Question

Use the unit circle to evaluate these expressions:

a. sin (17???? / 4 )
b. cos (19???? / 6 )
c. tan(450????)

Answer 1

a) We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to
`2\pi`:

`17 \pi/4 - 2\pi = (9\pi)/(4) -2\pi = \pi/4`

For this case we know that
`sin (\pi/4) = (√(2))/(2)`

So then
`sin((17 \pi)/(4)) = (√(2))/(2)`

b) We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to
`2\pi`:

`19 \pi/6 - 2\pi = (7\pi)/(6)`

For this case we know that
`cos (\pi/6) = (√(3))/(2)`

And we know that
`(7\pi)/(6)` is on the III quadrant since is equivalent to 210 degrees. And on the III quadrant the cosine is negative. So then
`cos((19 \pi)/(6)) = -(√(3))/(2)`

c) For this case that any factor of
`\pi` the sin function is equal to 0.

So from definition of tan we have this:

`tan (450\pi) = (sin(450 \pi))/(cos(450 \pi))= (0)/(cos(450\pi))= 0`

Explanation:

a. sin (17pi / 4 )

We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to
`2\pi`:

`17 \pi/4 - 2\pi = (9\pi)/(4) -2\pi = \pi/4`

For this case we know that
`sin (\pi/4) = (√(2))/(2)`

So then
`sin((17 \pi)/(4)) = (√(2))/(2)`

b. cos (19pi / 6 )

We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to
`2\pi`:

`19 \pi/6 - 2\pi = (7\pi)/(6)`

For this case we know that
`cos (\pi/6) = (√(3))/(2)`

And we know that
`(7\pi)/(6)` is on the III quadrant since is equivalent to 210 degrees. And on the III quadrant the cosine is negative. So then
`cos((19 \pi)/(6)) = -(√(3))/(2)`

c. tan(450pi)

For this case that any factor of
`\pi` the sin function is equal to 0.

So from definition of tan we have this:

`tan (450\pi) = (sin(450 \pi))/(cos(450 \pi))= (0)/(cos(450\pi))= 0`