Which is greater, #tan 1# or #tan^(-1) 1# ?
4 Answers
Explanation:
We find:
#tan 1 ~~ 1.5574#
#tan^(-1) 1 = pi/2 ~~ 1.5757#
So
Explanation:
Note that:
#d/(dx) tan x = sec^2 x >= 1#
for all real values of
If
#tan y = x#
So:
#(dy)/(dx) sec^2 y = 1#
#(dy)/(dx) = 1/(sec^2 y) = 1/(1+tan^2 y) = 1/(1+x^2)#
Hence:
#d/(dx) tan^(-1) x in (0, 1]# for all#x in RR#
So:
#d/(dx) (tan x - tan^(-1) x) = sec^2 x - 1/(1+x^2) >= 0#
for all
In addition note that
Hence
In particular note that for small
So:
#tan x > tan^(-1) x# for all#x in (0, pi/2)#
including
See below.
Explanation:
Note that near
The
Explanation:
Another approach. This one is pretty much all first semester of trig stuff as long as you've covered graphing tangent.
We know that
tangent is an increasing function and is continuous on the interval
Since
So