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Mathematics 9 Online
OpenStudy (anonymous):

Sin5pi Evaluate the trigonometric function using its period as an aid

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?

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!

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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