Articles of trigonometry

How do I find the angles of a triangle if I only have the lengths of the sides?

Is it possible to find the angles of a triangle if I only have its sides? If so, how can I achieve this? Regarding my knowledge of triangles: Whilst I was taught trigonometry a few years ago, I cannot for the life of me remember how to do things like use SOHCAHTOA to figure out […]

How to prove this trignometrical Identities?

The following two identities comes from my trigonometry module without any sort of proof, If $A + B + C = \pi $ then, $$\tan A + \tan B + \tan C = tan A \cdot tan B \cdot tan C$$ and, $$ \tan \frac{A}{2} \cdot \tan \frac{B}{2} + \tan \frac{B}{2} \cdot \tan \frac{C}{2} + […]

Trigonometric limit: $\lim_{x \to 0}\frac{1-\cos(ax)}{ax}=0$

In order to prove that $\displaystyle\lim_{x \to 0}\frac{1-\cos(ax)}{ax}=0$, with $a \ne 0$, I managed that $a=2$ and evaluated this limit: $$ \begin{align*} \quad \lim_{x \to 0}\frac{1-\cos(2x)}{2x}&= \lim_{x \to 0}\frac{1-(1-2\sin^2(x))}{2x}\\ &= \lim_{x \to 0}\frac{1-1+2\sin^2(x)}{2x}\\ &= \lim_{x \to 0}\frac{2\sin^2(x)}{2x}\\ &= \lim_{x \to 0}\frac{\sin^2(x)}{x}\\ &= \lim_{x \to 0} \frac{\sin(x)}{x} \cdot \sin(x)\\ &= 1 \cdot 0\\ &=0 \end{align*}$$ Can […]

Evaluating a trigonometric product $\prod_{n=1}^{\infty}\cos^2\left(\frac{1}{n^2}\right)$

I’m interested in finding a closed form for $$\prod_{n=1}^{\infty}\cos^2\left(\frac{1}{n^2}\right)$$ Wolfram Alpha confirms that it converges, but I can’t find any plausible closed forms. I’ve made some efforts to rewrite it in terms of stuff of the form $re^{i\theta}$ and make it a geometric series, but I think a more high powered solution may be needed. […]

Limit of a Rational Trigonometric Function $\lim_{x \to 0} \frac{\sin5x}{\sin4x}$

When solving a trigonometric limit such as: $$\lim_{x \to 0} \frac{\sin(5x)}{\sin(4x)}$$ we rework the equation to an equivalent for to fit the limit of sine “rule”: $$\lim_{x \to 0}\frac{\sin(x)}{x}=1$$ so, we move forward in such a manner as follows: $$=\lim_{x \to 0} \frac{\frac{5\sin(5x)}{5x}}{\frac{4\sin(4x)}{4x}}$$ $$=\frac{5}{4}\lim_{x \to 0} \frac{\frac{\sin(5x)}{5x}}{\frac{\sin(4x)}{4x}}$$ $$=\frac{5}{4}\cdot\frac{1}{1}$$ $$L=\frac{5}{4}$$ From that mindset, I am trying […]

Trigonometry : Simplify and find the value of $\tan\theta(1-\sec\frac{\theta}{2})(1-\sec\theta)(1-\sec2\theta)\dots(1-\sec2^{n-1}\theta)$ at n =1,2,3

Problem : Simplify and find the value of $\tan\theta(1-\sec\frac{\theta}{2})(1-\sec\theta)(1-\sec2\theta)\dots(1-\sec2^{n-1}\theta)$ at n =1,2,3 My approach : $\tan\theta(1-\sec\frac{\theta}{2})(1-\sec\theta)(1-\sec2\theta)…..(1-\sec2^{n-1}\theta)$ ….(i) $\tan\theta = \frac{2\tan\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}$ =$\frac{2\sin\frac{\theta}{2} \cos\frac{\theta}{2}}{\cos^2\frac{\theta}{2}- \sin^2\frac{\theta}{2}}$ Putting this value in (i) we get : $$ = \frac{2\sin\frac{\theta}{2} \cos\frac{\theta}{2}}{\cos^2\frac{\theta}{2}- \sin^2\frac{\theta}{2}}(1-\sec\frac{\theta}{2})(1-\sec\theta)(1-\sec2\theta)…..(1-\sec2^{n-1}\theta)$$ $$ =\frac{2\sin\frac{\theta}{2}}{\cos^2\frac{\theta}{2}- \sin^2\frac{\theta}{2}}(\cos\frac{\theta}{2} – \sin\frac{\theta}{2} )(1-\sec\theta)(1-\sec2\theta)…..(1-\sec2^{n-1}\theta)$$ $$ =\frac{2\sin\frac{\theta}{2}}{\cos\frac{\theta}{2} + \sin \frac{\theta}{2}}(1-\sec\theta)(1-\sec2\theta)…..(1-\sec2^{n-1}\theta)$$ Please suggest whether this is the right approach […]

Is there anyway to show $\left| {\frac{{\cos x – \cos y}}{{x – y}}} \right| \le 1$ other than taking derivatives?

The purpose is to show $\left| {\frac{{\cos x – \cos y}}{{x – y}}} \right| \le 1$ for any $x,y\in\Bbb{R}$. Taking partial derivatives with respect to $x,y$ respectively $\frac{{\partial \frac{{\cos x – \cos y}}{{x – y}}}}{{\partial x}} = 0$, $\frac{{\partial \frac{{\cos x – \cos y}}{{x – y}}}}{{\partial y}} = 0$ gives $\sin x={ – \frac{{\cos x […]

Formula for the angle of a line $y = mx$ as a function of $m$.

I was wondering if there was a way to calculate the angle made by a line $(\space y=mx)$ in the Cartesian plane using only $m$. I used the Pythagorean theorem in this figure: $$AO= \sqrt{AB^2+OB^2}=\sqrt{x^2+m^2x^2}=x \sqrt{1+m^2}$$ Now I know that $\alpha = \cos^{-1} (\cos \alpha) $. $$\cos \alpha = \frac{OB}{OA}=\frac{x}{x \sqrt{1+m^2}}=\frac{1}{\sqrt{1+m^2}}$$ $$\alpha = \cos^{-1} \left(\frac{1}{\sqrt{1+m^2}}\right)$$ […]

Trying to derive an inverse trigonometric function

I’d like to know how to derive these functions (I know the answers, I want to know how to get there) \begin{align*} f(x) &= \arcsin\left(\frac{x}{3}\right)\\ f(x) &= \arccos(2x+1)\\ f(x) &= \arctan(x^2)\\ f(x) &= \mathrm{arcsec}(x^7)\\ \end{align*} etc.

In the equation $x\cos(\theta) + y\sin(\theta) = z$ how do I solve in terms of $\theta$?

In the equation $$x\cos(\theta) + y\sin(\theta) = z,$$ how do I solve in terms of $\theta$? i.e $\theta = \dots$.