Question

How do you prove that the line y = x is a tangent to the curve y = sinx at the origin? Also, how do you deduce the values of m for which the line y = mx cuts the curve y = sinx in the interval 0<=x<=pi?

Answer 1

Please see the explanation below.

Explanation:

The equation of the curve is

`f(x)=sinx`

The derivative of this function is

`f'(x)=cosx`

The equation of a tangent to the curve at a point `(a, f(a))` is

`y-f(a)=f'(x)(x-a)`

When `a=0`

`y-0=cos(0)(x-0)`

`y=x`

This is the equation of the tangent to the curve at the origin.

`y=mx`

This will cut the curve in the interval `0<=x<=pi` only if

`0<=m<=1`

graph{(y-sinx)(y-x)(y-0)(y-0.5x)=0 [-1.747, 12.3, -3.496, 3.526]}