After doing a lot of proofs, I received feedback that some of these proofs may be questionable if we don't show our assumption of $\sin{x}$. We found out that this statement, although basic, posed challenges. Here it is:     Prove that $\left|\sin x\right|\leq|x|,\;\;\forall \;x\in \Bbb{R}.$   Note: This question is quite different from  because I am considering the whole of $\Bbb{R}.$ It is not a duplicate!
58695
$$\sin x&lt;x\,(0&lt;x&lt;\frac{\pi}{2})$$ In most textbooks, to prove this inequality is based on geometry illustration (draw a circle, compare arc length and chord ), but I think that strict proof should be based on analysis reasoning without geometry illustration. Who can prove it? Thank you very much.    ps:   By differentiation, monotonicity and Taylor formula, all are wrong, because $(\sin x)'=\cos x$ must use $\lim_{x \to 0}\frac{\sin x}{x}=1$, and this formula must use $\sin x&lt; x$. This is vicious circle.  If we use Taylor series of $\sin x$ to define $\sin x$, strictly prove $\sin x&lt;x$ is very easy, but how can we obtain geometry meaning of $\sin x$?
118439
Given $a, b \in \mathbb{R}$, prove that   $$\frac{|a + b|}{1 + |a + b|} \le \frac{|a|}{1 + |a|} + \frac{|b|}{1+|b|}$$  When does equality hold?  The only useful thing I could get (using the triangle inequality) is:  $$\begin{align}1 + \left|\left(a + \frac{ab}{2}\right) + \left(b - \frac{ab}{2}\right)\right| &amp;\le 1 + \left|a + \frac{ab}{2}\right| + \left|b - \frac{ab}{2}\right|\\ &amp;\le 1 + |a| + |b| + |ab|\\ &amp;=(1 + |a|)(1 + |b|)\end{align}$$  But it seems to be inapplicable! Any ideas?
20841
I encountered this question:  Is $\cos(x) \geq 1 - \frac{x^2}{2}$   for all $x$ in  $R$  ? I draw to myself the function graphs but i can't find the right way to answer this type of questions.
85736
$$\sin(a+b) = \sin(a) \cos(b) + \cos(a) \sin(b)$$  How can I prove this statement?
123466
I have to show that for every $x \in \mathbb{R}$ and every $n \in \mathbb{N}$ $$|\sin(nx)| \le n|\sin(x)|  $$ In the previous exercise, I have showed that $|\sin(x)|≤|x|$ with the use of the mean value theorem. I think that I cannot use this approach this time. I also tried to write $\sin(nx)$ as a series expansion but that doesn't work either.     Does anyone know how I can solve this?
7675
$$|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}$$   I have tried: $$|a\sin x+b \cos x|\leq |a+b|\leq \sqrt{a^2+b^2}$$  enough to prove: $$|a+b|\leq \sqrt{a^2+b^2}$$  But I can find how to continue from here
46356
How do I prove $\sin x$ is uniformly continuous on $\mathbb R$ with delta and epsilon?   I proved geometrically that $\sin x&lt;x$ and thus, $$|f(x_1)-f(x_2)|=|\sin x_1 - \sin x_2|\le|\sin x_1|+|\sin x_2|&lt;|x_1|+|x_2|$$   But this doesn't help me much finding a delta...   Thanks for any help!  P.S. I'm only at the beginning of calculus so I can't use many theorems and derivation  (because they haven't been regorously proven).
122086
I have to prove that $|\arctan x - \arctan y|\leq |x-y|$ for all real numbers $x$ and $y$. I know that it has something to do with the mean value theorem but I just don't know how to solve it. All help is appreciated  Thanks in advance :)
120944
I have been trying to prove the statement in the title, however I seem to get stuck at a certain point. Let $M$ be the maximal left-ideal of $R$. Then consider $M.r$ for $r \in R$. If $M.r \neq R$, we have that $M.r \subseteq M$. Now if this assumption would be true, we would have that $M$ is a right ideal as well. However when trying to prove that $M.r \neq R$ for all $r \in R$ I don't get very far. I know that if we assume equality, one would get $m.r=1$ for one $m \in M$. However I don't know how to go from here. Thank you in advance!
2419
Prove $||a|-|b|| \leq |a-b|$  I do it by making 4 cases :  Case 1 : $ a \geq 0$, $ b\geq 0$  So we have   $||a||-||b||  = |a - b|$ . hence proved  Case 2 :   $a &lt; 0 , b&lt; 0$  $|-a+b| = |-(a-b)|=|a-b|$  Case 3  $a &gt;0, b&lt;0$  $||a||-||b|| =|a+b| \leq |a|+|b| \leq a -b \leq |a-b| $  Case 4:  $a &lt;0 ,b&gt;0$  $|-a+b| = |-(a-b)| = |a-b|$  Is this OK ? Thanks
84717
Prove that $$\forall x,y \in \mathbb{R}:\vert\cos x- \cos y\vert\leq\vert x-y\vert.$$  My try : $ f(x) =\cos x$ and use the mean value theorem.
101328
Is it always right to claim that: $$\left|A - B\right| \le \left|A\right| + \left|B\right|$$  where $A, B \in \mathbb{R}$ ?
24299
Suppose $x \in (0,\frac{\pi}{2})$ and we want to prove $$x&lt;\frac{\sin x +\tan x}{2}$$I tried to prove it by taking $f(x)=\sin x+ \tan x -2x$ and show $f(x) &gt;0 ,when\space  x \in (0,\frac{\pi}{2})$ take f'$$f'=\cos x +1+\tan ^2 x-2\\=\tan^2 x-(1-\cos x)\\=\tan ^2 x-2sin^2(\frac x2)$$ I get stuck here ,because the last line need to be proved $\tan ^2 x&gt;2sin^2(\frac x2) ,when\space  x \in (0,\frac{\pi}{2})$ $\bf{Question}:$ Is there a geometrical method to prove the first inequality ? (or other idea)  Thanks in advance.    $\bf{Remark}: $I can see the function is increasing $when\space  x \in (0,\frac{\pi}{2})$ like below :
27603
Prove that   $$ \sin (x) \cdot \sin (2x) \cdot \sin(3x) &lt; \dfrac{9}{16} \quad \forall \ x \in \mathbb{R}$$      I thought about using derivatives, but it would be too lengthy.  Any help will be appreciated. Thanks.
121264
I want to prove that for each real number $x&gt;0$, the statement  $${\left(\frac{x+1 } {2}\right)}^{\!x+1} \leq x^x$$  is correct.   I’ve tried using mathematical induction methods, but haven’t found a solution. Although, using any method is appreciated.
111516
Is $$\sin^2(z) + \cos^2(z)=1$$ still true for all $z \in \Bbb{C}$? I've tried rewriting it using complex definitions of $\sin$ and $\cos$, and I don't see why it wouldn't, but the text I'm reading asks this question as if it shouldn't hold?
43818
I am taking a real analysis course. I have the following statement:     Prove that if  $a$ belongs to $\mathbb R$,  such that  $0\leq a &lt;  \epsilon$, for all $\epsilon &gt; 0$, then $a = 0$   I know how to prove it (using contradiction). Now, I was wondering, how I prove almost the same thing:     Prove that if  $a$ belongs to $\mathbb R$,  such that  $0\leq a \leq \epsilon$ for all $\epsilon &gt; 0$, then $a = 0$   Can someone show me how to do it? I am stuck with the equality to show a contradiction.  Note: this is not homework, just for my own knowledge.