# Tag Info

31

Behold the power of symmetry: $$\cos(\pi - x) = - \cos x\quad\text{and} \quad \sin (\pi - x) = \sin x,$$ therefore \begin{align} \int_0^\pi \cos x \log (\sin^2 x + 1) \,dx &= \int_0^\pi \cos (\pi - u)\log (\sin^2(\pi - u) + 1)\,du\\ &= - \int_0^\pi \cos u\log (\sin^2 u + 1)\,du, \end{align} hence the integral evaluates to $0$.

13

A sine or cosine wave has just one frequency and it called a pure wave for that reason. If you have a periodic function with a different shape you can Fourier analyze it and get some number of different frequencies. These are not pure waves because of the multiple frequencies.

8

A "pure" sine (or cosine) wave is a wave of the form $y(t)=c_1\sin(\omega t+\phi)$ or $y(t)=c_2\cos(\omega t+\phi)$. Essentially, it's a wave that may have been translated, scaled, or have its period modified, but at it's core it's still a sine or cosine wave. This is contrary to sums of various waves. A core result of a field known as Fourier Analysis is ...

8

This is very interesting. Apparently the interpretations of $\arccot(x)$ vary! While your math textbook assumes $\arccot(x)$ as the inverse of $\cot(x)$ on $\left(0, \pi\right)$, Symbolab (and Wolfram Alpha and other mathematics software) use $\left(-{\pi \over 2}, {\pi \over 2}\right]-\{0\}$. The result are two different versions of $\arccot(x)$, of ...

7

Let $\theta=\pi/4$ and evaluate.

6

Much easier: any rational function of $\sin$, $\cos$ will be periodic. Your function isn't.

5

There's a simpler reason: Every rational function in $\cos x, \sin x$ has (not necessarily minimal) period $2 \pi$, but $\cos(x^2)$ is not periodic.

5

You can use the lovely trigonometrical identity $$\lvert \sin{(x+iy)} \rvert^2 = \sin^2{x}+\sinh^2{y}$$ (use the angle-addition formulae, the relationships between the trigonometric and hyperbolic functions, and the Pythagorean identity): since $\sinh^2{y}>0$ unless $y=0$, the only zeros occur when $y=0$ and $x$ is a root of $\sin$, i.e. the usual real ...

5

The last equation reads $$d_0=-2\sin^2\left(\frac x2\right)=\cos(x)-1.$$ Then eliminating $d$ from the second and the third, $$d_k=c_{k+1}-c_k,\\ c_{k+1}-2c_k+c_{k-1}=2(\cos(x)-1)c_k.$$ If we assume that $c_k=\cos(kx)$, then after some simplification, $$\cos(kx+k)-2\cos(kx)+\cos(kx-x)=2\cos(kx)\cos(x)-2\cos(kx)\\=2(\cos(x)-1)\cos(kx).$$ Hence ...

5

The equation we want to solve is $$\sin(2x)-\tan(x)$$ You deduced correctly that we now have to solve $$2\sin(x)\cos(x)-\frac{\sin(x)}{\cos(x)}=0$$ which we can rewrite to $$2\sin(x)\cos(x)=\frac{\sin(x)}{\cos(x)}$$ or $$2\sin(x)\cos(x)^2=\sin(x)$$ Now, either $\sin(x)=0$ (in which case $x\in\{-180,0,180\}$, given that $x\in[-180,180]$), or we may divide ...

4

Notice, $$\sin 2x-\tan x=0$$ $$\frac{2\tan x}{1+\tan^2x}-\tan x=0$$ $$\tan x\left(\frac{1-\tan^2 x}{1+\tan^2x}\right)=0$$ $$\color{blue}{\tan x\cos 2x=0}$$ Now, solving for $x$, $$\tan x=0\iff x=n\cdot 180^\circ$$ where $n$ is any integer For given interval $[-180^\circ, 180^\circ]$, setting $n=-1, 0, 1$, one should get $$\color{blue}{x=-180^\circ, 0, ... 4 Trasform this into \cos x = \frac{1}{\sqrt{5}}. Now, you have an equation$$e^{ix}+e^{-ix}=\frac{2}{\sqrt{5}}$$or, with z=e^{ix},$$z^2-\frac{2}{\sqrt{5}}z+1=0$$Now, above is an algebraic equation, so z, the solution of this equation, must be algebraic. By Lindemann-Weirstrass theorem, if e^{ix} is algebraic, then ix must be transcendental, ... 4 Not really. Trigonometry in its limited, familiar form is not an area of active research. You will however find trigonometry all over various different fields of mathematical research which are active, just masked under different names and forms. At this point it is mostly applications or generalizations of what you and I would call trigonometry. ... 4 Assume, all the variables are real and positive so a>0:$$\int_{0}^{a}\frac{1}{\left(a^2+x^2\right)^{\frac{3}{2}}}\space\text{d}x=$$Substitute u=\arctan\left(\frac{x}{a}\right) and \text{d}x=a\sec^2(u)\space\text{d}u; This gives a new lower bound u=\arctan\left(\frac{0}{a}\right)=0 and upper bound ... 4 You can get them all from Euler's Identity: e^{i\theta} = \cos\theta + i\sin\theta. For instance, consider e^{i(\alpha+\beta)}. I'm going to rewrite it and modify both equations separately:$$e^{i(\alpha + \beta)} = e^{i\alpha}e^{i\beta}\cos(\alpha+\beta) + i\sin(\alpha+\beta) = ( \cos\alpha + i\sin\alpha)( \cos\beta + i\sin\beta)$$... 4 The text book is correct. Note that the limit of the argument is$$\lim_{x\to 1^{\pm}}\frac{x^2+1}{x^2-1}=\pm \infty$$and that for the Principal Values of the arccotangent, we have$$\lim_{x\to \infty}\arccot(x)=0$$and$$\lim_{x\to -\infty}\arccot(x)=\pi$$4 I strongly agree with Karl's and Björn's comments regarding the Latin orgin: spatium. See: Leonhard Euler, Mechanica sive motus scientia analytice exposita, Tomus I, Petropoli, 1736 : Propositio 4 [ page 13 ] Sit spatium AM, sive sit linea recta sive curva, =s, et celeritas, quam corpus habet in M sit c, quae erit functio quaedam ipsius s. Ab ... 4 On the Cartesian x,y plane, draw a ray from the origin (0,0) through the point (-12,5). This line makes an angle \theta with the positive x-axis (a negative angle, when measured in the counterclockwise direction). What is \tan(\theta)? How long is the segment from (0,0) to (-12,5)? What is \sin(\theta)? 4 Notice, the obtuse angle B (90^\circ<B<180^\circ) lies in second quadrant hence the value of \sin B is positive & is given as follows in terms of \tan B,$$\sin B=\left|\frac{\tan B}{\sqrt{1+\tan^2 B}}\right|=\left|\frac{\frac{-5}{12}}{\sqrt{1+\left(\frac{-5}{12}\right)^2}}\right|=\frac{5}{13}$$3 Now use double angle formulae to write \cos 2A in terms of \cos A, then turn any \sin^2 A into 1-\cos^2 A 3$$\int\cos x\ln(1+\sin^2x)\ dx=\ln(1+\sin^2x)\int\cos x\ dx-\int\left(\dfrac{d\ \ln(1+\sin^2x)}{dx}\int\cos x\ dx\right)dx=\sin x\cdot\ln(1+\sin^2x)-\int\dfrac{2\sin^2x\cos x}{1+\sin^2x}dx\int\dfrac{2\sin^2x\cos x}{1+\sin^2x}dx=\int\dfrac{2(1+\sin^2x-1)\cos x}{1+\sin^2x}dx=2\int\cos x\ dx-2\int\dfrac{\cos x}{1+\sin^2x}dx$$Set \sin x=u ... 3 Let y=\arcsin\dfrac{x-2}2\implies\sin y=\dfrac{x-2}2 As -\dfrac\pi2\le y\le\dfrac\pi2,\cos y\ge0 \cos y=+\sqrt{1-\left(\dfrac{x-2}2\right)^2}=\dfrac{\sqrt{4x-x^2}}2 \tan\left(\arcsin\dfrac{x-2}2\right)=\tan(y)=\dfrac{\sin y}{\cos y}=? 3 HINT Draw a geometric representation of \alpha = \arcsin((x-2)/2), i.e. a right-angled triangle with hypotenuse 2 and one side x-2. Mark where the angle \alpha would be. Now find the length of the 3rd side using Pythagorean Theorem and compute \tan \alpha directly from the triangle. 3 Link AC, BD and denote O as their intersection point. Since PQ \bot QR ,PQ//AC,AC = 2 PQ  and QR//BD,BD = 2QR\Rightarrow AC \bot BD,AC = 6,BD =8  then the quadrilateral is divided into to triangle ABC, ACD which share the same edge AC.Then the area of the quadrilateral equals to the sum area of these two triangle. Solve the problem using ... 3 \begin{array}{rcl} \sin 2\theta+\sin 4\theta &=& \cos \theta+\cos 3\theta \\ 2\sin 3\theta \cos \theta &=& 2\cos 2\theta \cos \theta \\ \cos \theta \, (\sin 3\theta-\cos 2\theta) &=& 0 \\ \cos \theta \, (3\sin \theta-4\sin^{3} \theta+2\sin^{2} \theta-1) &=& 0 \\ \cos \theta \, (1-\sin \theta)(4\sin^{2} \theta+2\sin ... 3 When dividing fractions you can make use of the fact that:$$\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\times\frac{d}{c}$$In your case, your first equation is most likely more ... 3 Your setup is highly overloaded. What we see is a cycloid \gamma and an extra line \ell. Kinematically \gamma is the superposition of a circular motion having radius R:=|S-C| and a forward motion in direction \vec v. If \ell is parallel to \vec v the set \gamma\cap\ell is either empty, or consists of infinitely many points which repeat ... 3 This is the same as maximizing f(x,y) = 2^x + 2^y subject to the constraint g(x,y) = x^2 + y^2 = 1. Using Lagrange multipliers, we look for points where g(x,y) = 1 and \nabla f(x,y) = ((\log 2)2^x,(\log 2)2^y) and \nabla g(x,y) = (2x,2y) are linearly dependent. This is equivalent to x2^{-x} = y2^{-y}. But since the function h(t) = t2^{-t} is ... 3 We may add \sin^2(0)=0, too. In that way, one is summing 180/4=45 terms of the form \sin^2(x) where x is uniformly and rather densely distributed over the angles (because the periodicity is 180 degrees), so one basically calculates 45 times the average value of \sin^2(x). The average value of \sin^2(x) is 1/2 so the result is ... 3 So, if \sin x = \cos (\frac{\pi}{2}-x), then$$\int_{x=0}^{\pi/2} \sqrt{1 - \sin x} \, dx = \int_{x=0}^{\pi/2} \sqrt{1 - \cos (\tfrac{\pi}{2} - x)} \, dx = \int_{u=0}^{\pi/2} \sqrt{1 - \cos u} \, du, the last equality due to the substitution $u = \pi/2 - x$, $du = - dx$. Now use the same method of solution, namely \$\sqrt{1 - \cos u} = \sqrt{2} \sin ...

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