Hello, Can someone please help me with this problem? Problem: csc(pi/2 - u) = sec u I understand that csc = 1/sin how do I factor ----- 1/sin(pi/2-u) That's where I'm basically stuck. Let me know, thanks!
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Originally Posted by l flipboi l Hello, Can someone please help me with this problem? Problem: csc(pi/2 - u) = sec u I understand that csc = 1/sin how do I factor ----- 1/sin(pi/2-u) That's where I'm basically stuck. Let me know, thanks! $\displaystyle csc \left(\frac{\pi}{2} - u \right) = \frac{1}{sin\left(\frac{\pi}{2} - u \right)}$ $\displaystyle sin(A+B) = sinAcosB + cosAsinB$ For $\displaystyle \theta = \frac{\pi}{2}$ $\displaystyle sin \theta = 1$ and $\displaystyle cos \theta = 0$
Last edited by e^(i*pi); Oct 31st 2009 at 01:30 PM. Reason: typo
Originally Posted by l flipboi l Hello, Can someone please help me with this problem? Problem: csc(pi/2 - u) = sec u I understand that csc = 1/sin how do I factor ----- 1/sin(pi/2-u) That's where I'm basically stuck. Let me know, thanks! $\displaystyle \sin\left(\frac{\pi}{2} - u\right) = \sin\left(\frac{\pi}{2}\right)\cos(u) - \cos\left(\frac{\pi}{2}\right)\sin(u)$ continue ...
So the law of distributing fractions is: a/b (c+d) = a / (bc+bd) ?
Originally Posted by l flipboi l So the law of distributing fractions is: a/b (c+d) = a / (bc+bd) ? don't complicate things ... just evaluate/simplify the expression $\displaystyle \sin\left(\frac{\pi}{2} - u\right) = \sin\left(\frac{\pi}{2}\right)\cos(u) - \cos\left(\frac{\pi}{2}\right)\sin(u) $ and finish the proof.
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