trig symmetry

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droodles

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trig symmetry

« on: February 05, 2008, 06:43:48 pm »

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$\frac {\pi}{2} < a < {\pi}$ with $cos (a) = -sin (b)$ where $0< b < \frac {\pi}{2}$ find a in terms of b

So the quadrant the first notation applies to is Q2, what do I do next?

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dcc

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Re: trig symmetry

« Reply #1 on: February 05, 2008, 06:51:44 pm »

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$a = \dfrac{\pi}{2} + b$

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