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25

Suppose that $x=\sin(10^\circ)$ and we want to prove the irrationality of $x$. Then we can use the Triple Angle Formula for $\sin$ to get $-4x^3+3x = \sin(30^\circ)=\frac12$; in other words, $x$ is a solution of the equation $-8x^3+6x-1=0$. But now that we have this equation we can use another tool, the Rational Root Theorem : any rational root $\frac pq$ ...


14

Since $\tan(\theta) = \dfrac{2t}{1-t^2}$, draw a picture using the fact that $\tan(\theta) = \frac{\mbox{opposite}}{\mbox{adjacent}}$ in a right triangle. From here, find the hypotenuse using the Pythagoren Theorem. Then, use the definition for $\sin(\theta)$ and $\cos(\theta)$ in a right triangle.


10

A rather remarkable identity is that, for any $\alpha$ and $\beta$, we can find a $\theta$ and $c$ such that: $$\alpha\sin(x)+\beta\cos(x)=c\sin(x+\theta).$$ To show this, we can expand the right-hand side by the angle-sum identity for sine: $$\alpha\sin(x)+\beta\cos(x)=c\sin(x)\cos(\theta)+c\cos(x)\sin(\theta)$$ and if we group coefficients of $\sin$ and ...


8

This statement is the same as proving $$\int_0^{2\pi} \frac{\sin^2 nx}{\sin^2 x} dx = 2\pi n.$$ Put $z=e^{ix}$ so that $dz = i e^{ix} \; dx = iz \; dx$ and use $$\sin x = \frac{e^{ix}-e^{-ix}}{2}$$ to get $$\int_{|z|=1} \frac{(z^n-1/z^n)^2}{(z-1/z)^2} \frac{1}{iz} dz = \int_{|z|=1} \frac{1}{z^{2n}} \frac{(z^{2n}-1)^2}{(z-1/z)^2} \frac{z}{iz^2} dz \\ = ...


7

Here is a cannon to shoot a fly. Rewrite the integral as \begin{equation} I(n):= \int_0^\pi\frac{1-\cos2nx}{1-\cos2x}\,dx\stackrel{2x\,\mapsto\, x}\Longrightarrow \frac{1}{2}\int_0^{2\pi}\frac{1-\cos nx}{1-\cos x}\,dx\tag{1} \end{equation} From my answer here, we have \begin{equation}\int_0^{2\pi}\frac{\cos mx}{p-q\cos x}\, ...


7

Draw a right triangle such that $\sin\theta = u$ with side lengths $u$, $\sqrt{1-u^2}$ and hypotenuse $1$. Then $$\cos(\arcsin(u)) = \cos\theta = \sqrt{1 - u^2}$$


6

Results used I will just state the following result as I do not wish to replicate Random Variable's brilliant work in this answer. $$\int^\frac{\pi}{4}_0\ln^2(\cos{x})\ {\rm d}x=\Im{\rm Li}_3(1-i)-\frac{\mathbf{G}}{2}\ln{2}+\frac{7\pi^3}{192}+\frac{5\pi}{16}\ln^2{2}$$ It is also quite easy to show that $$\sum^\infty_{n=1}\frac{H_n}{n^2}z^n={\rm ...


6

Such integral is related with the topological degree of a closed curve. We have: $$ I = \frac{1}{2}\int_{0}^{2\pi}\frac{\sin^2 x}{\|a-be^{ix}\|^2}\,dx = -\frac{i}{2}\oint_{\|z\|=1}\frac{\left(\frac{z-z^{-1}}{2}\right)^2}{(a-bz)(az-b)}\,dz$$ hence we just have to consider the residues of the last integrand function in $z=0$ and $z=\frac{b}{a}$ (since ...


5

Hint $1$: $\sin \theta \cdot \cos \theta = \dfrac{\sin 2\theta}{2}$ Hint $2$: $\sin \theta \cdot \cos \theta \leq \dfrac{\sin^2 \theta + \cos^2 \theta}{2} = \dfrac{1}{2}$. Choose the hint you like most...


5

Basically it is convenience and history. Mind you, there are also a number of trigonometric functions that were used historically but are no longer used, e.g. versine and haversine.


5

Using various trigonometric identities, you should be able to show the left hand side is equal to $2$. The ones you will need are: $$\cos(A+B)=\cos A\cos B-\sin A\sin B,\quad \cos(-A)=\cos A,\quad \cos(45^\circ)=\sin(45^\circ)=\frac1{\sqrt{2}}$$ as well as an understanding of what $\cot A$ means. The reason you have the answer of $2.000000\ldots$ (which I ...


5

$$\begin{align} 1-\cot{\left(x\right)} &=1-\frac{\cos{\left(x\right)}}{\sin{\left(x\right)}}\\ &=\frac{\sin{\left(x\right)}-\cos{\left(x\right)}}{\sin{\left(x\right)}}\\ &=-\sqrt{2}\frac{\sin{\left(\frac{\pi}{4}-x\right)}}{\sin{\left(x\right)}}.\\ \end{align}$$ Therefore, $$\begin{align} 1-\cot{\left(\frac{\pi}{4}-x\right)} ...


5

$$\lim\limits_{x\to\infty} \frac{x - \sin(x)}{x+\sin(x)}=\lim_{x\to\infty}\frac{1-\dfrac{\sin x}x}{1+\dfrac{\sin x}x}$$ Now $|\sin x|\le1$


5

Here's a straightforward algebraic approach: $$\sin\theta = 2\sin(\theta/2)\cos(\theta/2) = 2\tan(\theta/2).\cos^2(\theta/2)$$ $$ \ \ \ \ = \frac{2\tan(\theta/2)}{\sec^2(\theta/2)} = \frac{2\tan(\theta/2)}{1 + \tan^2(\theta/2)} = \frac{2t}{1+ t^2}$$ You can construct a similar argument for $\cos\theta$.


5

Use the fact that $$\arctan{\left ( \frac{p}{q} \right )} + \arctan{\left ( \frac{q}{p} \right )} = \frac{\pi}{2} $$ So all you need to do is arrange the sum to exploit the symmetry you need, i.e., $(p,q) \mapsto (q,p)$. To do this without repeating, start on the bottom/left axes; there are $9$ non-diagonal points each, so the sum along there is $9 ...


5

If you are allowed to use complex number, this integral can be integrated by repeat application of the Euler's formula $$e^{i\theta} = \cos(\theta) + i\sin\theta$$ Up to integration constant, the integral is equal to: $$\begin{align}\int -e^{\cos t}\sin(\sin t + t) dt &= - \int e^{\cos t}\Im\left[e^{i(\sin t + t)}\right] dt ...


5

Your idea is correct. So we have $$\frac{1}{1-\cos y}+\frac{1}{1+\cos y} = \frac{1+\cos y+ 1-\cos y}{(1+\cos y)(1-\cos y)} = \frac{2}{1-\cos^2 y}.$$ What do you know about $1-\cos^2 y$? Can you take it from here?


4

Consider the limit $$\lim_{n\to \infty} \frac{E_n}{n!}\left(\frac{\pi}{2}\right)^n$$ Using $E_{n} \sim \frac{(-1)^{(n-1)/2} 4^{n+1}}{n+1}B_{n+1}$ together with $B_{2n} \sim (-1)^{n+1}4\sqrt{\pi n}\left(\frac{n}{\pi e}\right)^{2n}$ and Stirlings approximation we get $$\lim_{n\to \infty} \frac{E_{2n-1}}{(2n-1)!}\left(\frac{\pi}{2}\right)^{2n-1} = ...


4

As other have said, that is the definition of derivative of $\cos x$ at $x = \pi/4$. To see why is that, set $h = x - \pi/4$, then you get $$\lim_{h \to 0} \frac{\cos(\pi/4 + h) - \cos(\pi/4)}h$$ So if you know that the derivative of $\cos x $ is $-\sin x$, you are done. On may argue though that this is like cheating; so how to prove the limit without ...


4

Note that $$e^{\cos t} \sin(\sin(t) + t) = e^{\cos t} \left( \sin(\sin t)\cos t + \cos(\sin t)\sin t \right)$$ which is equal to $$-\left(e^{\cos t}(\cos t)'\cos(\sin t) + e^{\cos t}(\cos(\sin t))'\right)$$ Got it from here?


4

For $\sin x\ne0,$ \begin{align} \frac{\sin5x}{\sin x} &= 5\cos^4x-10\cos^2x(1-\cos^2x)+(1-\cos^2x)^2\\ &= 16\cos^4x-12\cos^2x+1\\ &= \left(4\cos^2x-\frac32\right)^2+1-\frac94 \end{align} Since anything squared is $\ge0$, we have: \begin{align} 0 &\le \left(4\cos^2x-\frac32\right)^2\\ 1-\frac94 &\le ...


4

HINT $\sin(3x)=\sin(2x+x)$ use this and expand as a compound angle formula on right hand side and get a polynomial on RHS in terms of $\sin(x)$ for $\sin(3x) $ you will get $\sin(3 x)=3 \sin(x)-4 \sin^3(x)$ then sub $x=10^\circ$ and solve the equation as polynomial you will get answer for $\sin(10^\circ)$ in irrational form.


4

Hint. You may write, as $x$ tends to $+\infty$, $$ \arctan x=\frac{\pi}{2}-\arctan \frac{1}{x} $$ and use $$ \arctan u=u-\frac{u^3}{3}+\mathcal{O}(u^4), \quad u \rightarrow 0,$$ to obtain $$ (x^2+1)\arctan x=(x^2+1)\left(\frac{\pi}{2}-\arctan \frac{1}{x}\right)=\frac{\pi x^2}{2}-x+\frac{\pi }{2}+\mathcal{O}\left(\frac{1}{x}\right) $$ then $$ \begin{align} ...


4

Using Simplifying an Arctan equation, $$\arctan \frac{a}{b}+\arctan \frac{b}{a}=\frac\pi2$$ for $\dfrac ab>0$ We have such $\dfrac{10\cdot10}2$ pairs


4

It's the magic of $$\sin\left(\frac\pi4\right)=\cos\left(\frac\pi4\right)=\frac{1}{\sqrt2}$$ $$\cos x - \sin x=\sqrt2\left[\frac{\cos x-\sin x}{\sqrt2} \right]=\sqrt2\left[\frac{1}{\sqrt2}\cos x - \frac{1}{\sqrt2}\sin x\right]$$ Now use $$\sin(a-b)=\sin a\cos b-\cos a\sin b$$ with using $b=x$ and $a=\dfrac\pi4$


4

$$z_{\frac{\theta}2} = \cos\frac{\theta}2 + i \sin \frac{\theta}2 = (1+t^2)^{-\frac12}(1+it) $$ by deMoivre's theorem $$ z_{\theta} = z_{\frac{\theta}2}^2=(1+t^2)^{-1}\left((1-t^2)+2it\right) = \cos\theta+i\sin\theta $$ hence, equating real parts: $$ \cos \theta = \frac{1-t^2}{1+t^2} $$ and imaginary parts: $$ \sin \theta = \frac{2t}{1+t^2} $$ then taking ...


4

Recall that $\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$. If $\alpha$ is equal to $\beta$ then $\alpha+\beta$ is $2\alpha$, so we have $\tan(2\alpha)=\dfrac{2\tan\alpha}{1-\tan^2\alpha}$. Apply that to the case where $\alpha=\dfrac\theta2$. You get $$ \tan\theta = \frac{2\tan\frac\theta2}{1-\tan^2\frac\theta2} = ...


4

$$a^3+b^3=(a+b)^3-3ab(a+b)$$ $$\implies\cos^62x+\sin^62x=(\cos^22x)^3+(\sin^22x)^3=1-3\cos^22x\sin^22x$$ $$=1-\frac34(2\cos2x\sin2x)^2$$ $$=1-\frac34(\sin^24x)$$ $$=1-\frac34\cdot\frac{1-\cos8x}2$$ Finally $\cos y=\cos A\implies y=2m\pi\pm A$ where $m$ is any integer


4

Samurai, this is for the second time you post problems that relate to me. First you posted this question here which is exactly similar with my rated problem on Brilliant.org. I have raised objection to mods but they can do nothing since your post doesn't violate any rules here. Okay, fine. I can accept their reason. Now you post this question which I believe ...


4

This is not an answer, but my approach suggests that the answer is $$ I := \int_{0}^{1} \arctan^{2} x \log \left( \frac{x^{2}+1}{2x^{2}} \right) \, dx = \frac{19\pi^{3}}{192} + \frac{5\pi}{16}\log^{2}2 - 6 \Im \mathrm{Li}_{3}\left( \frac{1+i}{2} \right). $$ My approach is to write $$ I = \int_{0}^{1} \arctan^{2} x \log \left( \frac{x^{2}+1}{2} \right) ...



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