> 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
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!
2 answers
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
= 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
7 answers