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 ?
1 | initial version |
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 ?
2 | No.2 Revision |
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
3 | No.3 Revision |
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 $\arccos((1/2)\sqrt{3})$.
Thanks for your answer
4 | retagged |
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