# Pre-calculus...

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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.

• Pre-calculus... -

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

• Pre-calculus... -

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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