# Mixed Problems

• May 24th 2010, 09:08 AM
hughdini
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(30-X)=3śroot 2...finding the values between 0 and 360 degrees.
and
Prove sin(A+B)=sinAcosB+cosAsinB

Thank you guys!
• May 24th 2010, 10:13 AM
ebaines
For the first question you have:
$\displaystyle \sin(30-x) = \frac {3} {\sqrt 2}$

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.
• May 24th 2010, 10:29 AM
hughdini
Hey thanks buddy, is there a shorter way of prooving that identity?
• May 24th 2010, 10:33 AM
bigwave
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
• May 24th 2010, 10:48 AM
bigwave
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

$\displaystyle \sin{\left(A+B\right)} = \cos{\left(\frac{\pi}{2} -(A+B)\right)}$

$\displaystyle = \cos{\left(\left(\frac{\pi}{2} -A\right)-B\right)}$

$\displaystyle =\cos{\left(\frac{\pi}{2} -A\right)}\cos{B} +\sin{\left(\frac{\pi}{2} -A\right)}\sin{B}$

$\displaystyle =\sin{A}\cos{B}+\cos{A}\sin{B}$