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

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Student Comments

atyourservice | Student

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

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...

arccos(0)

arccos(0)

0 actually lies between quadrant 1 and 2, and the value there is pi/2, so the answer is...

`pi/2`