How do you evaluate #arctan((sqrt (3)) /3)#?

2 Answers
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

4
Apr 26, 2016

Answer:

#arc tan (sqrt3/3)= pi/6 in [0, pi]#. It has two values #pi/6 and (7pi)/6 in [0, 2pi]#. The general value is #npi+pi/6, n=0,+-1,+-2,+-3...#. ,

Explanation:

Thanks to Alan for the timely correction, over my seeing #pi/6 as pi/3#. Now, I am replacing #pi/3# by #pi/6#, everywhere.
#tan (pi/6) = 1/sqrt 3. tan ((7pi)/6)=tan(pi+pi/6)=tan (pi/6)=1/sqrt 3#.

If a is a solution in #[0, 2pi]# for #b = tan a#, then

#arc tan b = npi + a, n=0,+-1,+-2,+-3...#

Here, #b=(3/sqrt 3) = 1/sqrt 3 and a= pi/6 or (7pi)/6#.

Was this helpful? Let the contributor know!
1500
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

1
Alan P. Share
Apr 26, 2016

Answer:

#arctan(sqrt(3)/3)= pi/6=30^@#

Explanation:

Note that #sqrt(3)/3 = 1/sqrt(3)#
Here is a standard trigonometric triangle with this ratio for the tan:
enter image source here
Note that by definition the #arctan# function has a range of #[0,pi)#

Was this helpful? Let the contributor know!
1500