1. product to sum

If I have 10sin pi/6 cos pi/6 and need to move it to a sum. I am using 1/2[sin(u+v)+sin(u-v)], do I put the "10" in front of both sin? ex.
1/2[10sin(u+v)+10sin(u-v)]??

2. Originally Posted by dashreeve
If I have 10sin pi/6 cos pi/6 and need to move it to a sum. I am using 1/2[sin(u+v)+sin(u-v)], do I put the "10" in front of both sin? ex.
1/2[10sin(u+v)+10sin(u-v)]??
well yes..
$A\sin\alpha\cos\beta = A\left(\frac{1}{2}\left[\sin(\alpha+\beta) + \sin(\alpha+\beta)\right]\right) = \left(\frac{1}{2}\left[A\sin(\alpha+\beta) + A\sin(\alpha+\beta)\right]\right)$
since you can distribute it..

but i noticed that the angles for cos and sin is the same..
you can just use $\sin (2A) = 2\sin A\cos A$

3. thank you very much

4. Hello, dashreeve!

You're doing fancy Trig and you can't do simple algebra?

I have . $10\sin\frac{\pi}{6}\cos\frac{\pi}{6}$ .and need to move it to a sum.

I am using: $\sin u\cos v \;=\;\frac{1}{2}\bigg[\sin(u+v)+\sin(u-v)\bigg]$

Do I put the "10" in front of both sin?

. . $\frac{1}{2}\bigg[10\sin(u+v)+10\sin(u-v)\bigg]$

We have: . $10\cdot\underbrace{\sin(u)\cos(v)}$ . . . Leave the 10 "out front" !!

. . $10\cdot\overbrace{ \left(\frac{1}{2}\bigg[\sin(u + v) + \sin(u-v)\bigg]\right)}$ . . . . Got it?

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

By the way . . . $10\sin\frac{\pi}{6}\cos\frac{\pi}{6} \;=\;5\,\left(2\sin\frac{\pi}{6}\cos\frac{\pi}{6}\ right) \;=\;5\sin\frac{\pi}{3}$

5. I did get ti right, just slower than you guys (you're very smart). I need to think of it more in algebraic terms I guess. In essence I just distributed the 10 earlier than necessary and caused myself more work. Thanks again.