Question

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

Answer 1

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`.

Answer 2

`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`