As is it tangent line by using slope formula y-y1 = m(x-x1) given <span>a=pi/6=x1 </span>f(x)= sinx above equation can be wriiten as f(x)-f(a) = f’(a)(x – a) so L(x)=f’(a) (x-a) + f(a) f(x) = sinx =f(pi/6) =
1/2= y1 f’(x) = cosx f’(pi/6)=
3/2 = m now putting values L(x) = <span>√ </span>3/2( x-pi/6)+1/2 L(x) = (√ 3/2)x+6-pi√ <span>3/12</span>