
Mixed Problems
Hey i have some mixed problems i cant get my head around, i was wondering would any of you know the possible solutions to the following:
sin(30X)=3śroot 2...finding the values between 0 and 360 degrees.
and
Prove sin(A+B)=sinAcosB+cosAsinB
Thank you guys!

For the first question you have:
$\displaystyle
\sin(30x) = \frac {3} {\sqrt 2}
$
You know that the value of the sin function can vary between 1 and + 1 inclusive, yet if you evaluate $\displaystyle 3/\sqrt 2$ what do you get?
As for the proof of the sum of angles formula, see: Proof of the sum and difference formulas.

Hey thanks buddy, is there a shorter way of prooving that identity?

recall that when $\displaystyle \sin\theta = \frac{\sqrt{3}}{2}$
then $\displaystyle \theta = \frac{2\pi}{3} $, $\displaystyle \frac{\pi}{3}$
so in this case $\displaystyle \theta = 30  x$ if assuming that $\displaystyle 30 $ means $\displaystyle 30^o $ then you can calculate the values for x

to prove $\displaystyle \sin(A+B)=\sin{A}\cos{B}+\cos{A}\sin{B}$
one way to do this to use the identity
$\displaystyle
\sin{\theta} = \cos{\left(\frac{\pi}{2}  \theta\right)}
$then
$\displaystyle \sin{\left(A+B\right)} = \cos{\left(\frac{\pi}{2} (A+B)\right)} $
$\displaystyle = \cos{\left(\left(\frac{\pi}{2} A\right)B\right)}$
$\displaystyle =\cos{\left(\frac{\pi}{2} A\right)}\cos{B}
+\sin{\left(\frac{\pi}{2} A\right)}\sin{B}$
$\displaystyle =\sin{A}\cos{B}+\cos{A}\sin{B}$