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OpenStudy (anonymous):
Sin5pi
Evaluate the trigonometric function using its period as an aid
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OpenStudy (superdavesuper):
sin function has a period of 2 pi which means sin(a+2pi)=sin(a)
OpenStudy (superdavesuper):
so sin(5pi)=sin(pi+2pi+2pi)=...
OpenStudy (anonymous):
Uh.. Can you cancel things out??
OpenStudy (superdavesuper):
U can eliminate 2pi because the period is 2pi; meaning it repeats itself after 2pi :)
OpenStudy (anonymous):
Okay, well this didn't really get through to me... Mind doing a different example?
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OpenStudy (superdavesuper):
sin(a+2pi)=sin(a) n it doesnt matter what a is....
sin(5pi)=sin(3pi+2pi) n put a=3pi, u find sin(5pi)=sin(3pi) ok?
OpenStudy (anonymous):
Okay..
OpenStudy (superdavesuper):
sin(3pi)=sin(pi+2pi) so a=pi this time....sin(3pi)=?
OpenStudy (anonymous):
Sin(pi)??
OpenStudy (superdavesuper):
yup yup :) n sin(pi)=0 so heres ur answer!
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OpenStudy (anonymous):
Thanks xD buy is there a formula?
OpenStudy (superdavesuper):
hahahaa sine U ask for it, yes there is: sin(pi-a)=-sin(a) so in this case
sin(pi)=sin(pi-0)=-sin(0)=0
OpenStudy (anonymous):
What if a is negative??
OpenStudy (superdavesuper):
sin(pi-a)=-sin(a) is good for any a, + or -
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