Johnpeters
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If Sin1/2x= +/- (square root) ((1-cosx)/2)
and Cos1/2x= +/- (square root) ((1+cosx)/2)

Then how does tan1/2x become sinx/(1+cosx)?

Could someone explain the steps of the proof please. Thanks
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Smaug123
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(Original post by Johnpeters)
If Sin1/2x= +/- (square root) ((1-cosx)/2)
and Cos1/2x= +/- (square root) ((1+cosx)/2)

Then how does tan1/2x become sinx/(1+cosx)?

Could someone explain the steps of the proof please. Thanks
\sin(\frac{1}{2}x) = \pm \sqrt{\frac{1-\cos(x)}{2}}, \cos(\frac{1}{2}x) = \pm \sqrt{\frac{1+\cos(x)}{2}}.

Then \tan(\frac{1}{2}x) = \sqrt{\frac{1-\cos(x)}{1+\cos(x)}}. Multiply top and bottom by 1+\cos(x).
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Johnpeters
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(Original post by Smaug123)
\sin(\frac{1}{2}x) = \pm \sqrt{\frac{1-\cos(x)}{2}}, \cos(\frac{1}{2}x) = \pm \sqrt{\frac{1+\cos(x)}{2}}.

Then \tan(\frac{1}{2}x) = \sqrt{\frac{1-\cos(x)}{1+\cos(x)}}. Multiply top and bottom by 1+\cos(x).
Thanks a lot
EDIT: Don't you multiply the top and bottom by (squareroot) 1+cos(x)?
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the bear
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(Original post by Smaug123)
\sin(\frac{1}{2}x) = \pm \sqrt{\frac{1-\cos(x)}{2}}, \cos(\frac{1}{2}x) = \pm \sqrt{\frac{1+\cos(x)}{2}}.

Then \tan(\frac{1}{2}x) = \sqrt{\frac{1-\cos(x)}{1+\cos(x)}}. Multiply top and bottom by 1+\cos(x).
good explanation.

just wondering why the "plus or minus" is not on the tan ?
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Smaug123
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(Original post by Johnpeters)
Thanks a lot
EDIT: Don't you multiply the top and bottom by (squareroot) 1+cos(x)?
Yes, sorry - I meant "multiply the fraction which is inside the square root", not "multiply the whole expression". Sorry.
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Smaug123
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(Original post by the bear)
good explanation.

just wondering why the "plus or minus" is not on the tan ?
Consider when tan(x/2) is positive - it's exactly when x is between 0 and pi, or 2pi and 3pi, or…
Consider when sin(x) is positive - it's exactly when x is between 0 and pi, or 2pi and 3pi, or…
And 1+cos(x) is always positive.
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the bear
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(Original post by Smaug123)
Consider when tan(x/2) is positive - it's exactly when x is between 0 and pi, or 2pi and 3pi, or…
Consider when sin(x) is positive - it's exactly when x is between 0 and pi, or 2pi and 3pi, or…
And 1+cos(x) is always positive.
thanks for that
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