos(x/2))^2 we get
    cosx=[1-(tan(x/2))^2]/[1+(tan(x/2))^2]
    now cross multiply
    [1+(tan(x/2))^2]cosx=1-(tan(x/2))^2
    cosx+cosx(tan(x/2))^2=1-(tan(x/2))^2
    cosx(tan(x/2))^2+(tan(x/2))^2=1-cosx
    [(tan(x/2))^2][cosx+1]=1-cosx
    (tan(x/2))^2=(1-cosx)/1+cosx
    or
    (1-cosx)/1+cosx=(tan(x/2))^2
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