How do you find the equation of the line tangent to the graph of y=sin x at the point where x=pi/3?

1 Answer
Jim G.
Feb 4, 2017

#y=1/2x+1/6(3sqrt3-pi)#

Explanation:

The equation of the tangent in #color(blue)"point-slope form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y-y_1=m(x-x_1))color(white)(2/2)|)))#
where m represents the slope and # (x_1,y_1)" a point on the line"#

#color(orange)"Reminder " m=dy/dx" at x = a"#

#y=sinxrArrdy/dx=cosx#

#"At "x=pi/3: dy/dx=cos(pi/3)=1/2#

#"and " y=sin(pi/3)=sqrt3/2#

#rArrm=1/2" and " (x_1,y_1)=(pi/3,sqrt3/2)#

#rArry-sqrt3/2=1/2(x-pi/3)#

#rArry=1/2x+1/6(3sqrt3-pi)#

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