how to find the tangent line at (0,0) for the graph of y=sinx and is the slope of that line positive or negative and why?

y=x

goes from lower left to upper right with slope 1

just look at the graph of y=sinx and its symmetry to see this

Gradient of any tangent line to the function y = sin(x):

y' = cos(x).

When x = 0, cos(0) = 1.

Equation of the line passing through (0,0), with gradient equal to 1:

y - 0 = 1 * (x - 0).

y = x (Answer).

Graph: https://www.desmos.com/calculator/h3oo7kgxy3

Differentiate

y = SinX

dy/dx = CosX (The slope)

When x = 0

dy/dx = Cos(0) = >

dy/dx = 1 (The slope of the curve Sine at x = 0 )

Since Cos(0) = (+)1 then the slope is positive

Hence the tangent line is

y - 0 = 1( x - 0 )

y = x The tangent line of the Sine curve at ( 0,0)

y = sin x

y' = cos x

the slope of tangent line where x = 0:

y' = cos 0 = +1

Equation of the tangent line:

y - 0 = 1(x - 0)

y = x

tsk...can't differentiate ? y ' = cos x , at x = 0 we have y' = 1 > 0 and T line is : y - sin 0 = { 1 } [ x - 0 ]