y=cos(x−5π/2)
amplitude = 1
period = 2π
If we had the simple curve y = cos x
the quarter points would be
(0,1) , (π/2, 0) , (π, -1), (3π/2 , 0) , and repeating at (2π, 1)
but we shifted our basic cosine curve 5π/2 units to the right, so our points are:
(0 + 5π/2, 1), (π/2 + 5π/2, 0) , (π + 5π/2, -1) .....
or
(5π/2, 1) , (3π , 0) , (7π/2, -1) , ....
remember that the period is 2π
so we can add or subtract 2π to get more or other points
(5π/2 - 2π,1) , (3π - 2π, 0) , (7π/2, -1) , ....
(π/2, 1) , (π,0), (3π/2 , -1), ...
ahhhh , but that is the same as y = sin(x)
check:
http://www.wolframalpha.com/input/?i=y%3Dcos(x%E2%88%925%CF%80%2F2)
Determine the quarter points.
y=cos(x−5π/2)
(Type an exact answer, using π as needed. Use integers or fractions for any numbers in the expression.)
- Please show step by step work. Show how you're getting the numbers/answers because I want to learn it. Thank You.
3 answers
THANK YOU SOOO MUCH REINY!! GOD BLESS YOU. YOU'RE AMAZING!
ahhhh , but that is the same as y = sin(x)
:)
:)
0 answers