Mixed Problems

May 2010
4
0
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!
 
Jun 2008
1,389
513
Illinois
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.
 
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May 2010
4
0
Hey thanks buddy, is there a shorter way of prooving that identity?
 
Nov 2009
717
133
Wahiawa, Hawaii
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
 
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Nov 2009
717
133
Wahiawa, Hawaii
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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