Question

How do you find the equation of the tangent line to the curve `f(x) = sin cos(x)` at x = pi/2?

Answer 1

For this periodic-wave-function, with period `2pi, (pi/2. 0)` is a tangent-crossing-curve point of inflexion. The equation of the tangent is `x + y = pi/2`.

Explanation:

`f'(x) = - sin x cos (cos x)`.
`f'(pi/2)=-sin (pi/2) cos (cos( pi/2))=(-1)(1)=-1`
`f''(x)=-cos x cos(cos x)-sin x (-sin(cos x)(-sin x)`.
`f''(pi/2)=0`.

So, for this periodic-wave-function, with period `2pi, (pi/2. 0)` is a tangent-crossing-curve point of inflexion.

The slope of the tangent is `-1`.

The equation of the tangent is `x + y = pi/2`.

An idiosyncrasy of this graph is that the amplitude is not 1. It is sin(1 radian)=0.84147, nearly.

This is evident from `|cos x|<=1, and so, |f(x)|<=|sin( 1 radian )|<=0.841471`.

One wave of this periodic graph is given for `x in [0, 2pi]`.

Answer 2

To find the equation of a line, we need a point and a slope.

`f(x) = sin(cos(x))`

Find the `y` coordinate
of the point on the curve where `x=pi/2`.

`y = f(pi/2) = sin(cos(pi/2)) = sin(0)=0`.

So the point is `(pi/2,0)`

Find the slope
of the tangent line at `x=pi/2`.

Use the derivative. Use the chain rule to differentiate:

`f'(x) = cos(cos(x))(d/dx(cos(x))) = cos(cos(x))(-sin(x))`

At `pi/2`, the slope of the tangent line is

`m = f'(pi/2) = cos(cos(pi/2))(-sin(pi/2))`

`= cos(0)(-1) = (1)(-1) = -1`

Find an equation of the line with slope `m=-1` through the point `(pi/2,0)`

There are various answers possible.

`y-0 = -1(x-pi/2)`

`y = -x+pi/2`

`x+y=pi/2`