Asked by Anonymous
How do you figure out trigonometric functions that aren't on the unit circle?
example: sin(-27pi/6)
example: sin(-27pi/6)
Answers
Answered by
MathMate
sin(-27pi/6)
= sin(-(27/6)π ±2kπ)
Adjust the value of k so that the expression within the parentheses is within 0 to 2π.
In this particular case, putting k=+3 will give
sin(-27pi/6) = sin((3/2)π)
= sin(-(27/6)π ±2kπ)
Adjust the value of k so that the expression within the parentheses is within 0 to 2π.
In this particular case, putting k=+3 will give
sin(-27pi/6) = sin((3/2)π)
Answered by
Sean
When k is an integer
sin(x) = (sin x + 2k*pi)
set k=2 and...
sin(-27pi/6) = sin(-27pi/6 + 2*2*pi)
= sin(-27pi/6 + 24pi/6)
= sin(-pi/2)
= -1
sin(x) = (sin x + 2k*pi)
set k=2 and...
sin(-27pi/6) = sin(-27pi/6 + 2*2*pi)
= sin(-27pi/6 + 24pi/6)
= sin(-pi/2)
= -1
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