If i know $\displaystyle \sin x$ then how can we find $\displaystyle \sin(\frac{x}{2})$

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- Jul 24th 2009, 02:19 AM #1

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- Jul 24th 2009, 02:36 AM #2

- Jul 24th 2009, 02:38 AM #3

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- Jul 24th 2009, 02:42 AM #4

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- Jul 24th 2009, 02:47 AM #5

- Jul 24th 2009, 02:48 AM #6

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- Jul 24th 2009, 02:59 AM #7

- Jul 24th 2009, 03:00 AM #8
Is there 22.5 in a trig table ? -_-

Note that since 45 is in the first quadrant, 45/2 is also in the first quadrant.

We know that $\displaystyle \sin x=2\cos\tfrac x2\sin\tfrac x2$

So $\displaystyle \sin^2x=4\cos^2\tfrac x2\sin^2\tfrac x2$

Let $\displaystyle T=\sin^2\tfrac x2$

Then $\displaystyle \sin^2x=4(1-T)\cdot T=4T-4T^2$

This is a quadratic equation. In your situation, we have :

$\displaystyle 4T^2-4T-\frac 12=0$

Solve for T, by remembering it's positive (because it's a square)

And then take its square root, by remembering that it must be positive (because x/2 is in the first quadrant)

- Jul 24th 2009, 03:00 AM #9

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