Mixed Problems

**hughdini** · May 24th 2010, 09:08 AM

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)=3÷root 2...finding the values between 0 and 360 degrees.
and
Prove sin(A+B)=sinAcosB+cosAsinB

Thank you guys!

**ebaines** · May 24th 2010, 10:13 AM

For the first question you have:
$ \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 $3/\sqrt 2$ what do you get?

As for the proof of the sum of angles formula, see: Proof of the sum and difference formulas.

**hughdini** · May 24th 2010, 10:29 AM

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

**bigwave** · May 24th 2010, 10:33 AM

recall that when $\sin\theta = \frac{\sqrt{3}}{2}$
then $\theta = \frac{2\pi}{3}$ , $\frac{\pi}{3}$

so in this case $\theta = 30 - x$ if assuming that $30$ means $30^o$ then you can calculate the values for x

**bigwave** · May 24th 2010, 10:48 AM

to prove $\sin(A+B)=\sin{A}\cos{B}+\cos{A}\sin{B}$

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

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

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

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

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

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