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    Again where 0°≤x≤360°

    cos²(x/2)=1+sin(x/2)
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     \displaystyle \cos^2(x/2)=1-\sin^2(x/2)

    Use this identity to form a quadratic in  \sin(x/2)
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    (Original post by B_9710)
     \displaystyle \cos^2(x/2)=1-\sin^2(x/2)

    Use this identity to form a quadratic in  \sin(x/2)
    i have come to

    1-sin²(x/2)=1+sin(x/2)
    sin²(x/2)+sin(x/2)=0
    sin(x/2)(sinx+1)=0

    where did i go wrong?
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    (Original post by thefatone)
    i have come to

    1-sin²(x/2)=1+sin(x/2)
    sin²(x/2)+sin(x/2)=0
    sin(x/2)(sinx+1)=0

    where did i go wrong?
    Should be  \sin(x/2)+1 .
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    (Original post by B_9710)
    Should be  \sin(x/2)+1 .
    but i took sin(x/2) out as a factor? or does that not work?
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    (Original post by thefatone)
    i have come to

    1-sin²(x/2)=1+sin(x/2)
    sin²(x/2)+sin(x/2)=0
    sin(x/2)(sinx+1)=0

    where did i go wrong?
    \displaystyle \sin^2 \frac{x}{2} + \sin \frac{x}{2} = \sin \frac{x}{2} \left(\sin \frac{x}{2} + 1\right)
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    (Original post by Zacken)
    \displaystyle \sin^2 \frac{x}{2} + \sin \frac{x}{2} = \sin \frac{x}{2} \left(\sin \frac{x}{2} + 1\right)


    damn i forgot, i'm being an idiot again, thanks a bunch guys
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    (Original post by Zacken)
    \displaystyle \sin^2 \frac{x}{2} + \sin \frac{x}{2} = \sin \frac{x}{2} \left(\sin \frac{x}{2} + 1\right)
    why doesn't the value for x = 180° work?
    when i draw the sin graph i get 3 values for where sin=0
    logically if i sub in 180 they don't work

    what about the other one aswell?
    where sin x= -1 there's only one value which is 270°
    of course if i sub it in 270° won't work but why?
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    (Original post by thefatone)
    why doesn't the value for x = 180° work?
    when i draw the sin graph i get 3 values for where sin=0
    logically if i sub in 180 they don't work

    what about the other one aswell?
    where sin x= -1 there's only one value which is 270°
    of course if i sub it in 270° won't work but why?
    Remember it's \frac{x}{2}, so so if you put 180 degrees in, you'll get sin(180/2) = sin 90 = 1. You need x/2 = 0, 180, 360, etc... not x.
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    (Original post by Zacken)
    Remember it's \frac{x}{2}, so so if you put 180 degrees in, you'll get sin(180/2) = sin 90 = 1. You need x/2 = 0, 180, 360, etc... not x.
    i guess logic and subbing values is the way to go on this one then cheers
    you can also guarantee i'll make a trig mk6 thread xD
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    (Original post by thefatone)
    i guess logic and subbing values is the way to go on this one then cheers
    you can also guarantee i'll make a trig mk6 thread xD
    No, you have sin a = 0 when a = 0, 180, 360. Now let a = x/2. Then x/2 = 0, 180, 360. So x = 0, 360, 720.

    sin a = -1, when a = 270. Now let a = x/2. Then x/2 = 270 so x = 2 * 270, etc...
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    (Original post by Zacken)
    No, you have sin a = 0 when a = 0, 180, 360. Now let a = x/2. Then x/2 = 0, 180, 360. So x = 0, 360, 720.

    sin a = -1, when a = 270. Now let a = x/2. Then x/2 = 270 so x = 2 * 270, etc...
    ah that's what i was looking for thanks so much, i think over the holiday i would've peeled my brain from my skull if i didn't have any help from all u guys at tsr
 
 
 
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