Hot answers tagged

7

$\sin(x)$ is an odd function. $\sin(-x) = -\sin(x)$ for all $x$. It is $\cos$ that is even.


6

Your form, $\sqrt{\frac{2-\sqrt3}{4}}$, is already close: $$\begin{align*} \sin15^\circ &=\sqrt{\frac{2-\sqrt3}{4}}\\ &= \sqrt{\frac{4-2\sqrt3}{8}}\\ &= \sqrt{\frac{1-2\sqrt3+3}{8}}\\ &= \sqrt{\frac{(1-\sqrt3)^2}{8}}\\ &= \frac{\sqrt3 -1 }{2\sqrt2} \end{align*}$$


6

By definition, $\sin tA=\frac1{2i}(e^{itA}-e^{-itA})$. So you need only to show that $$t\mapsto e^{itA},\quad t\in\Bbb R$$ is continuous. (For the $e^{-itA}$ part, just note that it is the inverse of $e^{itA}$ and that taking inverse is a continuous map in the invertible matrix space. [Or rather, we don't even need to bother with taking inverse, thanks to ...


5

Let me show the first identity. By the double-angle identity: \begin{align} & \sin^2(10^\circ) - \sin^2(20^\circ) - \sin^2(40^\circ) \\ = & \frac{1 - \cos(20^\circ)}{2} - \frac{1 - \cos(40^\circ)}{2} - \frac{1 - \cos(80^\circ)}{2} \\ = & -\frac{1}{2} - \frac{1}{2}(\cos(20^\circ) - \cos(40^\circ) - \cos(80^\circ)).\\ \end{align} So to show the ...


5

Looking at derivatives of the functions $\sin(\alpha t) e^{-\lambda t}$ and $\cos(\alpha t) e^{-\lambda t}$, it's reasonable to try for an antiderivative of the form $$ F(t) = (A \sin(\alpha t) + B \cos(\alpha t)) e^{-\lambda t} $$ where $A$ and $B$ are constants. Differentiating this, $$ F'(t) = ((-A \lambda - B \alpha) \sin(\alpha t) + (A \alpha - B ...


5

Recall that $$\cos 3x=4\cos^3 x- 3\cos x$$ $$\implies 2(4\cos^3 12^{\circ}-3\cos 12^{\circ})=2\cos36^\circ$$ Using $$\cos 36^\circ=\frac{1+\sqrt{5}}4$$ We get the that $$8\cos^3 12^{\circ}-6\cos 12^{\circ}=\frac{1+\sqrt{5}}2=\phi$$


5

Hint The general term can be written as $$\tan^{-1}\frac{1}{r^2+3/4}$$ $$=\tan^{-1}\frac{r+1/2-(r-1/2)}{(r-1/2)(r+1/2)+1}$$ $$=\tan^{-1}(r+1/2)-\tan^{-1}(r-1/2)$$


5

This is because $\lim_{x\to a}f(x)=b$ is equivalent to $\lim_{n\to\infty}f(x_n)=b$ for every sequence $(x_n)_{n\in\Bbb N}$ with $x_n\to a$.


5

A neat way to think of this is by noticing that $$ \cos(2a) = 1 - 2\sin^2(a) $$ Hence, $$ \sin^2(a) = \frac{1 - \cos(2a)}{2} $$ So, it's the graph of $\cos(a)$ flipped, "sped up" by a factor of 2, raised up by 1 unit above the $y$-axis, and then finally shrunk by a factor of $2$ along the $y$-axis. WolframAlpha plot for reference


4

Your list seems to be ok. Although I would rather take the series definition $e^x = \sum\limits_{n=0}^\infty \frac {x^n}{n!} $. This gives you a definition for exponentials of complex numbers right from the beginning. Your limit definition for Euler's number follows easily. But of course your way works fine too. As for $\pi$, a "standard" definition is ...


4

Explicitly, with the choice $$u = \cos \alpha t, \quad du = -\alpha \sin \alpha t \, dt, \\ dv = e^{-\lambda t}, \quad v = -\lambda^{-1} e^{-\lambda t},$$ the first integration by parts gives $$I = \int e^{-\lambda t} \cos \alpha t \, dt = -\frac{1}{\lambda} e^{-\lambda t} \cos \alpha t - \frac{\alpha}{\lambda} \int e^{-\lambda t} \sin \alpha t.$$ Then, the ...


4

$$ \underbrace{\frac {2\tan\alpha}{1-\tan^2\alpha} = \tan(2\alpha)}_\text{double-angle formula} = \underbrace{\frac{4n^2}{4n^4-1} = \frac{2\left( \dfrac 1 {2n^2} \right)}{1 - \left(\dfrac 1{2n^2}\right)^2}}_{\begin{smallmatrix} \text{This gets us a “1'' where we} \\ \text{need it in the denominator.} \end{smallmatrix}}. $$ We have ...


4

The "arc" refers to an arc of a unit circle. Thus in the following picture, $\arcsin(y)$ is the length of the blue arc.


4

Let $DB$ intersect $FG$ at point $K$. Note that since $F$ is the center of the square, $K$ is the midpoint of $FG$. Also, since $FI:HK=1:\sqrt{3}$, this implies that $FK:JK=1:\sqrt{3}$. Thus $JK=\frac{\sqrt{3}}{1+\sqrt{3}} \times \frac{FG}{2}$. Thus $b:c=CI:JG=\frac{1}{2}:\frac{\sqrt{3}}{1+\sqrt{3}} \times \frac{1}{2}+\frac{1}{2}$, which is not the ...


4

Hint $$\lim_{x\to \pi}\frac{\cos(x)-1}{x-\pi}=\lim_{x\to \pi}\frac{\cos(x)-\cos(\pi)}{x-\pi}=...$$


3

You have $$\tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}=\frac{4n^2}{4n^4-1}$$ Rearranging this gives $$\tan^2\alpha+\tan\alpha\left(2n^2-\frac{1}{2n^2}\right)-1=0$$ $$\Rightarrow\left(\tan\alpha+2n^2\right)\left(\tan\alpha-\frac{1}{2n^2}\right)=0$$ Hence $$\tan\alpha=-2n^2 \text{or} \frac{1}{2n^2}$$


3

The definition of $\pi$ is based on the route you take to define the circular functions $\sin x, \cos x$. The traditional approach based on the circle (that's why the name circular functions) is rigorous/intuitive/fruitful/easy. Many believe that the geometric definition based on circle is not rigorous, but I have shown in this answer that it is a fully ...


3

Thanks to @N.F.Taussig for pointing out some errors. Suppose you have a regular pentagon, with vertices $\{A,B,C,D,E\}$ (labelled cyclically so $A$ is neighbor to $B$ and $E$). Let us take each side length to be $1$. Let $X$ be the length of a diagonal, say $AC$. Let's first compute $X$. To do it, let $P$ be the intersection of $AC$ and $BE$. Now we ...


3

I'm going to start with the fact that $\sin x=\frac 1 2$ since you told me you knew that in the comments. Once you have that, using a calculator, you should get $\sin^{-1} \frac 1 2=30^\circ$. Now, you know that $\sin(180^\circ-x)=\sin x$, so you also have the answer $180^\circ-30^\circ=150^\circ$. Again, no graphing required, but you still have to use an ...


3

$$\sum_{cyc}\|z_1-z_2\|^2 = 3\left(\|z_1\|^2+\|z_2\|^2+\|z_3\|^2\right)-\|z_1+z_2+z_3\|^3 \leq \color{red}{87}$$ hence we just need to prove that there is some triangle $ABC$ with centroid $G$ such that $AG=2,BG=3,CG=4$ to prove that the previous inequality is tight. But $AG^2=4,BG^2=9,CG^2=16$ give the squared lengths of the medians, then the squared ...


3

For the first one, notice that $\cos\left(\frac{1}{x}\right)\sin^4(x)$ is bounded and $\lim_{x\to0}x^2=0$. For the second one, use the product rule and you can use Taylor series (or l'Hôpital's rule): ...


3

I'm not sure that this falls into the category of "intuitive explanation" but the general phenomenon you are considering is a consequence of the commutativity of composition of continuous functions and limits. If you let $f(x) = \frac{\sin x}x$, then for any continuous function $g$ with $g(0)=0$ we have $\lim_{x \rightarrow 0} f \circ g(x) =1$.


3

Using the sine of the difference you can see that $$ \sin(-x) = \sin (0-x) = \sin 0 \cos x - \cos 0 \sin x = -\sin x $$ so sine is indeed odd. Using a similar technique, you can show cosine is even as well.


3

Completing a $3$-by-$3$ grid of squares, and assigning their diagonals length $1$, gives this: Here, we see the classical geometric mean construction that tells us $$\frac{|\overline{PA}|}{|\overline{PC}|} = \frac{|\overline{PC}|}{|\overline{PA^\prime}|} \qquad\to\qquad \frac{a}{1} = \frac{1}{a+1} \tag{$\star$}$$ This relation says exactly that $a = ...


3

Use Cauchy-Schwarz inequality: $$\left(a+b\sqrt{2}\sin{x}+c\sin{2x}\right)^2\le (a^2+b^2+c^2)(1+2\sin^2x+\sin^22x)$$ $$\left(a+b\sqrt{2}\sin{x}+c\sin{2x}\right)^2\le 100\cdot(1+2\sin^2x+\sin^22x)$$ $$|a+b\sqrt{2}\sin{x}+c\sin{2x}|\le 10\cdot\sqrt{1+2\sin^2x+\sin^22x}$$ $$1\le1+2\sin^2x+\sin^22x\le \frac{13}{4}$$ Then $$-5\sqrt{13}\le ...


3

Use $$\sin\left(x\right)=\cos\left(\frac{\pi}{2}-x\right)$$ $$\cos (\pi+x)=-\cos x$$ The result follows.


3

Slightly more general answer. Let $R$ and $B$ be the lengths of the red and blue lines respectively. If the radius of the circles is $r$, then we have the equations $$R=2r$$ since $R$ is the diameter of one of the circles, and $$B+r=\sqrt{r^2+(2r)^2}=r\sqrt5$$ since $B+r$ is the hypotenuse of a right triangle with legs of length $r$ and $2r$. Hence ...


3

With $A$ the origin, rotate everything so that $AG$ lies on the $x$-axis. Drop a perpendicular from $H$ to meet the $x$-axis at $J$. Extend segment $CE$ to meet segment $HJ$ at point $K$. The key to the proof is the fact that that angle $\angle HEK$ has measure $18^\circ$; this can be deduced after examining the angles in the regular pentagon. Picture (not ...


2

Series method: $$ E_n:=\int_0^\infty t^n e^{-\lambda t}dt=n!\lambda^{-n-1} $$ and $$ \sum_{n=0}^\infty \frac{(-1)^n a^{2n}t^{2n}}{(2n)!} = \cos(at) $$ so $$ \int_0^\infty \cos(at)e^{-\lambda t}dt = \sum_{n=0}^\infty \frac{(-1)^na^{2n}E_{2n}}{(2n)!} =\sum_{n=0}^\infty (-1)^n a^{2n} \lambda^{-1-2n} =\frac{\lambda}{a^2+\lambda^2} $$ But of course integration ...


2

You surely know that $z=\cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5}$ is the root in the first quadrant of $z^5-1=0$ so it satisfies $$ z^4+z^3+z^2+z+1=0 $$ that can also be written as $$ z^2+\frac{1}{z^2}+z+\frac{1}{z}+1=0 $$ or, by noting that $z^2+1/z^2=(z+1/z)^2-2$, $$ \left(z+\frac{1}{z}\right)^2+\left(z+\frac{1}{z}\right)-1=0 $$ Therefore, $$ ...



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