11 votes
Accepted

Why does the limit $\lim _{a \rightarrow \infty} \frac{\int_0^a \sin ^4 x d x}{a}$ not exist?

The limit does exist. A slightly more general statement is as follows: Theorem. Suppose $f : \mathbb{R} \to \mathbb{C}$ is locally integrable (i.e., integrable on any finite subinterval of $\mathbb{...
Sangchul Lee's user avatar
5 votes
Accepted

Proof that $\cos(x) = \cos\left(\frac{\pi}{2}\right)\implies x = \frac{\pi}{2} + k\pi$

$$ \begin{align*} \frac{\pi}{2}+k\pi&=\begin{cases}\frac{\pi}{2}+2n\pi &\text{if $k=2n$}\\-\frac{\pi}{2}+2n\pi&\text{if $k=2n-1$}\end{cases} \end{align*}. $$
Vincent Batens's user avatar
4 votes

Proof that $\cos(x) = \cos\left(\frac{\pi}{2}\right)\implies x = \frac{\pi}{2} + k\pi$

Maybe listing them for various $k$ will convince you: $$\color{red}{ \frac{\pi}{2}+2k\pi=\dots-\frac{7\pi}{2},-\frac{3\pi}{2},\frac{\pi}{2},\frac{5\pi}{2},\frac{9\pi}{2},\dots}$$ and $$\color{blue}{-\...
b00n heT's user avatar
  • 16.3k
3 votes

Eliminating $\theta$ from $x^2+y^2=\frac{x\cos3\theta+y\sin3\theta}{\cos^3\theta}$ and $x^2+y^2=\frac{y\cos3\theta-x\sin3\theta}{\cos^3\theta}$

Revising and extending my comment to @Maverick's answer ... (For completeness, I'll derive the result of that answer. Since I suggested the approach in a comment to the question, I don't believe this ...
Blue's user avatar
  • 75.2k
3 votes

Is it possible to adjust the sine function to pass through specific points?

With so few points, there will be many ways to do this. However you will have to decide if the results "look sinusoidal enough" for you. One way is to include higher odd harmonics one by one....
uhoh's user avatar
  • 1,949
2 votes

Probability of Rectangle in Unit Circle - Explain Solution Please

Starting with your equations 1 you can square both sides leaving you with $b^2 \le \sin^2(\theta)$, and $a^2 \le \cos^2(\theta)$. Recall that $\cos^2(\theta) + \sin^2(\theta) = 1$ So we can add $\cos^...
Greg Hughes's user avatar
2 votes

Finding the perimeter of an equilateral triangle given height

The height of the equilateral triangle splits the triangle into 2 30$^\circ$-60$^\circ$-90$^\circ$ triangles. In these special right triangles, the hypotenuse is twice the length of the short leg, ...
John's user avatar
  • 138
2 votes
Accepted

Does $\tan^{-1}\left(\frac{x}{y}\right)+\tan^{-1}\left(\frac{y}{x}\right)=\text{sgn}(xy)\frac{\pi}{2}$?

From the post you linked, we know that $$\arctan(t)+\arctan\left(\frac{1}{t}\right)=\frac{\pi}{2}\text{sgn}(t),\quad\forall t\in\mathbb{R}\setminus \{0\}$$ Substituting $t=\frac{x}{y}$ with $y\neq 0$, ...
Davide's user avatar
  • 316
2 votes

$\left( 1 - |\alpha| \right)^{2} \leq |1 - 2\alpha\cos(x) + \alpha^{2}| \leq \left( 1 + |\alpha| \right)^{2}$ with $\alpha \in \mathbb{D}_{1}$

Gary's comment resolves the upper bound. For the lower bound, the idea is similar to Mittens'; you can use Euler's formula for $e^{ix} = \cos x + i \sin x$ to write $$ \left|1 - 2\alpha \cos x + \...
Ahmad Barhoumi's user avatar
2 votes

Eliminating $\theta$ from $x^2+y^2=\frac{x\cos3\theta+y\sin3\theta}{\cos^3\theta}$ and $x^2+y^2=\frac{y\cos3\theta-x\sin3\theta}{\cos^3\theta}$

Squaring and adding gives us $$2(x^2+y^2)^2=\frac{x^2+y^2}{\cos^6\theta}\Rightarrow x^2+y^2=\frac{1}{2}\sec^6\theta$$ Putting in original equations we obtain $x\cos3\theta+y\sin3\theta-\frac{1}{2}\sec^...
Maverick's user avatar
  • 8,932
2 votes

Showing $\tan x = \frac{\sin\alpha\sin y}{1 - \sin\alpha\cos y}$, given $\sin x = \sin \alpha\sin (x + y)$

Divide $$\sin x = \sin \alpha\sin x\cos y + \sin\alpha\cos x\sin y$$ by $\cos x$ to get $$\tan x = \sin \alpha\tan x\cos y + \sin\alpha\sin y$$ and solve for $\tan x$
jjagmath's user avatar
  • 17.7k
2 votes

The triangle $ABC$ is right-angled in $A$. Prove that the inequality $(1-\sin B)(1-\sin C)\leq \frac{{(\sqrt{2}-1)}^2}{2}$ holds.

Hint : It is a right angle triangle so, $B + C = 90^{\circ}$. So, this can be re-written as $$(1-\sin B)(1-\cos B)$$ I hope you will be proceed from to get maxima at $B=45^{\circ}$ and then the final ...
Mahendra Varma's user avatar
2 votes

The triangle $ABC$ is right-angled in $A$. Prove that the inequality $(1-\sin B)(1-\sin C)\leq \frac{{(\sqrt{2}-1)}^2}{2}$ holds.

As it was pointed out in other answers and comments, this amounts to determining the maximum value of $f(x)=(1-\sin x)(1-\cos x)$ for $x \in [0, \frac{\pi}{2}]$. Since $f$ is differentiable, the ...
PierreCarre's user avatar
  • 20.9k
2 votes
Accepted

The triangle $ABC$ is right-angled in $A$. Prove that the inequality $(1-\sin B)(1-\sin C)\leq \frac{{(\sqrt{2}-1)}^2}{2}$ holds.

WLOG the hypotenuse is 1. Let the legs be $a,b$. Denote $ab = x$. $$(1-a)(1-b) = \frac{a^2b^2}{1+a+b+ab} \le \frac{a^2b^2}{1+2\sqrt {ab}+ab} = \left(\frac{x}{1+\sqrt x}\right)^2$$ Now, consider: $$\...
D S's user avatar
  • 3,576
1 vote

Solve the equation $\arcsin\bigg(\dfrac{x+1}{\sqrt{x^2+2x+2}}\bigg)-\arcsin\bigg(\dfrac{x}{\sqrt{x^2+1}}\bigg)=\dfrac{\pi}{4}$

The reason behind this situation is that $\sin(\pi/2 - \theta) = \sin(\pi/2 + \theta)$, even though $\pi/2 - \theta \neq \pi/2 + \theta$. Given this, $$\sin(\arcsin(\frac{3}{\sqrt{10}})) = \sin(\frac{\...
Mohammad 's user avatar
1 vote

Proof that $\frac{d^a}{dx^a}\sin(x) = \sin(x+\pi a/4)$ iff these two infinite series are equivalent?

Your operator is linear and homogeneous, i.e. : $$ \frac{d^a}{dx^a}f(\lambda x)=\lambda^a \frac{d^af}{dx^a}(\lambda x) $$ Therefore, your claim is equivalent to proving that $\exp$ is fixed by your ...
LPZ's user avatar
  • 2,540
1 vote
Accepted

Explanation to how the length between 2 centers is adjusted as angle changes

TL DR Yes, for any given angle you will need to adjust the x axys. Intuition Start with the two centers aligned on the x axys. If you move the center up the new distance between the two centers will ...
Marco's user avatar
  • 1,824
1 vote
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Interpretation of change in direction cosines of a variable line: Pythagorean theorem for small angles?

Without loss of generality, we take the fixed point to be the origin. Let $L$ and $L'$ be two close lines in ${\mathbb R}^3$ passing through the origin with angle $\delta\theta$ between them. Let $P=L\...
Three aggies's user avatar
  • 3,773
1 vote
Accepted

Showing $\tan x = \frac{\sin\alpha\sin y}{1 - \sin\alpha\cos y}$, given $\sin x = \sin \alpha\sin (x + y)$

You're doing it right in the middle part--you're almost there! Right after where you say "giving...", just before "Despite many attempts...", multiply out the right side, collect ...
MPW's user avatar
  • 43.4k
1 vote

The triangle $ABC$ is right-angled in $A$. Prove that the inequality $(1-\sin B)(1-\sin C)\leq \frac{{(\sqrt{2}-1)}^2}{2}$ holds.

$\sin B = \dfrac{b}{a} , \sin C = \dfrac{c}{a} $, so $ (1 - \sin B) (1 - \sin C) = \dfrac{(a - b)(a - c)}{a^2}$ We can take $a = 1$ without loss of generality because we talking about angles here not ...
of course's user avatar
  • 21k
1 vote

Is it possible to adjust the sine function to pass through specific points?

What about piecewise-defined functions? If those work for you, you can use the periodic version of interpolation by cubic splines. That would most likely satisfy the condition you want on the second ...
Lorenzo Pompili's user avatar

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