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Math Help - [SOLVED] How can i show this...trig identities??

  1. #1
    Junior Member maths900's Avatar
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    [SOLVED] How can i show this...trig identities??

    How can i show that

    2-\sqrt{2+2cos\theta}=4sin^2(\theta/4)
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by maths900 View Post
    How can i show that

    2-\sqrt{2+2cos\theta}=4sin^2(\theta/4)

    Formula

     <br />
\sqrt{\frac{1+cos(\theta)}{2}} = cos(\frac{\theta}{2})<br />

     <br /> <br />
\sqrt{\frac{1-cos(\theta)}{2}} = sin(\frac{\theta}{2})<br />


    Now its better if you try it and ask incase of further troubles
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  3. #3
    MHF Contributor

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    Quote Originally Posted by maths900 View Post

    How can i show that 2-\sqrt{2+2cos\theta}=4sin^2(\theta/4)
    this is not an identity! for example it's not true if \theta = 2\pi. in order to make it an identity we need \frac{\theta}{2} to be in the quadrant 1 or 4, where cosine is positive. the proof then is very easy.

    just use the identities: 1+\cos(2x) = 2 \cos^2 x and 1-\cos(2x)=2\sin^2x.
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  4. #4
    Moo
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    Hello,
    Quote Originally Posted by maths900 View Post
    How can i show that

    2-\sqrt{2+2cos\theta}=4sin^2(\theta/4)
    \cos(2x)=1-2 \sin^2(x)
    Hence 2 \sin^2(\theta/4)=1-\cos(\theta/2)

    So 4 \sin^2(\theta/4)=2-2\cos(\theta/2)

    Then \cos(2x)=2 \cos^2(x)-1 \Rightarrow 2\cos(\theta)=(2 \cos^2(\theta/2))^2-2 \Rightarrow (2\cos^2(\theta/2))^2=2+2 \cos(\theta)
    ---> 2 \cos(\theta/2)=\pm \sqrt{2+2 \cos(\theta)}

    Now you have to see in which quadrant \theta/2 is...


    Quote Originally Posted by ADARSH View Post
    Formula

     <br />
\sqrt{\frac{1+cos(\theta)}{2}} = cos(\frac{\theta}{2})<br />

     <br /> <br />
\sqrt{\frac{1-cos(\theta)}{2}} = sin(\frac{\theta}{2})<br />


    Now its better if you try it and ask incase of further troubles
    A cosine or a sine can be negative...
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  5. #5
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    Hello, maths900!


    We need the double-angle identities:

    . . \sin^2\!A \:=\:\frac{1-\cos2A}{2}

    . . \cos^2\!A \:=\:\frac{1+\cos2A}{2}



    Show that: . 2-\sqrt{2+2\cos\theta}\;=\;4\sin^2\!\left(\tfrac{\th  eta}{4}\right)

    The left side is:

    . . 2 - \sqrt{2(1 + \cos\theta)}

    . . =\;2-\sqrt{4\left(\frac{1+\cos\theta}{2}\right)}

    . . =\;2 - \sqrt{4\cos^2\!\tfrac{\theta}{2}}

    . . =\;2-2\cos\tfrac{\theta}{2}

    . . =\;2\left(1 - \cos\tfrac{\theta}{2}\right)<br />

    . . = \;4\left(\frac{1-\cos\frac{\theta}{2}}{2}\right)

    . . =\;4\sin^2\!\tfrac{\theta}{4}

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  6. #6
    Junior Member maths900's Avatar
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    Thanks everyone, thought i'd have to use the double angle identities but i'm not as good with them as you guys are, thank you
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