16 дек. 2015 г. · FYI: Here's a trigonograph (picture proof), where you would take x=A+B and y=A−B (so that A=(x+y)/2 and B=(x−y)/2). |
9 окт. 2011 г. · The inequality |sinx−y2|<|x−y2| is well known, while |cosx+y2|≤1 is even more well known. The first inequality is sharp if x−y≠0. |
26 янв. 2017 г. · Trigonometric Theorem Proof: If sinx=siny, then, x=nπ+(−1)ny. I am working on trigonometric theorems from this book. I am referring to Theorem ... |
28 апр. 2020 г. · The way to make this rigorous is via the Mean Value theorem. For any x<y∈R there is a c∈(x,y) s.t. sin(x)−sin(y)=cos(c)(x−y). |
17 мая 2014 г. · Hint 1: Notice that the derivative of f(x)=sin(x) is bounded and apply the mean value theorem. |
21 апр. 2016 г. · How to use mean value theorem to prove the inequality |sinx−siny|≤|x−y| for all x,y∈R? So let us set f(x)=sinx ... |
20 июл. 2019 г. · We know that sinx=siny implies that x=2kπ+y or x=2kπ+π−y. If we want to solve sinx=sinx using this method, it gives x=2kπ+x or x=2kπ+π−x but ... |
2 апр. 2019 г. · Let A and B be as you defined. Then sin(A+B)=sin(x+y2+x−y2). Evaluate this and use the given identities. |
20 июл. 2022 г. · Let x<y. Without loss of generality assume that sinx≤siny. The |siny−sinx|=siny−sinx=∫yxcostdt=(y−x)−∫yx(1−cost)dt==(y−x)−12∫yxsin2(t/2)dt. |
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