Posted by **Isis** on Monday, May 20, 2013 at 5:48pm.

How do you remove the 2 pi in this problem?

csc 19pi/4 remove 2 pi -> (8pi/4) -> 11pi/4 -> 3pi/4

>> i really don't know how do they the 2 pi twice so it would become 8pi.

Please help me!

- Pre-calculus... -
**Isis**, Monday, May 20, 2013 at 5:50pm
> i really don't know how do they REMOVE the 2 pi twice so it would become 8pi.

Please help me!

>>sorry, please help me

- Pre-calculus... -
**Reiny**, Monday, May 20, 2013 at 6:36pm
19π/4

= 4π + 3π/4

then

csc (19π/4)

= csc (4π + 3π/3)

since csc 19π/4

= cos 19π/4 + i sin 19π/4

= cos (4π + 3π/4) + i sin (4π + 3π/4)

since both the cosine and the sine function have periods of 2π , adding or subtracting multiples of 2π will give us the same answer

so let's "remove " 4π

= cos 3π/4 + i sin 3π/4

= csc (3π/4)

check:

csc 19π/4 = -1/√2 + i 1/√2

csc 3π/4 = -1/√2 + i 1/√2

I really have no idea what

csc 19pi/4 remove 2 pi -> (8pi/4) -> 11pi/4 -> 3pi/4

is supposed to mean.

unless the did this:

19π/4 - 2π = 11π/4

11π/4 - 2π = 3π/4 , which is what I did above.

the 8π/4 does not belong, it could just be a typo

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