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In my maths textbook, a geometric proof for an inequality relating to trigonometric functions: $$ \cos x < \frac {\sin x}{x} < 1 $$ for $0 < |x| < \frac {\pi}{2}$ is given.

However, I was thinking if there's an alternative algebraic proof for the same inequality using the ranges of $\sin x$ and $\cos x$.

We can write: $$ 0 < \sin x < 1$$ and $$ 0 < \cos x < 1$$ The second inequality can be further modified to be written as: $$ 0 < x \cdot \cos x < x$$ if we assume that $ x > 0 $

But going by this, we need to prove further that: $$ x \cdot \cos x < \sin x < x $$ Can we arrive at this step directly using the two previous inequalities? If so, how? If not, what should the intermediaries be? Or is my approach entirely wrong?

In case my approach is wrong, please provide the correct algebraic proof.

PS: If someone would like to see the geometric proof, it's here.

If possible, do not prove using calculus.

archthegreat
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  • For the left inequality, see https://math.stackexchange.com/q/1789562/42969 – Martin R Mar 11 '23 at 07:13
  • For the right inequality, see https://math.stackexchange.com/q/1725574/42969 – Martin R Mar 11 '23 at 07:15
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    In a sense, the question depends upon what your starting definitions are for sine and cosine. They are often introductorily defined geometrically and hence such proofs are geometric in nature. However, an "algebraic proof" would likely be premised on alternative definitions. Calculus renders the inequality simple to prove, but then you can ask how you wish to obtain the derivatives of various trignometric functions absent any geometric proofs of various trignometric limits. – Golden_Ratio Mar 11 '23 at 07:24
  • @Golden_Ratio I'd like not to use calculus for the proof. I've edited the question to include that, but what do you exactly mean by 'alternative definition'? Is the unit circle definition of trig functions not enough? – archthegreat Mar 11 '23 at 07:33
  • @MartinR thanks for the proof of the right inequality. – archthegreat Mar 11 '23 at 07:35
  • @archthegreat The unit circle definition is geometric in nature, so how do you suggest we prove the result without geometry? Some property or another implicit in your proof will be geometric. – Golden_Ratio Mar 11 '23 at 07:36
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    Well, you can not expect an algebraic proof if you define the trigonometric functions geometrically. – Vivaan Daga Mar 11 '23 at 07:37
  • @Golden_Ratio That is exactly my question. How do you prove the inequality algebraically? If there's a need for an alternative definition, what is it? – archthegreat Mar 11 '23 at 07:42
  • @archthegreat For instance, they can be defined by an infinite series. See here. – Golden_Ratio Mar 11 '23 at 07:49
  • Some authors (most notably Georgi E. Shilov; see his book Elementary Real and Complex Analysis) take the inequality in question as part of the definition of the sine and cosine. – Ramen Nii-chan Mar 11 '23 at 08:31
  • There's no way you are going to prove it using only the inequalities $0<\sin x<1$ and $0<\cos x<1$. After all, if you have two numbers $a$ and $b$, and all you know about them is that they belong to the interval $(0,1)$, then there's absolutely nothing you can say about the sizes of $a$ and $b$ relative to each other. – Hans Lundmark Mar 11 '23 at 11:36
  • (And by the way, $0 < \sin x < 1$ isn't true if $0 < |x| < \pi/2$ and $x<0$.) – Hans Lundmark Mar 11 '23 at 11:37

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