`arccos(cos((7pi)/2))` Use the properties of inverse trigonometric functions to evaluate the expression.

atyourservice | Student


this could be thought of as 3 whole rounds `(6pi/2) ` plus` pi/2`

as `pi/2` is between [-1,1] we don't need to do anything. the arccos is the inverse of cos, so  

`arccos(cos(7pi/2)) =cos(7pi/2)`

`7pi/2` is the same as `pi/2` as shown above, and cos `(pi/2) = 0 ` (because on the unit circle pi/2 has the coordinates (0,1), and the x is the cos)

`arccos(0) = 90^@` or `pi/2`

iamtall14 | Student
Arccosine and cosine are inverses of each other, so arccos(cos(7pi/2)) would be equal to cos(7pi/2). The answer is 0. So...
0 actually lies between quadrant 1 and 2, and the value there is pi/2, so the answer is...