which is larger cos(sin x) or sin(cos x) ? does this depend on x?

Question

i need to know the process on how to do it

Answer 1

Cos (sin x) is always greater than sin (cos x). This may be seen most simply by graphing the function cos (sin x) - sin (cos x), but if you want to be truly rigorous about it, here's a proof:

Lemma 1: -√2 ≤ cos x + sin x ≤ √2
Proof: Since f(x) = cos x + sin x is periodic with period 2π, it suffices to show that this inequality holds for the interval [-π, π] (or any other interval of length 2π). Since this is a continuous function, it has a maximum and a minimum on said interval by the extreme value theorem. Further, since it is differentiable, this must occur either on the endpoints, or at a point where the derivative is 0. The derivative is -sin x + cos x, which is 0 precisely where sin x = cos x, which on the interval [-π, π] happens only at π/4 and -3π/4. Evaluating at these 4 points, we see that:
f(π) = f(-π) = -1
f(π/4) = 1/√2 + 1/√2 = √2
f(-3π/4) = -1/√2 - 1/√2 = -√2
Thus, √2 and -√2 are the maxima and minima, respectively, on f(x) on the interval [-π, π], and by periodicity, on the whole real line. Q.E.D.

Lemma 2: -√2 ≤ cos x - sin x ≤ √2
Proof: follows from lemma 1, as this is just f(-x).

Main theorem: cos (sin x) - sin (cos x) has no zeros on the whole real line.
Proof: Suppose that for some x, cos (sin x) - sin (cos x) = 0. Then we have

cos (sin x) = sin (cos x)
sin (π/2 - sin x) = sin (cos x) (by the fact that cos x = sin (π/2-x))

And so:

π/2 - sin x = cos x ± 2πk OR π/2 - sin x = π - cos x ± 2πk, where k is some integer. In the first case:

π/2 - sin x = cos x ± 2πk
π/2 ± 2πk = sin x + cos x
Now if k≥0, then the left hand side is greater than or equal to π/2. But π/2 > √2, and √2 ≥ sin x + cos x by lemma 1, so π/2 + 2πk > sin x + cos x, and the equality does not hold. Conversely, suppose k<0. Then k≤-1 (since k is an integer), and the left hand side is less than or equal to -3π/2. -3π/2< -√2 ≤sin x + cos x, so π/2 + 2πk < sin x + cos x, and the equality cannot hold. Thus, sin x + cos x ≠ π/2 ± 2πk

Now consider the second case:

π/2 - sin x = π - cos x ± 2πk
cos x - sin x = π/2 ± 2πk

But this is impossible, because as established above, either π/2 ± 2πk > √2 or π/2 ± 2πk < -√2, and -√2 ≤ cos x - sin x ≤ √2 by lemma 2, so cos x - sin x ≠ π/2 ± 2πk. Thus in both cases we wind up with an equality which is everywhere false, meaning the original equality, cos (sin x) - sin (cos x) = 0, cannot hold either.

Corollary: for all real x, cos (sin x) > sin (cos x)
Proof: Let g(x) = cos (sin x) - sin (cos x). g(0) = 1 - sin 1 > 0. We know that for all x, either g(x) > 0, g(x) < 0, or g(x) = 0. However, by our main theorem, it is impossible that g(x) = 0. It is also impossible that g(x) < 0, since g(x) is continuous, so by the intermediate value theorem, there would exist some c between 0 and x where g(c) = 0, which is impossible. Therefore, for all x, g(x) > 0. And cos (sin x) - sin (cos x) > 0 for all x implies cos (sin x) > sin (cos x) for all x. Q.E.D.

Answer 2

Cos(sin x) is bigger. If X = 90 , sin 90 is 1 and the cos of 1 is 0.99
Cos 90 is 0 sin of that is 0. If you make X = 9 then cos(sin x) is 0.99 and sin(cos x) is 0.017. If you make the 9 a -9 it wil give you exactly the same answers.

Answer 3

sin(x) and cos(x) are functions of "waves" and so because they are waves they oscillate back and forth between "-1" and "1". So it all depends on what "x" is to determine which is bigger. But just remember they are just one and the same thing but they just start off at different areas on the wave (90 degrees to be exact)...for example at "0" the sine wave is at "0" but the cosine wave is at its peak at "1"...they are just at different points on the wave. These are beautiful functions that become very useful in 3D work such as video game design e.g. hidden surface removing....

Answer 4

An XL us 'often' an 18-20, however it somewhat relies upon on the decrease of the blouse. if it is equipped, it is going to likely be a touch tight. even if that's a looser decrease, you should be high-quality. solid success!

Answer 5

I graphed it and found much to my surprise that

cos(sin(x)) > sin(cos(x))

for any x.

Answer 6

Cos(sin(x)) is larger.

Answer 7

cos(sinx) > sin(cosx)
It doesn't depend on x.

Proof.
Let f(x) = cos(sinx) - sin(cosx)
f'(x) = -sin(sinx)cosx - cos(cosx)(-sinx)
Solve f'(x) = 0 for all critical points,and find f(min),
f(min) = 0.107

Answer 8

Neither, they are just offset by 90 degrees. and/or yes it is dependent on x.

Answer 9

It depends on x.

Answer 10

it depends on x....it coulbe negative or positive in the equation...that would make the diffrence