# Mixed Problems

#### hughdini

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

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.

hughdini

#### hughdini

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

#### 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

hughdini

#### 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

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

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

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

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

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