Then from $\sin(x+2\pi)=c\sin(x+\pi)=c^2\sin x=\sin x$, we see that $\sin$ is a periodic continuos function, hence bounded. (Admittedly, this cannot be expanded to $e^{-x^2}$ in any way) Share

There are six main trigonometric functions, namely sin θ, cos θ, tan θ, cot θ, tan θ, cosec θ, and sec θ. We know that sine function is the ratio of the perpendicular and hypotenuse of a right-angled triangle. The domain and range of trigonometric function sine are given by:

2021年10月18日 · it follows that sinx is a real function. Thus sin2x ≥ 0. From Sum of Squares of Sine and Cosine‎, we have that cos2x + sin2x = 1. Thus it follows that: From Ordering of Squares in Reals and the definition of absolute value, we have that: The result follows.

We know that -10\leq sin x \leq 5000 −10 ≤ sinx ≤ 5000. Thus Sin x is a bounded function. There can be infinite m and M. Minimum value of sinx is -1 and maximum value is 1. Thus glb=-1 and lub=1. What is the least upper bound of \ { x \} {x}? Notation: \ { \cdot \} …

I've been tasked with finding the upper and lower bounds of the element: $A = sin (\frac {\pi.n} {2n+3}) | n\in\mathbb {N}$ I think I have found the upper bound by doing: $\lim_ {n\to +\infty} sin (\f...

2020年3月21日 · I'm trying to find the boundaries of $Sin^2(x)$. This is my procedure: For any $\pi/2\le x\le 3\pi/2$ I apply the function $sin(x)$ which for that interval it would be decreasing. So, $1 \ge si...

2 Boundedness: Unlike polynomial functions (with the exception of constant functions), the function sin x and cos x have both a maximum value and a minimum value. For both functions, the maximum value of the function is 1, and the minimum value is ¡1. Both functions take all values between ¡1 and 1, so the range of both functions is ¡1 · y · 1.

2024年8月5日 · Playing with desmos I have accidentally noticed that the sequence of partial sums. {∑n=1N sin(sin(n)): N ≥ 1} {∑ n = 1 N sin (sin (n)): N ≥ 1} is bounded. However, I did not succeed in proving this statement. My main idea was to show that there exists some constant C> 0 C> 0 such that.

NOTE: Now there are some serious discrepancies between Sin, Cos, and Tan. The way to think of this is that even if is not in the range of tan 1 (x), it is always in the right

2013年6月28日 · Hint: Consider the function $f$ given by $f(x)=\sin (2x)-x$ for all $x$ in $[0,\pi/4]$. Differentiate it.