What is the exact value of #cos(tan^-1 (2/3))#?

Question

9301 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-15T14:14:15

#tan^-1(2/3)# is a #t# in #(- pi/2, pi/2)# with #tan t = 2/3#

#t# is in quadrant 1 or 2, so #cost# is positive.

The problem has become:
#tan t = 2/3# and #cost > 0# .Find #cost#.

Use your favorite method for solving such problems.

It's hard to draw and label a triangle or unit circle here, so I'll use the identities:

#cost = 1/sect# and #tan^2t + 1= sec^2t#

#sec^2t = (2/3)^2 + 1 =4/9 + 1 = 13/9#

So, #sect = sqrt13/3# (Remember #cost > 0#, so #sect > 0# as well)

#cost = 3/sqrt13 = (3sqrt13)/13#