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  1. #1
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    help me out

    If i know $\displaystyle \sin x$ then how can we find $\displaystyle \sin(\frac{x}{2})$
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  2. #2
    Member SENTINEL4's Avatar
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    Maybe one way is that if you have sinx=y , where y is a number, find from the tables for which x you have y.
    Then if you find x, find x/2 and from the tables again find sin(x/2)
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    If i know sin x = y from the trig table but i dont have sin (x/2), then????
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    If you know $\displaystyle \sin{x}$ then you know $\displaystyle x$, if you know $\displaystyle x$ you know $\displaystyle \frac{x}{2}$ , if you know $\displaystyle \frac{x}{2}$ then you know $\displaystyle \sin{\frac{x}{2}}$
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  5. #5
    Member SENTINEL4's Avatar
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    Quote Originally Posted by pomp View Post
    If you know $\displaystyle \sin{x}$ then you know $\displaystyle x$, if you know $\displaystyle x$ you know $\displaystyle \frac{x}{2}$ , if you know $\displaystyle \frac{x}{2}$ then you know $\displaystyle \sin{\frac{x}{2}}$

    Exactly
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  6. #6
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    ok....
    i know sin 45 = 1/sqrt 2
    but i dont know sin 22.5
    ?????
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  7. #7
    Member SENTINEL4's Avatar
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    By using trig tables you should be able to find it.
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  8. #8
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    Quote Originally Posted by SENTINEL4 View Post
    By using trig tables you should be able to find it.
    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)
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    Quote Originally Posted by jashansinghal View Post
    ok....
    i know sin 45 = 1/sqrt 2
    but i dont know sin 22.5
    ?????
    OK I understand what you are asking now, you want to express $\displaystyle \sin{\frac{x}{2}}$ in some exact form, in terms of $\displaystyle \sin{x}$ ? In that case you need to use the half angle formula:

    $\displaystyle 2 \sin^2{\frac{x}{2}} = 1 - \cos{x}$
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