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.
Please help me!

  • Pre-calculus... -

    > 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... -

    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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