tangents are drawn from the origin to the curve y=sinx, then their point of contact lies on the curve _________
done , look back at your earlier post
To find the point of contact of the tangents drawn from the origin to the curve y = sin(x), we need to determine the x-coordinate(s) where the tangents intersect the curve.
First, let's find the equation of the tangent line. The equation of a line in slope-intercept form is given by:
y = mx + c
where m is the slope of the line, and c is the y-intercept.
We know that the origin is the point (0, 0), so the line passes through (0, 0). Let's find the slope of the tangent line.
The derivative of y = sin(x) with respect to x gives us the slope of the curve at any given point. Differentiating sin(x) gives us:
dy/dx = cos(x)
At the point of contact, the slope of the tangent line is equal to the slope of the curve at that point. So, we set the derivative equal to the slope:
m = cos(x)
Since the line passes through the origin, the equation of the tangent line becomes:
y = cos(x)x
Now, we need to find the x-coordinate(s) at which the tangent line intersects the curve y = sin(x). To do this, we equate the equations of the curve and the tangent line and solve for x:
sin(x) = cos(x)x
It is not possible to find the exact analytical solution for this equation. However, using numerical methods or algebraic approximations, we can estimate the values of x at which the equation is satisfied.
Therefore, the point(s) of contact lies on the curve y = sin(x), but the exact x-coordinate(s) cannot be determined precisely without further calculations.