Without any software and approximations prove that $\sec(52^{\circ})-\cos(52^{\circ})>1$

Question

Without any software and approximations prove that $$\sec(52^{\circ})-\cos(52^{\circ})>1$$

We can use some known trig values like $18^{\circ}$,$54^{\circ}$,etc

My try:

I considered the function: $$f(x)=\sec(x)-\cos(x)-1,\: x\in \left (0, \frac{\pi}{3}\right)$$

We have the derivative as: $$f'(x)=\sec x\tan x+\sin x >0$$ so $f$ is Monotone increasing.

So we have: $$f(52^{\circ})>f(45^{\circ})=\frac{1}{\sqrt{2}}-1$$ but not able to proceed

The inequality is equivalent to $\cos(52^{\circ}) < (\sqrt 5 - 1)/2$, which is pretty sharp. — Martin R, Sep 22 '22 at 14:49
If it reduces to what @MartinR said, we can expand out the taylor series of cosine and manually chug a bunch of terms to prove the bound. As for the square root of 5, you can do a similar thing I guess with taylor series. It would truly be quite brute force but given enough time and assuming no mistakes this is plausible ig — Max0815, Sep 22 '22 at 15:24
Is it possible to show that $\dfrac{\cos52^{\circ}}{\cos72^{\circ}}<2$? — insipidintegrator, Sep 22 '22 at 15:30
You can almost figure $\cos(52°)$ by hand by applying the sum-of-angle and half-angle identities to 45° and 60° to get $\cos(52.5°) = \sqrt{\frac{4 + \sqrt2 - \sqrt6}{8}} \approx 0.608761$. — Dan, Sep 22 '22 at 22:04
@Dan I had the same idea by approximating it by $\cos(51^\circ)$ but it already weakens the inequality too much apparently. — dezdichado, Sep 22 '22 at 22:31
@dezdichado: Yeah, what we need to find is a constructible angle that's between $\arccos(\frac{-1+\sqrt5}{2}) \approx 51.827292°$ and 52°. For example, 51.9375°. — Dan, Sep 22 '22 at 22:41
How is $51.9375$ degrees constructible, @Dan? I thought constructible angles are $2\pi / n$ where $ n = 2^k 3^a 5^b 17^c 257^d 65537^e$, where $k\in\mathbb{Z}$ and $a,b,c,d,e\in{0,1}$. But $360/51.9375 = 1920/277$. — Benjamin Wang, Sep 23 '22 at 20:07
@BenjaminWang: Don't forget that all integer multiples of constructible angles are constructible. $51.9375° = \frac{277 \times 3°}{2^4}$. — Dan, Sep 23 '22 at 22:59
I think that a canonical answer should use (i) trigonometric identities (triple angle, double angle, half angle etc.), and (ii) degrees whose trigonometric function values can be expressed in radical form ($30^\circ, 45^\circ, 60^\circ, 18^\circ$ etc.; integer degrees are preferred; simple radical forms are preferred), and (iii) simple manipulations (for example, simplifying the expression, squaring both sides), and (iv) all numbers used are in radical form (optional; for example, for $\pi$, we use the approximate value 3.14159). So, I think @eyeballfrog's answer belongs to this kind. — River Li, Sep 25 '22 at 00:18

user · Accepted Answer · 2022-09-23T19:51:02.127

12

We have that for $\theta\in(0,\pi/2)$, $f(\theta)=\sec \theta -\cos \theta$ is an increasing function and

$$\sec \theta-\cos \theta =1 \implies \cos \theta = \frac{\sqrt 5-1}2 =\frac 1\varphi$$

that is $\theta$ is an angle of a Kepler_triangle, precisely $\theta=51.83°$ (almost equal to the slope of the Great Pyramid of Giza), therefore

$$\sec (52°)-\cos (52°) >1$$

(credit)

Assuming we don't know the value for Kepler angle, according to the following triangles

we have

$$\beta - \theta > \sqrt{(\sin \beta - \sin \theta)^2+(\cos \beta - \cos \theta)^2}>0.0379 \;\text{rad}>2° \implies \theta <52°$$

edited Sep 23 '22 at 19:51

answered Sep 22 '22 at 21:03

user

162,563

Wow nice, i got a few thoughts now...will update the post soon – Ekaveera Gouribhatla Sep 23 '22 at 01:20
But without a calculator, how do you prove that the Kepler angle is less than 52°? – Dan Sep 23 '22 at 18:38
1

@Dan The idea here is to assume this as a "well known" result, as for the value for $e$ or $\pi$. However I understand your point. I'm going to post also another answer which use the same idea but without the knowledge of the value for Kepler angle. – user Sep 23 '22 at 19:11
@Dan I've added the idea as an appendix here. – user Sep 23 '22 at 19:51
Nice work, user. (+1) – nonstudent Sep 23 '22 at 20:03

eyeballfrog · Answer 2 · 2022-09-23T00:34:07.513

8

For angles less than 60 degrees, $\cos(3x)$ is a decreasing function of $\cos(x)$. So apply triple angle identity twice to $\cos(52)<(\sqrt{5}-1)/2$ to get the equivalent formulation $\cos(72) > (7033 -3145\sqrt{5})/2$. Since $\cos(72)=(\sqrt{5}-1)/4$, the inequality reduces to $521/233 < \sqrt{5}$, which is true, though just barely.

edited Sep 23 '22 at 00:34

answered Sep 23 '22 at 00:28

eyeballfrog

23,253

1

(+1) It is very nice. – River Li Sep 23 '22 at 03:14
2

I believe this should be the canonical answer. – dezdichado Sep 23 '22 at 05:16
I agree, this is a very nice answer too! – user Sep 23 '22 at 19:52

River Li · Answer 3 · 2022-09-25T01:04:07.670

Update:

Remarks: We can calculate $\cos (3 \cdot 52^\circ)$ in radical form. Then use $\cos 3u = 4\cos^3 u - 3\cos u$ to prove $\cos 52^\circ < \frac{\sqrt 5 - 1}{2}$. This is based on @eyeballfrog's idea.

We need to prove that $\cos 52^\circ < \frac{\sqrt 5 - 1}{2}$.

Letting $a = \cos 52^\circ$ and $b = \frac{\sqrt 5 - 1}{2}$. Using $a > \cos 60^\circ = 1/2$ and $b > 1/2$, we have $4a^2 + 4ab + 4b^2 - 3 > 12\cdot (1/2)^2 - 3 = 0$. Using $$a - b = \frac{(4a^3 - 3a) - (4b^3 - 3b)}{4a^2 + 4ab + 4b^2 - 3},$$ it suffices to prove that $4a^3 - 3a < 4b^3 - 3b$ or $$4\cos^3 52^\circ - 3\cos 52^\circ < 4\left(\frac{\sqrt 5 - 1}{2}\right)^3 - 3\cdot \frac{\sqrt 5 - 1}{2} = \frac{5\sqrt 5 - 13}{2} $$ or (using $\cos 3u = 4\cos^3 u - 3\cos u$) $$\cos 156^\circ < \frac{5\sqrt 5 - 13}{2}.$$

We have $$\cos 156^\circ = \cos (120^\circ + 36^\circ) = -\frac12 \cos 36^\circ - \frac{\sqrt 3}{2} \sin 36^\circ.$$ Using $\cos 36^\circ = \frac{1 + \sqrt 5}{4}$ and $\sin 36^\circ = \frac{\sqrt{10 - 2\sqrt 5}}{4}$, we have $$\cos 156^\circ = - \frac{1 + \sqrt 5 + \sqrt{30 - 6\sqrt 5}}{8}.$$ (Note: Using $\sin 36^\circ = \cos 54^\circ$ and $\sin 36^\circ = 2\sin 18^\circ \cos 18^\circ$ and $\cos 54^\circ = 4\cos^3 18^\circ - 3 \cos 18^\circ$, we have $2\sin 18^\circ \cos 18^\circ = 4\cos^3 18^\circ - 3\cos 18^\circ$ or $4\sin^2 18^\circ + 2\sin 18^\circ - 1 = 0$ which results in $\sin 18^\circ = \frac{\sqrt 5 - 1}{4}$.)

It suffices to prove that $$- \frac{1 + \sqrt 5 + \sqrt{30 - 6\sqrt 5}}{8} < \frac{5\sqrt 5 - 13}{2}$$ or $$\sqrt{30 - 6\sqrt 5} > 51 - 21\sqrt 5$$ or $$2136\sqrt 5 > 4776$$ which is true.

We are done.

Some thoughts:

Denote $x = \frac{\pi}{180}$.

First, we have $$\cos 28 x = \cos (4 \cdot 52 x - \pi) = - \cos (4 \cdot 52x) = - 8\cos^4 52 x + 8 \cos^2 52x - 1. \tag{1}$$

Second, we have $$\sin 6x = \cos (3\cdot 28x) = 4\cos^3 28 x - 3 \cos 28 x. \tag{2}$$

Third, we have $$\sin 6x = - \frac{1 + \sqrt 5}{8} + \frac18\sqrt{30 - 6\sqrt 5}. \tag{3}$$

From (2) and (3), we solve the cubic equation to get $\cos 28 x$ (closed form).

Then, from (1), we obtain the closed form of $\cos 52x$.

score 1 · Answer 4 · answered Sep 23 '22 at 05:04

Thanks to "user" for giving me thought to complete the proof. Here is the proof:

We are aiming to prove $\sec(52^{\circ})-\cos(52^{\circ})>1$. Let $\phi$ be the Golden ratio and $\psi$ be its reciprocal..

Consider $$f(x)=\sec x-\cos x,\:\:0<x<\frac{\pi}{2}$$ Its evident that $f$ is Monotone increasing. Also $$\begin{aligned} & f(t)=1 \\ \Rightarrow & \sec t-\frac{1}{\sec t}=1 \\ \Rightarrow & \sec t=\phi \\ \Rightarrow & t=\sec ^{-1}(\phi) \end{aligned}$$ We know that by taylor's series: $$\sin ^{-1}(x)=x+\frac{x^3}{6}+\frac{3 x^5}{40}+\cdots,|x| \leqslant 1$$ Using the fact that: $$\sec ^{-1}(v)=\cos ^{-1}\left(\frac{1}{v}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{1}{v}\right)$$ So we have $$t=\sec ^{-1}(\phi)=\frac{\pi}{2}-\left(\psi+\frac{\psi^3}{6}+\frac{3 \psi^5}{40}+p\right)$$ Where $p>0$. Also let $$t_0=\frac{\pi}{2}-\left(\psi+\frac{\psi^3}{6}+\frac{3 \psi^5}{40}\right)$$ We know that $\psi$ satisfies $$\psi^2+\psi-1=0$$ We have the following results which can be easily derived: $$\begin{aligned} \psi^2 &=1-\psi \\ \psi^3 &=2 \psi-1 \\ \psi^5 &=5 \psi-3 \\ \Rightarrow & \psi+\frac{\psi^3}{6}+\frac{3 \psi^5}{40}=\frac{205 \psi-47}{120} . \end{aligned}$$ We have: $$\begin{aligned} & t_0=\frac{\pi}{2}-\left(\psi+\frac{\psi^3}{6}+\frac{3 \psi^5}{40}\right) \\ \Rightarrow & t_0=\frac{\pi}{2}-\left(\frac{205 \psi-47}{120}\right) \\ \Rightarrow & \frac{52 \pi}{180}-t_0=\frac{205 \psi-47}{120}-\frac{19 \pi}{90}=\frac{615 \psi-(141+76 \pi)}{360}>0 \end{aligned}$$ Hence we have : $$\begin{aligned} & t_0<\frac{52 \pi}{180} \\ \Rightarrow & t=t_0-p<\frac{52 \pi}{180} \end{aligned}$$ Finally we have $$\begin{aligned} 1 &=f(t)<f\left(\frac{52 \pi}{180}\right) \\ & \Rightarrow \sec (52^{\circ})-\cos (52^{\circ})>1 . \end{aligned}$$

I will be happy if someone proves $\frac{615 \psi-(141+76 \pi)}{360}>0$. I used approx values which i am not supposed to. — Ekaveera Gouribhatla, Sep 23 '22 at 05:08

Barackouda · Answer 5 · 2022-09-24T12:05:37.570

Hint .

Can you show the inequality for $10\leq x\leq 65$:

$$f\left(x\right)=x\left(\sec\left(x\cdot\frac{\pi}{180}\right)-\cos\left(\frac{\pi}{180}\cdot x\right)\right)> \frac{2(x-9)^{3}-2(x-9)^{2}+15(x-9)+75}{3000}$$

?

Some other hint :

Define :

$$h(x)=\left(\sec\left(x\cdot\frac{\pi}{180}\right)-\cos\left(\frac{\pi}{180}\cdot x\right)\right),p(x)=\frac{2(x-9)^{3}-2(x-9)^{2}+15(x-9)+75}{3000}$$

Then we have for $10<x<65$ :

$$h''(x)>0,p''(x)>0$$

So we have using strong convexity for $x\in[45,52]$ :

$$xh(x)\geq x\left(h'\left(45\right)\left(x-45\right)+h\left(45\right)+\frac{h''\left(45\right)}{2}\left(x-45\right)^{2}\right)>p(x)$$

If $0<a<1$ and $0\leq x\leq 2a\pi$ then :

$$1-\cos\left(x\right)-\left(\frac{\sin\left(a\pi\right)}{a\pi}\right)^{2}\cdot\frac{x^{2}}{2}\geq 0$$

See [1] for a reference .

Now a lemma using the concavity of $\sin(x)$ on $(0,\pi/2)$ :

We have :

$$0<\frac{\sin\left(\frac{\pi}{6.25}\right)-\sin\left(\frac{\pi}{6}\right)}{\frac{\pi}{6.25}-\frac{\pi}{6}}-\frac{\sqrt{3}}{2}$$

A second lemma :

$$\pi<\frac{185}{100}\sqrt{3}$$

Using lemma 1 and 2 with $a=1/6.25$ and $x=52\cdot\frac{\pi}{180}$ for the first inequality we got :

$$\left(\frac{6.25}{\pi}\left(\frac{\sqrt{3}}{2}\left(\frac{1.85\sqrt{3}}{6.25}-\frac{1.85\sqrt{3}}{6}\right)+\frac{1}{2}\right)\right)^{2}\cdot\frac{\left(52\cdot\frac{\pi}{180}\right)^{2}}{2}=1934881/512000<1-\cos\left(52\cdot\frac{\pi}{180}\right)$$

Wich is sufficient to show the claim proposed in the comment @MartinR.

Reference :

[1] Becker,Michael and Lawrence E.Stark An extremal inequality for the Fourier coefficients of positive cosine polynomials ,Univ. Beograd .Publ. Elektrotehn Fak. Ser . Mat ., No.577-No.588 (1977)57-58.

As a convex function, $h$ satisfies the inequality $h(x) \ge h(45)+h'(45)(x-45)$. But is is not obvious to me why $h(x) \ge h(45)+h'(45)(x-45) + h''(45)(x-45)^2/2$ is true. – And where does the last estimate $> p(x)$ come from? — Martin R, Sep 23 '22 at 11:04

user · Answer 6 · 2022-09-24T12:55:16.747

I propose another way using the idea of golden ratio with triple-angle formula.

We have that for $\theta\in(0,\pi/2)$, $f(\theta)=\sec \theta -\cos \theta$ is an increasing function and

$$\sec \theta-\cos \theta =1 \implies \cos \theta = \frac{\sqrt 5-1}2 =\frac 1\varphi$$

Since $\theta < \beta =54°$ we have that

$$\theta < \alpha =52° \iff \cos \theta<\cos \alpha \iff \cos (3\theta)> \cos (3\alpha)$$

with

$$\cos(3\theta)= 4\cos^3 \theta-3\cos \theta = \frac4{\varphi^3}-\frac3{\varphi}=5\varphi-9$$

and using golden gnomon properties

$$\cos(3\alpha)=\cos( 4\beta-60°)=\cos(4\beta)\cos(60°)+\sin(4\beta)\sin(60°)=$$

$$=-\cos(36°)\cos(60°)-\sin(36°)\sin(60°)=-\frac{\varphi}4-\frac{\sqrt 3}2\sqrt{\frac 3 4 -\frac \varphi 4}$$

which leads to

$$\cos (3\theta)> \cos (3\alpha)\iff \frac{\sqrt 3}2\sqrt{\frac 3 4 -\frac \varphi 4}>9-\frac{21}4 \varphi \iff \varphi >\frac{144}{89}$$

which is true, indeed $F_{11}=89$ and $F_{12}=144$ are two consecutive Fibonacci numbers and their ratios approximate $\varphi$ such that

$$\frac{F_{n+1}}{F_{n}}<\varphi $$

for $n$ odd, therefore $\theta <52°$ and the given inequality holds.

Without any software and approximations prove that $\sec(52^{\circ})-\cos(52^{\circ})>1$

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