# Need help with a couple trig problems!

• Jul 20th 2013, 09:24 PM
coolioschmoolio
Need help with a couple trig problems!
Hello everyone I was wondering if you could help me out with a couple trigonometry problems. For the first one I attempted it and got stuck. For the second one, are you supposed to set the real part on left to the real part on right, and the imaginary part on the left to the imaginary part on the right?
e.g. on left side: real part = sin(x) imaginary part = cos(y). On right side: real part =-cos(x) imaginary part = -1
So if I were to set them equal to each other it would look like:
real parts: sin(x) = -cos(x) and imaginary parts: cos(y) = -1
answer for real parts would be 3pi/4 or 7pi/4 . Except with the given range, it would have to be 7pi/4. (x =7pi/4)
answer for imaginary parts would be pi. (y = pi)
Is this correct?
• Jul 20th 2013, 10:15 PM
Prove It
Re: Need help with a couple trig problems!
\displaystyle \begin{align*} \frac{7\pi}{4} \end{align*} is NOT in the region \displaystyle \begin{align*} -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \end{align*}. What would be the equivalent angle to \displaystyle \begin{align*} \frac{7\pi}{4} \end{align*} in that region?

For the first question, surely you can at least see \displaystyle \begin{align*} \sin{(2x)} = \frac{2}{5} \end{align*}. How could you go from here?
• Jul 22nd 2013, 12:22 PM
coolioschmoolio
Re: Need help with a couple trig problems!
Thanks for the help I appreciate it very much. The algebra part of my brain kept telling me to set the first equation to 0 in order to solve but I realized how easy it was once you directed me a little bit. I'll be sure not to make the same mistake in the future. As for #2, luckily I checked this forum right before class and saw that the correct answer is -pi/4 according to the range. Correct?
Thanks again for the help it is greatly appreciated. :)
• Jul 22nd 2013, 06:45 PM
Prove It
Re: Need help with a couple trig problems!
Yes \displaystyle \begin{align*} -\frac{\pi}{4} \end{align*} is correct :)