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

Question

Ezekiel Alexander · Accepted Answer

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

Evelyn Agard · Answer

undefined To find the equation of the line tangent to the graph of y = sin x at the point where x = π/3, we need to determine the slope of the tangent line and the coordinates of the point of tangency.

First, we find the derivative of y = sin x, which is dy/dx = cos x.

Next, we substitute x = π/3 into the derivative to find the slope of the tangent line at that point. So, the slope is dy/dx = cos(π/3) = 1/2.

Now, we need to find the y-coordinate of the point of tangency. Substituting x = π/3 into the original equation y = sin x, we get y = sin(π/3) = √3/2.

Therefore, the point of tangency is (π/3, √3/2).

Using the point-slope form of a linear equation, y - y₁ = m(x - x₁), where (x₁, y₁) is the point of tangency and m is the slope, we can substitute the values to find the equation of the tangent line.

Thus, the equation of the line tangent to the graph of y = sin x at the point where x = π/3 is y - √3/2 = 1/2(x - π/3).