Can you help me with this complex equation simplifying
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HallsofIvy
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There appears to be something, like parentheses, cut off. Ignoring that, the expression (not an equation) is \(\displaystyle \frac{sin(\theta)+ i cos(\theta)}{sin(\theta)- i cos(\theta)}\).
"Rationalize the denominator"! That is, multiply both numerator and denominator by the "complex conjugate" of the denominator, \(\displaystyle sin(\theta)+ i cos(\theta)\):
\(\displaystyle \frac{sin(\theta)+ i cos(\theta)}{sin(\theta)+ i cos(\theta)}\frac{sin(\theta)+ i cos(\theta)}{sin(\theta)+ i cos(\theta)}\)\(\displaystyle = \frac{sin^2(\theta)+ 2isin(\theta)cos(\theta)- cos^2(\theta)}{sin^2(\theta)+ cos^2(\theta)}\).
Of course, \(\displaystyle sin^2(\theta)+ cos^2(\theta)= 1\) so this is \(\displaystyle sin^2(\theta)- cos^2(\theta)+ 2i sin(\theta)cos(\theta)\). If you remember the "double angle" formulas from trigonometry, \(\displaystyle sin^2(\theta)- cos^2)(\theta)= -cos(2\theta)\) and \(\displaystyle 2sin(\theta) cos(\theta)= sin(2\theta)\), then you can write this as \(\displaystyle f(\theta)= -cos(2\theta)+ i sin(2\theta)\).
"Rationalize the denominator"! That is, multiply both numerator and denominator by the "complex conjugate" of the denominator, \(\displaystyle sin(\theta)+ i cos(\theta)\):
\(\displaystyle \frac{sin(\theta)+ i cos(\theta)}{sin(\theta)+ i cos(\theta)}\frac{sin(\theta)+ i cos(\theta)}{sin(\theta)+ i cos(\theta)}\)\(\displaystyle = \frac{sin^2(\theta)+ 2isin(\theta)cos(\theta)- cos^2(\theta)}{sin^2(\theta)+ cos^2(\theta)}\).
Of course, \(\displaystyle sin^2(\theta)+ cos^2(\theta)= 1\) so this is \(\displaystyle sin^2(\theta)- cos^2(\theta)+ 2i sin(\theta)cos(\theta)\). If you remember the "double angle" formulas from trigonometry, \(\displaystyle sin^2(\theta)- cos^2)(\theta)= -cos(2\theta)\) and \(\displaystyle 2sin(\theta) cos(\theta)= sin(2\theta)\), then you can write this as \(\displaystyle f(\theta)= -cos(2\theta)+ i sin(2\theta)\).