Use Euler's identity to prove trig identity

[Guest]

Hey guys I need some advice about a problem from an exam I recently took.
The question was: Use Eulaer's Identity to prove the trig identity:
2sin(1/2 (A+B))*cos(1/2 (A-B)) = sinA + sinB

This was my solution:



I got 0 out of 16 points for that, which I think is completely unfair. I understand I was supposed to use cosx=1/2 * (e^ix + e^-ix) and the other one for sinx instead, but I couldn't think of those durning the exam, so I just used e^ix=cosx + isinx. The proof ended up working out fine, but I did one questionable step which was crossing out the "Im" on the fourth line. Isn't this still somewhat right?


Also, I got points taken off for the problem under it because I wrote a domain as {x: sinx > 0}. Is this not an acceptable way to write a domain?

 

[stapel]

Your writing is not very clear in the image you posted. The following is my best guess as to what you mean for exercise(?) (5):

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\(\displaystyle \L e^{i\theta}\, =\, \cos{(\theta)}\, +\, i\sin{(\theta)}\)

\(\displaystyle \L \sin{(A)}\, +\sin{(B)}\, =\, Im\{e^{iA}\}\, +\, Im\{e^{iB}\}\)

\(\displaystyle \L 2Im\{e^{i\left(\frac{A}{2}\, +\, \frac{B}{2}\right)}\}\, Re\{e^{i\left(\frac{A}{2}\, -\, \frac{B}{2}\right)}\}\, =\, Im\{e^{iA}\, +\, e^{iB}\}\)

\(\displaystyle \L Re\{e^{i\left(\frac{A}{2}\, -\, \frac{B}{2}\right)}\}\, =\, \frac{Im\{e^{iA}\, +\, e^{iB}\}}{2Im\{e^{i\left(\frac{A}{2}\, +\, \frac{B}{2}\right)}\}}\)

\(\displaystyle \L Re\{e^{i\left(\frac{A}{2}\, -\, \frac{B}{2}\right)}\}\, =\, \frac{e^{iA}\, +\, e^{iB}}{2\left(e^{i\left(\frac{A}{2}\, +\, \frac{B}{2}\right)} \right)}\)

. . . . .\(\displaystyle \L =\, \frac{1}{2}\left(\frac{e^{iA}}{e^{i\left(\frac{A}{2}\, +\, \frac{B}{2}\right)}}\. +\, \frac{e^{iB}}{e^{i\left(\frac{A}{2}\, +\, \frac{B}{2}\right)}} \right)\)

. . . . .\(\displaystyle \L =\, \frac{1}{2}\left( e^{i\left(\frac{A}{2}\, -\, \frac{B}{2}\right)}\, + \, e^{i\left(\frac{B}{2}\, -\, \frac{A}{2}\right)}\right)\)

. . . . .\(\displaystyle \L =\, \frac{1}{2}\left( e^{i\left(\frac{A}{2}\, -\, \frac{B}{2}\right)}\, +\, e^{-i\left(\frac{A}{2}\, -\, \frac{B}{2}\right)}\right)\)

. . . . .\(\displaystyle \L =\, \frac{1}{2}\left( 2\,Re\{e^{i\left(\frac{A}{2}\, -\, \frac{B}{2}\right)}\}\right)\)

\(\displaystyle \L Re\{e^{i\left(\frac{A}{2}\, -\, \frac{B}{2}\right)}\}\, =\, Re\{e^{i\left(\frac{A}{2}\, -\, \frac{B}{2}\right)}\}\)

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For exercise(?) (6), you appear to have posted the following:

\(\displaystyle \L f(x)\,= \,\sin{(x)};\,\mbox{domain is } \Re\,\mbox{ and range is } \[-1,\,1]\)

\(\displaystyle \L g(x)\,=\,\ln{(x)};\, \mbox{domain is } (0,\,+\infty)\,\mbox{ and range is } \[-\infty,\,+\infty]\)

\(\displaystyle \L (g\,\circ\,f)(x)\,=\, \ln{(\sin{(x)}}\)

\(\displaystyle \L \mbox{domain is all }x\mbox{ such that }\sin{(x)}\,>\,0\)

\(\displaystyle \L \mbox{range is the interval}\, \[-\infty,\,0\]\)

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Please reply with confirmation or clarification. Also, please specify what the original questions were, what the instructions were, and how the above solutions relate.

Thank you.

Eliz.

 

[Guest]

Yes that seems right, thanks for typing it out for me.
The original Question was:
Use Euler's identity to prove the trig identity:

\(\displaystyle 2sin(1/2 (A+B))*cos(1/2 (A-B)) = sinA + sinB\)

I believe my work was correct, but my professor gave me no credit. I'd like to know if my solution is at least partially correct.


As for Question 6, I just wanted to know if writing a domain as {x: sin(x) > 0} is an acceptable form, or if you must take the extra step to write out the values for which sin(x)>0.