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I need to prove : $\sin^2(x)\leq\ {x^2}$

My Attempt: (I quickly drew the graph of the functions and got the result, but after a lot of trial I am unable to prove it algebraically).

How can we prove this algebraically?

The Graph:

enter image description here

Rasputin
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    That is surely wrong, and your graph is not that of the sine function. Do you mean $\sin^2(x) \le x^2$? – Martin R Nov 11 '20 at 16:45
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    That's not the graph of $\sin x$. And your inequality is not true. – B. Goddard Nov 11 '20 at 16:45
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    It is well-known that $|\sin(x)| \le |x|$, see for example https://math.stackexchange.com/q/1456393/42969. – Martin R Nov 11 '20 at 16:50
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    This is calculus but. The derivative of $\sin x$ is $\cos x$ and so the derivative of $\sin x$ at $0$ is $\cos 0 = 1$. However the derivative of $x^2$ is $2x$ and so the derivative of $x^2$ at $0$ is $2*0=0$. So $\sin x$ is increasing at $0$ at a faster rate then $x^2$ and as $\sin 0 =0^2$, it must be that $\sin x$ "overtakes" $x^2$ imediately and there is a section above $0$ where $\sin x> x^2$. In fact the area is large. $\sin \frac {\pi}4 =\frac {\sqrt 2}2\approx .707$ whereas $\frac {pi^2}{4^2}\approx .617$. – fleablood Nov 11 '20 at 16:54
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    @MartinR The Typing Mistake Has Been Corrected. – Rasputin Nov 11 '20 at 17:30

2 Answers2

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By the mean value theorem, $$\forall x\in\Bbb R\quad\exists c\in\Bbb R\quad\sin x-\sin0=(x-0)\cos c$$ hence $$\sin^2x=x^2\cos^2c\le x^2.$$

Anne Bauval
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I would split the inequality up into different cases. $$ \sin^2 x \leq x^2 $$ For $x\leq-1$ and $x\geq1$, the inequality is certainly true, as the range of $\sin^2 x$ is $[-1,1]$, while $x^2\geq1$ when $x \leq -1$ and $x \geq 1$. When $0\leq x < 1$, we know that $x \geq \sin x$ because $\sin x$ has a gradient strictly less than $1$. Since $x \geq \sin x$, it follows that $x^2 \geq \sin^2 x$.* Also, $\sin^2(-x)=\sin^2(x)$ and $(-x)^2 = x^2$, meaning that the case $-1 < x<0$ is identical.

*It is not true in general that if $a>b$ then $a^2>b^2$. However, if $a$ and $b$ are both positive, then this statement is always true.

Joe
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