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10

Start by linearizing $\sin^{4}(x)$ : $$ \forall x \in \mathbb{R}, \, \sin^{4}(x) = \frac{1}{8}\Big( \cos(4x) - 4\cos(2x) + 3 \Big). $$ This leads to : $$ \sum_{j=1}^{m} \sin^{4}\Big( \frac{j\pi}{2m} \Big) = \frac{1}{8}\Bigg( \sum_{j=1}^{m} \cos\Big( \frac{2j\pi}{m} \Big) - 4 \sum_{j=1}^{m} \cos\Big( \frac{j\pi}{m} \Big) + 3m \Bigg). $$ The sums ...


7

You can't take the $x$ out of the first integral on the right side because $x$ is the integration variable and is not a constant. You would have to handle it differently like this: Let $u = 1+x^2$, and can you continue to the finish?


7

Use $$\tan{(3x)}=\tan{x}\tan{(60^{\circ}+x)}\tan{(60^{\circ}-x)}$$ and let $x=20^\circ$.


6

Suppose your parabola is given by $$y=a x^2+b x + c$$ Its derivative is $y'=2 a x + b$. A vector perpendicular to your parabola at the point $(x,y)$ is thus $$(-2 a x - b,1)$$ Let us normalize this vector $$\vec{n}=\frac{(-2 a x - b,1)}{\sqrt{(2 a x + b)^2+1}}$$ The curves at distance $d$ from your parabola (green in your figure) are thus $$(x,a x^2+b x + ...


6

Just so this question can be answered: You originally did the integration by parts incorrectly: $u=x,\ \ \ \ \ dv=\frac{1}{1+x^2}dx$ $du=dx,\ \ \ \ \ v=\int\frac{1}{1+x^2}dx$ $$\int\frac{x}{1+x^2}dx=x\int\frac{1}{1+x^2}dx-\int\left(\int\frac{1}{1+x^2}dx \right)dx$$


5

The first thing you are going to want is the cosine function. If $y=s(x)$ is the solution to $y^{\prime\prime}(x)+y(x)=0$, $y(0)=0$, $y^{\prime}(x)=1$, then let $c(x)=y^{\prime}(x)$. Then $$c^{\prime\prime}(x)+c(x)=s^{\prime\prime\prime}(x)+s^{\prime}(x)=\frac d{dx}(s^{\prime\prime}(x)+s(x))=0$$ So $c(x)$ is a solution to the same differential equation with ...


5

Your integral is related with the Fourier cosine transform of $\frac{1}{\sqrt{1+x}}$, that can be computed through Fresnel integrals, but is not an elementary function. For instance: $$\begin{eqnarray*} \int_{1}^{M}\cos(t\sqrt{x^2-1})\,dx &=& \frac{1}{2}\int_{0}^{\sqrt{M^2-1}}\frac{\cos(tu)}{\sqrt{u+1}}\,du\\ &=& ...


4

Suppose you have an $u>1$ such that both $u^n+u^{-n}$ and $u^{n+1} + u^{-n-1}$ are rational. Then $u^n$ satisfies the equation with rational coefficients $(T - u^n)(T-u^{-n}) = T^2 - (u^n + u^{-n})T + 1 = 0$, so $u^n$ is algebraic and its only possible conjugate (over $\Bbb Q$) is $u^{-n}$. Similarly, $u^{n+1}$ is algebraic and its only possible ...


4

Notice that the term $1/x^2$ is the negative derivative of the function's argument $1/x$. Set $u=1/x$, $du = -1/x^2$. $-du=1/x^2 dx$. The new integral is of $-\csc^2(u)du$. Integrate that to get $cot(u) + C$, and substitute $u$ back into the expression to get $\cot(1/x) + C$.


4

This writes as $\;\cos2\theta=\dfrac{\sqrt 3}2=\cos\dfrac\pi6$. Now you have to know that $\cos x=\cos\alpha\iff\theta\equiv\pm\alpha\pmod2\pi$, and similarly: $\sin x=\sin \alpha\iff\begin{cases}\theta\equiv\alpha\pmod{2\pi},\\\theta\equiv\pi-\alpha\pmod{2\pi}, \end{cases}$ $\tan x=\tan\alpha\iff x\equiv\alpha\pmod\pi.$ So here, you have ...


4

HINT: Use Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $ and $\sin(A+B)=\sin(\pi-C)=?$ and Sine Law


3

Hint: As $\cot x\ne0,$ multiply both sides by $\tan^2x,$ $$1-\tan x -\tan^2x=0\iff\tan x=1-\tan^2x\iff\tan2x=2$$


3

$\cot x \cdot (\tan x - \sin x \cdot \cos x)$ = $\frac{1}{\tan x} \cdot (\tan x - \sin x \cdot \cos x)$ = $(1 - \sin x \cdot \cos x \cdot \frac{\cos x}{\sin x})$ = $1 - \cos ^2x$ = $\sin ^2x$


3

With $\zeta=e^{2\pi\mathrm i/p}$ as Gerry suggested, \begin{align} \sum_{l=0}^{\left\lfloor\frac mp\right\rfloor}\binom m{lp} &=\sum_{j=0}^m\binom mj\frac1p\sum_{k=0}^{p-1}\zeta^{jk}\\ &=\frac1p\sum_{k=0}^{p-1}\sum_{j=0}^m\binom mj\zeta^{jk}\\ &=\frac1p\sum_{k=0}^{p-1}\left(1+\xi^k\right)^m\\ ...


3

Set: $a=\cos \theta$, $b=\sin\theta$. Then: $(a-b)^2 + (a+b)^2 = a^2 -2ab + b^2 + a^2 + 2ab + b^2 = 2a^2 + 2b^2 = 2(a^2 + b^2)=2(\sin\theta^2 + \cos\theta^2)=2.$


3

If $x$ is both inside and outside of trigonometric functions, don't look for an explicit solution in ̲g̲e̲n̲e̲r̲a̲l̲ ̲c̲a̲s̲e̲. You should use numerical methods in such cases. If you just look for the answer and the solution method is not important to you, here is the wolfram link showing that there is a root $x\approx5.2$ in this interval. First note ...


3

They require you to use the identities: $$\sin(\alpha\pm \beta)=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta$$ $$\cos (\alpha\pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta$$ From these one can infer the double-angle formulae: $$\sin 2\alpha = 2\sin \alpha\cos \alpha$$ $$\cos 2\alpha = \cos^2 \alpha -\sin^2 \alpha = 2\cos^2 \alpha -1 ...


3

Better to ask what $-x$ might mean itself in my opinion. An angle of $-x$ can be viewed as a clockwise rotation whereas $x$ is anticlockwise. Now in a unit circle you can record the $\cos$ of an angle as the horizontal distance on the $x$ axis and so intuitively the right triangle formed by rotating clockwise will have the same horizontal distance as one ...


3

hint: Square both sides: $A^2(1-\sin^2\theta) =A^2\cos^2 \theta = B^2 + 2B\sin \theta + \sin^2 \theta$, and you simplify this to get a quadratic equation in $\sin \theta$. Can you finish it ?


3

Since: $$ \cos(x)=\prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right) $$ we have: $$ \cosh(x)=\prod_{n\geq 0}\left(1+\frac{4x^2}{(2n+1)^2\pi^2}\right)\leq \exp\left(x^2\sum_{n\geq 0}\frac{4}{(2n+1)^2 \pi^2}\right)=e^{x^2/2}. $$ In a similar fashion, from: $$ \sinh(x) = x\prod_{n\geq 1}\left(1+\frac{x^2}{n^2 \pi^2}\right) $$ we get: $$ \frac{\sinh ...


2

You have $$A\cos\theta-\sin\theta=B$$ Now write the left hand side as $$R\cos(\theta+\alpha)=R\cos\theta\cos\alpha-R\sin\theta\sin\alpha$$ Therefore $$R\cos\alpha=A$$ and $$R\sin\alpha=1$$ Hence $$R=\sqrt{A^2+1}$$ and $$\tan\alpha=\frac 1A$$ Can you solve it now?


2

Let $\triangle{ABC}$ be the triangle with $AB=7,BC=8,CA=9$, and let $D$ be the apex. Also, let $R$ be the circumradius of the triangle, and let $O$ be the circumcenter. Let $O'$ be the point on the plane on which the triangle exists such that $DO'$ is perpendicular to the plane. Then, having that $\triangle{DO'A}\cong\triangle{DO'B}\cong\triangle{DO'C}$ ...


2

Let $\theta=\sin^{-1}x$. Then, $$x=\sin\theta$$ $$x=\cos\left(\frac{\pi}2-\theta\right)$$ $$\cos^{-1}x=\cos^{-1}\left(\cos\left(\frac{\pi}2-\theta\right)\right)$$ Since $\theta=\sin^{-1}x$, we have $\theta\in\left(\frac{\pi}2,\frac{3\pi}2\right)$. $$\frac{\pi}2-\theta\in(-\pi,0)$$ ...


2

As others have noted, this is really just a deformation or change-of-coordinates problem. Our ellipse is defined in a rectangular cartesian coordinate system, and we want to map it to another system where coordinates are distance along the parabola and normal distance away from the parabola. These kinds of deformations are common in high-end CAD systems. ...


2

The map $$s\mapsto\left\{\eqalign{x_0(s)&:={\rm arsinh}\, s \cr y_0(s)&:=1-\sqrt{1+s^2}\cr}\right.\qquad(-\infty<s<\infty)$$ maps the $s$-axis isometrically onto the catenary $$\gamma:\qquad y=1-\cosh x=-{1\over2}x^2-{1\over24}x^4+?x^6\ ,$$ which is in the intended range a very good approximation to the parabola $y=-{1\over2}x^2$. One computes ...


2

Yes it is. You have just to sum $\pm a$ in teh argument of the upper cosine $$f(x)=\frac{\cos(x-a+a)}{\cos(x-a)}$$ and use the expression for the sum of angles $$f(x)=\frac{\cos(x-a)\cos(a)-\sin(x-a)\sin(a)}{\cos(x-a)}=\cos(a)-\tan(x-a)\sin(a)$$


2

It is the case $m=4$ of the general formula: $$ \prod_{k=1}^m\tan\left(\frac{k\pi}{2m+1} \right)=\sqrt{2m+1} $$ see: https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Identities_without_variables


2

For $x<0$ set $x=\sin z$. Then you have $$\arcsin(\sqrt{1-x^2})=\arcsin(\cos z)= \frac{1}{2}\pi - \arccos(\cos z)$$ $$\arccos(\sin z)= \frac{1}{2}\pi - \arccos(\cos z)$$ $$\arcsin x= \arcsin (\sin z)= z$$ $$\mathrm{arccot}(\frac{\sqrt{1-x^2}}{x})=\mathrm{arccot}(\frac{\cos z }{\sin z}) =\mathrm{arccot}(\cot z) = \pi + z$$ (Note: The last expression would ...


2

hint: $(a-b)^2 = a^2 - 2ab+b^2, (a+b)^2 = a^2 + 2ab+b^2$, and $\sin^2 \theta + \cos^2 \theta = 1$. Use the above formulas with $a = \sin \theta, b = \cos \theta $.


2

The radius of the circle you've constructed is close to $3/4$ of the length of the starting segments, but not exactly. For ease of calculation let's assume the starting segments are $2$ units long. Apply Pythagoras' theorem to find that the yellow segment has length $\sqrt 3$ and the blue segment has length $b:=\frac12(\sqrt{15}-\sqrt 3)$. The ratio of these ...



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