# Tag Info

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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