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23

Here's a friendly equilateral triangle: The sides are all of the same length - let's say $a$. The angles are all the same too, and since the angles must add up to $180^\circ$, we conclude that the three angles in the equilateral triangle are equal to $180^\circ/3=60^\circ$. Now we do something sneaky. We draw a line all the way down from the top ...


13

\begin{eqnarray} \frac{2m}{m^4+m^2+2}&=&\frac{2m}{m^4+2m^2-m^2+1+1}\\ &=&\frac{2m}{(m^2+1)^2-m^2+1}\\ &=&\frac{2m}{(m^2+m+1)(m^2-m+1)+1}\\ &=&\frac{(m^2+m+1)-(m^2-m+1)}{(m^2+m+1)(m^2-m+1)+1} \end{eqnarray} so almost done! $$ \arctan(a)-\arctan(b)=\arctan\left(\frac{a-b}{1+ab}\right) $$ \begin{eqnarray} ...


11

To see that $\sin(x) \approx x$ for small $x$ all you have to do (without using the Taylor series) is look at the graph: You can see that $\sin x = x$ when $x = 0$, and since the gradient of the graph is approximately 1 for $-0.5<x<0.5$, $\sin x$ increases approximately at the same rate as $x$ does. This leads to the result that $\sin x \approx x$ ...


9

Because if you split an equilateral triangle in half you get a 30-60-90 triangle.


8

Observe that $$ \left|f\left(\tfrac{1}{5}\right)-f\left(\tfrac{1}{10}\right)\right|\leqslant \frac{|f'(\xi)|}{10} $$ for some $\xi\in[\tfrac{1}{10},\tfrac{1}{5}]$ by the mean value theorem. And it is straightforward to check that $|f'(\xi)|\leqslant 1$.


7

Method $1$: Recall that $\sin(2x) = 2\sin(x)\cos(x)$. From what you have we have $$\sin(x) \cos(x) = \dfrac13 \implies \sin(2x) = \dfrac23$$ Since $x$ lies in the first quadrant, $2x$ lies in the first or second quadrant. Hence, we have $$2x = \arcsin\left(\dfrac23\right) \text{ or }\pi-\arcsin\left(\dfrac23\right)$$ This gives us that $$x = \dfrac12 ...


6

False if $x$ and $y$ are really arbitrary. Consider $a=b=\frac12$ and $x=0$ and $y=3\pi$: \begin{align} a\sin x+b\sin y&=\frac12 \sin(3\pi) = 0 \\ \sin(ax+by) &= \sin\left(\frac{3\pi}2\right) = -1 \end{align} It is true if you restrict to the concave region of sine (e.g. $0\le x,y\le\pi$) and $a,b\ge0$, which it is a special case of Jensen's ...


5

Both $\cos$ and $\sin$ have derivative bounded by $1$ in absolute value; therefore both satisfy a Lipschitz condition with constant $1$, and so does any $\circ$-composition of these functions.


5

You have $\sin x= x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \cdots$ and for small $x$ only the first term is significant. Similar expressions for small $x$ are $\cos x \approx 1- \dfrac{x^2}{2}$ and $\tan x \approx x$.


5

Maybe the following will help. It's not an answer, but it's too long for a comment. Suppose the given distance is $d$. Then $L$ must be simultaneously tangent to three spheres of radius $d$, one centered at each of the given points. Consider the case where the distances between some pair of the points is $D>2d$. The radius $d$ spheres can be nestled ...


4

Let me briefly explain the case of $3$ dimension: Find the plane contains those $3$ points. In this plane, find the circumcenter of the triangle formed by these $3$ points, then the straight line perpendicular to this plane and passing through the circumcenter is the required line.


4

Hint. Assume $a>0,\,x+a>0$. Integrating by parts gives $$\int \sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) dx=x\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) -\int x\left( \frac1{2x} \frac{\sqrt{ax}}{x+a}\right)dx \tag1 $$ and the last integral is easy to evaluate $$\int \frac{\sqrt{x}}{x+a}\:dx=2\int \frac{u^2}{u^2+a}\:du\quad (\sqrt{x}=u). ...


4

$$\frac{1+\sin x}{1-\sin x}-\frac{1-\sin x}{1+\sin x}=\frac{(1+\sin x)^2-(1-\sin x)^2}{(1-\sin x)(1+\sin x)}=\frac{4\sin x}{\cos^2 x}$$ Also please read this to learn how to write math expression.


4

The map: $$ \varphi: x \to \sin(\cos x) $$ is a contraction of the interval $[0,1]$. Since: $$ \sup_{x\in[0,1]} \left|\varphi'(x)\right| = -\varphi(1) = \cos(1)\sin(\sin 1)=0.7216\ldots<\frac{3}{4}\tag{1}$$ it follows that: $$ ...


3

We have $$\sum_{k} c_k\sin(\theta+\phi_k) = \sum_k\left(c_k\cos(\phi_k)\sin(\theta)+c_k\sin(\phi_k)\cos(\theta)\right) = a\sin(\theta) + b\cos(\theta)$$ where $a=\sum_k c_k\cos(\phi_k)$ and $b=\sum_k c_k\sin(\phi_k)$. The maximum of $a\sin(\theta) + b\cos(\theta)$ is $\sqrt{a^2+b^2}$. Hence, the maximum value is $$\sqrt{\left(\sum_k c_k\cos(\phi_k)\right)^2 ...


3

Hints: Fill in details $$\frac{1-\cos^2x}{\cos x}=\frac32\implies 2\cos^2x+3\cos x-2=0\implies \cos x=\begin{cases}\frac{-3+\sqrt{25}}{4}=\frac12\\{}\\\frac{-3-\sqrt{25}}{4}=-2\end{cases}$$ Second "solution" is impossible (why?), so we're left only with $$\cos x=\frac12\iff x=\pm60^\circ+360^\circ k\;,\;\;k\;\;\text{an integer}$$ What do you have then ...


3

Letting $f(x)=\left|\sin(x)\right|$, first show that $f(x+\pi)=f(x)$ for all $x$, then use that to show that for any $u$: $$\int_{u}^{u+\pi} f(x)\,dx = \int_{0}^{\pi} f(x)\,dx = \int_0^{\pi} \sin(x)\,dx$$


3

This statement follows from the theorem: If $BC$ is the hypotenuse of a right-angled triangle $\triangle ABC$, it follows that the median $AM$ (which corresponds to the hypotenuse) is $AM = \dfrac{BC}{2}$. Try to apply some basic geometry to the triangles, which are created.


3

Expanding on @labbhattacharjee HINT: Since $$ \frac{\sin(A-B)}{\sin(A+B)}=\frac{\sin(A)\cos(B)-\sin(B)\cos(A)}{\sin(A)\cos(B)+\sin(B)\cos(A)}=\frac{5}{7} \ \ \ \ (1) $$ Using Componendo & Dividendo Now, is easy to prove "Componendo & Dividendo": $$ \frac{a}{b}=\frac{c}{d} \ \Rightarrow \frac{a+b}{a-b}=\frac{c+d}{c-d} $$ Then $$ ...


3

Assuming that $x\in(0,1)$, we have: $$\sin\left(\arcsin x-\arctan\frac{2}{x}\right)= x\cdot\cos\arctan\frac{2}{x}-\sqrt{1-x^2}\cdot\sin\arctan\frac{2}{x}$$ by the sine addition formulas and the Pythagorean theorem in the form $\cos\arcsin x=\sqrt{1-x^2}$. Since: $$\cos\arctan\frac{2}{x} = \frac{x}{\sqrt{4+x^2}},\qquad ...


3

Why would someone learn the values of trigonometric functions? When I was child, we usually didn't use calculator in the elementary school but we used a table. As I was nine years old I asked my mother how did the editor fill this tables. She told me, that he might have constructed triangles and measured their sides. Of course she was wrong but her idea ...


3

$$1-(3x-4x^3)^2=1-9x^2+24x^4-16x^6$$ $$=1-x^2-8x^2(1-x^2)+16x^4(1-x^2)=(1-x^2)(1-8x^2+16x^4)$$ $$=(1-x^2)(1-4x^2)^2$$ Now $3-12x^2=3(1-4x^2)$ $\implies\dfrac{3-12x^2}{1-(3x-4x^3)^2}=\dfrac{3(1-4x^2)}{\sqrt{1-x^2}|1-4x^2|}$ Now $|1-4x^2|=+(1-4x^2)\iff1-4x^2\ge0\iff-\dfrac12\le x\le\dfrac12$ Again, $\arcsin(3x-4x^3)=3\arcsin x\iff-\dfrac\pi2\le3\arcsin ...


2

Think of the geometric interpretation of $\sin\theta$ and $\theta$ (the one using the unit circle). $\sin\theta$ is the straight-line length from $(\cos\theta,\sin\theta)$ to the $x$-axis. $\theta$ is the curved length along the circle from that point to the $x$-axis. When $\theta$ is small, we're considering a small section of the circle, and a very small ...


2

You have to learn fundas. For $\sin x $ to be positive $ 0< x < \pi$. For $\cos x $ to be positive, $ -\pi/2 < x < \pi/2.$ For other quadrants they are negative. For $ 2 x $ it is same situation. Bringing in derivative is next step in learning fundas. The derivative is negative for reducing functions, another way of saying it is that its ...


2

Consider that your rectangle drawn with perspective (using a vanishing point) forms a triangle when the sides are extended to the vanishing point. If you move your rectangle closer to the vanishing point the resulting triangle is similar to the original one. So the "rectangle" after moving it will be similar to the original. The distance from the front ...


2

It isn't equal to $8$ for every $\alpha$. For example, let $\alpha=\frac{2\pi}3$, then value of your expression is $0$.


2

We take $u=\tan\left(\frac{x}{2}\right) $, $du=\frac{1}{2}\sec^{2}\left(\frac{x}{2}\right)dx $, then we use the substitutions $\sin\left(x\right)=\frac{2u}{u^{2}+1} $, $\cos\left(x\right)=\frac{1-u^{2}}{u^{2}+1} $ and $dx=\frac{2du}{u^{2}+1} $. So we have ...


2

$$\int \frac{1}{1+\cos x+\sin x}dx=\int \frac{1}{\frac{2}{2}(1+cos x)+2\cos(x/2)\sin(x/2)}dx$$ $$=\int \frac{1}{2\cos^2(x/2)+2\cos(x/2)\sin(x/2)}dx$$ $$=\int \frac{1}{2\cos^2(x/2)(1+\tan(x/2))}dx$$ $$=\int \frac{.5\sec^2(x/2)}{1+\tan(x/2)}dx=\log(1+\tan(x/2))+C$$


2

$$\cos\dfrac x2=\pm\dfrac1{\sqrt2}\iff\cos x=2\cos^2\dfrac x2-1=0$$ $x=(2m+1)\dfrac\pi2$ where $m$ is any integer


2

Your solution of $x/2 = \pm \pi/4 + 2 \pi k$ is equivalent only to $cos(x/2) = 1/\sqrt2$. For $cos(x/2) = -1/\sqrt2$, $x/2 = \pm 3 \pi /4 + 2 \pi k$. Combining them, we get $x/2 = \pm \pi /4 + \pi k$ ,or $x/2 = \pi /4 + \pi k/2$ ,or $x = \pi /2 + \pi k$ PS: Try imagining/drawing angles in Quadrants of Cartesian coordinates if you are confused in any ...



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