$\Delta ABC$ in the figure below: $\angle 1+\angle 2=\angle 3+\angle 4,\quad$ $E\in AB,\; D\in AC,\; F=BD\cap CE,$ $BD=CE$. Prove: $AB=AC$ The exact version figure should look like: This problem should be a little more difficult than the Steiner-Lehmus Theorem.

Assuming $$\cos(36^\circ)=\frac{1}{4}+\frac{1}{4}\sqrt{5}$$ How to prove that $$\tan^2(18^\circ)\tan^2(54^\circ)$$ is a rational number? Thanks!

I’ll give a proof of the following expansion: $$\frac{\sin x}{x} = \prod_{i=1}^{\infty} \cos \frac{x}{2^i}$$ $${\sin x} = 2 \cos \frac{x}{2}\sin \frac{x}{2}$$ $${\sin x} = 2^2 \cos \frac{x}{2}\cos \frac{x}{4}\sin \frac{x}{4}$$ $$ {\sin x} = 2^3 \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8} \sin\frac{x}{8} $$ $$ {\sin x} = 2^k \cos \frac{x}{2} \cos \frac{x}{4} \cdots\cos \frac{x}{2^k} \sin\frac{x}{2^k} $$ […]

I’m trying to solve $$\int_{-\infty}^{\infty}{\frac{1}{(4+x^2)\sqrt{4+x^2}} \space dx}$$ By substituting $x=2\tan{t}$. I get as far as: $$\int_{x \space = -\infty}^{x \space = \infty}{\frac{1}{(4+(\underbrace{2\tan{t}}_{x})^2)\sqrt{4+(\underbrace{2\tan{t}}_{x})^2}} \cdot \underbrace{2(1+\tan^2t) \space dt}_{dx}} = \dots$$ $$\dots = \frac{1}{4} \cdot \int_{t \space = -\infty}^{t \space = \infty}{\frac{1}{\sqrt{1+\tan^2t}} \space dt}$$ Now what? Have I done anything wrong? I don’t see how I could continue […]

During calculations I came across the following identity: $$M+2(1-m) = \sum_{l=1}^{M-1} \frac{\cos\left(2\pi\frac{(2-m)\,l}{M}\right) – \cos\left(2\pi\frac{m\,l}{M}\right)}{2(1-\cos\left(2\pi\frac{l}{M}\right)}, \quad \forall m\in \{2,\dots, M\}, M\in \mathbb{N}$$ I cannot see why this rather complicated sum should give such a simple expression. Does anyone know trig-tricks to simplify the sum? Can one already see intuitively that the result is linear in $m$ […]

Let $u = \arctan(x)$, hence $x = \tan(u)$ for $u$ belongs in $(-\frac\pi2, \frac\pi2)$. Since $u$ belongs in $(-\frac\pi2, \frac\pi2)$, we consider $\sin(u)$ where $u$ belongs in $(-\frac\pi2, \frac\pi2)$. I used the unit circle to determine that the hypotenuse is $\sqrt{x^2 + 1}$ and got an answer $\sin(u) = \frac{x}{\sqrt{x^2 + 1}}$ when I consider […]

Once upon a time, when Wikipedia was only three-and-a-half years old and most people didn’t know what it was, the article titled functional equation gave the identity $$ \sin^2\theta+\cos^2\theta = 1 $$ as an example of a functional equation. In this edit in July 2004, my summary said “I think the example recently put here […]

Do you know any method to calculate $\cos(6^\circ)$ ? I tried lots of trigonometric equations, but not found any suitable one for this problem.

This question already has an answer here: Trigonometry problem 2 answers

Let $f$ be the function of domain $\mathbb{R}$ defined by $$f(x) = \cos x – \sin x$$ Show that $f$ had period of $2\pi$. How do you find the period of a function that has several trigonometric functions like this one? I tried converting them all to sine: $$f(x) = \sqrt{1-\sin^2x}-\sin x$$ But what do […]

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