finding the derivative of sin(pi-x)

Question

 

Find the equation of the tangent to the curve y=sin(pi-x) at the point (pi/6 ,1/2), in exact form?

 

Answer 1

finding the derivative of sin(pi-x)

 

Find the equation of the tangent to the curve y=sin(pi-x) at the point (pi/6 ,1/2), in exact form?

 y = sin(p-x)

y'=cos(p-x)(-1)= - cos(p-x) = m ( slope)

m = - cos(p - p/6) = - cos(-5p/6)

m = - cos(5p/6)= - (- sqrt3/2)= (sqrt3)/2

y - y(1) = m(x - x(1))

y - 1/2 = sqrt3/2*(x - p/6)

y = xsqrt3/2 + 1/2 - p*sqrt3/12