How do you solve #tan(arccos((sqrt2)/2))#?

2 Answers
Jul 18, 2016

Ans.#=1#.

Explanation:

Recall that #arccosx=theta, |x|<=1 iff costheta=x, theta in [o,pi]#

Hence, #arccos(sqrt2/2)=pi/4#, and, so,

#tan{arccos(sqrt2/2)}=tan(pi/4)=1#.

Jul 18, 2016

#tan[arccos(sqrt(2)/2)] =1#

Explanation:

#arccos((sqrt(2))/2)# returns the angle #theta# where we have:

Tony B

#cos(theta)=("adjacent")/("hypotenuse") =sqrt(2)/2#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
But we need the tangent and this means we need to know the length of the opposite (h).

#=>h=sqrt(2^2-2)=sqrt(2)#

So #tan[arccos(sqrt(2)/2)] ->tan(theta) = h/sqrt(2)=sqrt(2)/sqrt(2) = 1#