Daniel B. answered • 05/12/21

A retired computer professional to teach math, physics

You wrote

"the sin x graph gives the slope of -cos x at every point".

You could have also chosen to say

"the sin x graph gives the slope of 1-cos x at every point".

Or you could have chosen any other constant besides 1.

But I think the choice of 1 illustrates your point the best.

Notice that 1-cos(π) = 2, which is the area under the sin curve between 0 and π.

The function F(x) = 1-cos(x) is most illustrative because F(0) = 0.

That mean, that F(x) represents the accumulated area as you keep increasing

your range starting from 0.

In contrast, the function 100-cos(x) would represent the accumulated area + 99.

And the function -cos(x) represents the accumulated area - 1.

------------------------------------------------

Yes, you are right, we are just subtracting the y-values.

Let me try to explain by answering three questions.

Only Question #2 is the one you asked.

Question #1:

"What does it say on the graph of 100-cos x, the RHS i.e (100-cos (pi))-(100-cos(0)).

On 100-cos x graph how does it express that we have calculated area?"

Answer:

The function 100-cos(x) records how we accumulate area under sin(x) when going from 0 to π.

But at the beginning, when x = 0, we have been given a head start of 99.

So by the time we get to π, 100-cos(π) records not only the gain between 0 and π, but also the

head start of 99.

The difference (100-cos (π))-(100-cos(0)) gives us the net gain between 0 and π,

conveniently cancelling out the initial head-start.

Question #2:

"What does it say on the graph of -cos x, the RHS i.e (-cos (pi))-(-cos(0)).

On -cos x graph how does it express that we have calculated area?"

Answer:

The function -cos(x) records how we accumulate area under sin(x) when going from 0 to π.

But at the beginning, when x = 0, we have been given a handicap of -1.

So by the time we get to π, -cos(π) records not only the gain between 0 and π, but also the

handicap of -1.

The difference (-cos (π))-(-cos(0)) gives us the net gain between 0 and π,

conveniently cancelling out the initial handicap.

Question #3:

"So in what way is -cos(x) special."

Answer:

The function -cos(x) is special only in that it has the integration constant of 0.

The function 1-cos(x) is special in a different way: It actual records the area under sin(x)

without any head start or handicap.

The reason is that 1-cos(0) = 0.

Having the initial value of 0 makes the function special in that the area under the curve is simply

1-cos(π).

The function -cos(x) would be equally special if we wanted the area under sin(x) starting from π/2.

Does this help?

Rahul A.

Yes but what does it say on the graph of -cos x, the RHS i.e (-cos (pi))-(-cos(0)). On -cos x graph how does it express that we have calculated area. When we plot the graph of -cos x(the integral graph) we see that we are just subtracting the y-values i.e (-cos (pi))-(-cos(0)). Am I right?05/14/21