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lim h--> 0 (sin (pi/6+h) - sin pi/6)/ h
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numerator becomes zero
this is derivative d[sin[x]]--> cos[x] cos[pi/6]= sqrt[3]/2
Use sin(A+B)=sinAcosB+cosAsinB (sin(pi/6+h) - sin(pi/6))/h =(sin(pi/6)cos(h)+cos(pi/6)sin(h)-sin(pi/6))/h =(1/2)cos(h) + (sqrt(3)/2)*sin(h)-(1/2) So lim h->0 cos(h)=1, sin(h)/h=1 the limit becomes sqrt(3)/2
This is just the first principles approach to calculate the derivative of sin(x), which should give cos(x).
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