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Mathematics 7 Online
OpenStudy (rizwan_1122):

lim x-->0 (1-cosx)/sin2x

rvc (rvc):

do u know L hopital rule?

OpenStudy (rizwan_1122):

didn't know L hopital rule?

OpenStudy (trojanpoem):

lim of sinax/x = a , same with cos... lim 1-1+2sin^2(x/2) /sin2x = 2 sin^2(x/2) / sin2x 2sin(x/2) * sin(x/2) /sin2x * x/x 2sin(x/2) sin(x/2)/x / sin2x/x 2sin(0) * 0.5 / 2 = 0

OpenStudy (trojanpoem):

I doubt you can read that xD upside down.

OpenStudy (rizwan_1122):

i dosnt understand

OpenStudy (trojanpoem):

Use half angle rule , you know it ? cosx = 1- 2sin^2(x/2) That way you will eliminate the 1. Now divide over x in denominator and nominator lim of sinax/x = a so you will get 0.5 / 2 but you have sin^2 so you will replace one of them with 0

OpenStudy (trojanpoem):

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