Sine
Let R be a random number such that 0 ≤ R ≤ 1.
Then 0 ≤ R x ≤ x.
Since sin(0) = 0, we can simply conclude that 0 ≤ sin(R x) ≤ sin x, right?
However, we know that -1 ≤ sin x ≤ 1. Does this fact invalidate the claim that 0 ≤ sin(R x) ≤ sin x? Let's take a look at sin x vs sin(R x) graphically.
From the graph, we can clearly see that 0 ≤ sin (Rx) ≤ sin x. Let's take a look at the region x ∈ [0, 2π].
In the region x ∈ [0, π/2], we can clearly see that 0 ≤ sin (Rx) ≤ sin x.
As we explore the next region x ∈ [0, π], we still get positive values for sin (Rx). However, since R takes the value between 0 and 1, we get values 0 ≤ sin (Rx) ≤ 1.
Case: if R = 0.5 and x = π, sin(Rx) = 1.
When x ∈ [π/2, 3 π/2], the value of sin x declines below 0. However, since R is random, sin(Rx) takes the values sin x to 1.
Note that when x = 3π/2, sin(x) = -1.
At the final sub interval x ∈ [3π/2, 2 π], sin (Rx) takes on all values in the interval [-1, 1].
To summarize:
When 0 ≤ x ≤ π/2, 0 ≤ sin (Rx) ≤ sin x.
When π/2 ≤ x ≤ π, 0 ≤ sin (Rx) ≤ 1.
When π ≤ x ≤ 3π/2, sin x ≤ sin (Rx) ≤ 1.
When 3π/2 ≤ x ≤ 1, -1 ≤ sin (Rx) ≤ 1.
Interesting that we can't just say 0 ≤ sin (Rx) ≤ sin x for all x.
Cosine
We can make the similar argument that even though:
0 ≤ R ≤ 1, 0 ≤ R x ≤ x, and cos(0) = 1,
The statement 1 ≤ cos(Rx) ≤ cos x is not true for all x. Looking the graph cos x vs cos (Rx):
Looking at the sub-intervals of [0, 2π]:
When 0 ≤ x ≤ π/2, cos x ≤ cos (Rx) ≤ 1.
When π/2 ≤ x ≤ π, π/2 ≤ cos (Rx) ≤ 1.
When π ≤ x ≤ 3π/2, -1 ≤ cos (Rx) ≤ 1.
When 3 π/2 ≤ x ≤ π, -1 ≤ cos (Rx) ≤ 1.
When random numbers are involved with functions, interesting things happen.
sin (Rx) cos (Rx) vs. 1/2 sin (2Rx)
Using the trig identity:
(sin x) * (cos x) = 1/2 * sin(2x)
Let R represent a random number function which returns a random number between 0 and 1. On a calculator, each time the random number function is called, a different random number is returned.
I will leave this as an open question: what happens at each interval [0, π/2], [π/2, π], [π, 3π/2], and [3π/2, 2 π]? I'll end this blog entry with a graphing comparison.
Note the range of sin(Rx) cos(Rx) is [-1, 1].
Stay safe everyone! Eddie
This blog is property of Edward Shore. 2012
