# help me out

• Jul 24th 2009, 03:19 AM
jashansinghal
help me out
If i know $\sin x$ then how can we find $\sin(\frac{x}{2})$(Wondering)(Thinking)(Itwasntme)
• Jul 24th 2009, 03:36 AM
SENTINEL4
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)
• Jul 24th 2009, 03:38 AM
jashansinghal
If i know sin x = y from the trig table but i dont have sin (x/2), then????
• Jul 24th 2009, 03:42 AM
pomp
If you know $\sin{x}$ then you know $x$, if you know $x$ you know $\frac{x}{2}$ , if you know $\frac{x}{2}$ then you know $\sin{\frac{x}{2}}$
• Jul 24th 2009, 03:47 AM
SENTINEL4
Quote:

Originally Posted by pomp
If you know $\sin{x}$ then you know $x$, if you know $x$ you know $\frac{x}{2}$ , if you know $\frac{x}{2}$ then you know $\sin{\frac{x}{2}}$

Exactly (Wink)
• Jul 24th 2009, 03:48 AM
jashansinghal
ok....
i know sin 45 = 1/sqrt 2
but i dont know sin 22.5
?????
• Jul 24th 2009, 03:59 AM
SENTINEL4
By using trig tables you should be able to find it.
• Jul 24th 2009, 04:00 AM
Moo
Quote:

Originally Posted by SENTINEL4
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 $\sin x=2\cos\tfrac x2\sin\tfrac x2$

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

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

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

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

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, 04:00 AM
pomp
Quote:

Originally Posted by jashansinghal
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 $\sin{\frac{x}{2}}$ in some exact form, in terms of $\sin{x}$ ? In that case you need to use the half angle formula:

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