Find the equation of the tangent line to the curve y=2sinx at the point ((pi/6), 1). The equation of the tangent line can be written in y=mx+b form, where m=sqrt3.

What is b?

To find the equation of the tangent line to the curve y=2sinx at the point ((pi/6), 1), we can use the derivative of the function.

The derivative of y=2sinx can be found by applying the chain rule. The derivative of sinx is cosx, and the derivative of 2 is 0, so the derivative of y=2sinx is dy/dx=2cosx.

Now, to find the slope of the tangent line at the point ((pi/6), 1), we substitute x=pi/6 into the derivative:

dy/dx = 2cos(pi/6) = 2(sqrt(3)/2) = sqrt(3)

Given that the slope of the tangent line (m) is sqrt(3), we can write the equation of the line in the form y=mx+b, where m=sqrt(3).

Substituting the coordinates of the point ((pi/6), 1) into the equation, we have:

1 = sqrt(3)(pi/6) + b

To find b, we can solve this equation for b:

1 = (sqrt(3)/6)pi + b

To isolate b, we move (sqrt(3)/6)pi to the other side of the equation:

b = 1 - (sqrt(3)/6)pi

Therefore, the value of b is 1 - (sqrt(3)/6)pi.

Axel