# 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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