As is it tangent line
by using slope formula
y-y1 = m(x-x1)
given a=pi/6=x1
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) = √ 3/2( x-pi/6)+1/2
L(x) = (√ 3/2)x+6-pi√ 3/12
