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No simplification is done to invert trigonometric functions ?

I noticed that cos(pi/6) gives 1/2sqrt(3) as expected but arccos(1/2sqrt(3)) gives arccos(1/2sqrt(3)). Is it a missing feature or is there an option that you must call explicitely to obtain arccos(1/2sqrt(3))=pi/6 ?

No simplification is done to invert trigonometric functions ?

I noticed that cos(pi/6) gives 1/2sqrt(3) as expected but arccos(1/2sqrt(3)) gives arccos(1/2sqrt(3)). Is it a missing feature or is there an option that you must call explicitely to obtain arccos(1/2sqrt(3))=pi/6 ?

Thanks for your answer

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No simplification is done to invert trigonometric functions ?

I noticed that cos(pi/6) $\cos(\pi/6)$ gives 1/2sqrt(3) $(1/2)\sqrt{3}$ as expected but $\arccos((1/2)\sqrt{3})$ gives $\arccos((1/2)\sqrt{3})$.

but arccos(1/2sqrt(3)) gives arccos(1/2sqrt(3)).

Is it a missing feature or is there an option that you must call explicitely to obtain arccos(1/2sqrt(3))=pi/6 $\arccos((1/2)\sqrt{3})=\pi/6$ ?

Thanks for your answer

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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