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| 2011-03-13 10:27:57 +0200 | asked a question | No simplification is done to invert trigonometric functions ? I noticed that $\cos(\pi/6)$ gives $(1/2)\sqrt{3}$ as expected but $\arccos((1/2)\sqrt{3})$ gives $\arccos((1/2)\sqrt{3})$. Is it a missing feature or is there an option that you must call explicitely to obtain $\arccos((1/2)\sqrt{3})=\pi/6$ ? Thanks for your answer |
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