compute arcos(- radical 3 / 2) and also ( -1/ radical 3)
2007-08-18 18:28:05 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
In the principal domains:
arcos(- radical 3 / 2) = 5pi/6
arctan( -1/ radical 3) = -pi/3, I assumed it was arctangent.
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Ideas: Determine the reference angle first, and then adjust the answer to the right quadrant.
2007-08-18 20:23:58 · answer #1 · answered by sahsjing 7 · 0⤊ 0⤋
Imagine a right triangle
| x | = â3 / 2 , | y | = ½ , hypotenuse = 1
in each quadrant . ( Note : Ï = 180° )
â
, â
£ : cos ( ± 30° ) = ( â3 / 2 ) / 1 = â3 / 2
â
¡ , â
¢ : cos ( 180° ± 30° ) = ( - â3 / 2 ) / 1 = - â3 / 2
150° = arc cos ( - â3 / 2 )
â
: tan ( 30° ) = ( ½ ) / ( â3 / 2 ) = 1 / â3
â
¡ : tan ( 180° - 30° ) = ( ½ ) / ( - â3 / 2 ) = - 1 / â3
â
¢ : tan ( 180° + 30° ) = ( - ½ ) / ( - â3 / 2 ) = 1 / â3
â
£ : tan ( - 30° ) = ( - ½ ) / ( â3 / 2 ) = - 1 / â3
2007-08-18 21:27:08 · answer #2 · answered by Zax 3 · 0⤊ 0⤋
arccos(-â3 / 2) = 5Ï/6 or 7Ï/6. check the definition to see if the 7Ï/6 is allowed. usually arccos is the inverse of cos restricted to [0,Ï], which would leave out 7Ï/6.
2007-08-18 18:50:35 · answer #3 · answered by Philo 7 · 0⤊ 0⤋
150 and 210 degrees
not sure about the second one
are you sure that it isn't arctan??
2007-08-18 18:52:51 · answer #4 · answered by Rousey 2 · 0⤊ 0⤋