Articles of trigonometry

Integrals of the form ${\large\int}_0^\infty\operatorname{arccot}(x)\cdot\operatorname{arccot}(a\,x)\cdot\operatorname{arccot}(b\,x)\ dx$

I’m interested in integrals of the form $$I(a,b)=\int_0^\infty\operatorname{arccot}(x)\cdot\operatorname{arccot}(a\,x)\cdot\operatorname{arccot}(b\,x)\ dx,\color{#808080}{\text{ for }a>0,\,b>0}\tag1$$ It’s known$\require{action}\require{enclose}\texttip{{}^\dagger}{Gradshteyn & Ryzhik, Table of Integrals, Series, and Products, 7th edition, page 599, (4.511)}$ that $$I(a,0)=\frac{\pi^2}4\left[\ln\left(1+\frac1a\right)+\frac{\ln(1+a)}a\right].\tag2$$ Maple and Mathematica are also able to evaluate $$I(1,1)=\frac{3\pi^2}4\ln2-\frac{21}8\zeta(3).\tag3$$ Is it possible to find a general closed form for $I(a,1)$? Or, at least, for $I(2,1)$ or $I(3,1)$?

Help with this trigonometry problem

Prove the given identity. $$\left(\sin\frac {9\pi}{70}+ \sin\frac {29\pi}{70} – \sin\frac {31\pi}{70}\right) \left(\sin\frac {\pi}{70}-\sin\frac {11\pi}{70} – \sin\frac {19\pi}{70}\right) =\frac {\sqrt {5} -4}{4}$$ Please help, I could not gather enough ideas, even on how to get to the first step. I thought of using the transformation formula (from sum to product) but did not find a fruitful […]

Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle

Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle This question came up in a miscellaneous problem set I have been working on to refresh my memory on several topics I did earlier this year. I have tried changing $4\sin(A)\sin(B)\sin(C)$ to $$4\sin(B+C)\sin(A+C)\sin(A+B)$$ by making substitutions by reorganizing $A+B+C=\pi$. I then did the same thing […]

Understanding imaginary exponents

Greetings! I am trying to understand what it means to have an imaginary number in an exponent. What does $x^{i}$ where $x$ is real mean? I’ve read a few pages on this issue, and they all seem to boil down to the same thing: Any real number $x$ can be written as $e^{\ln{x}}$ (seems obvious […]

Indefinite Integral with “sin” and “cos”: $\int\frac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $

Indefinite Integral with sin/cos I can’t find a good way to integrate: $$\int\dfrac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $$

Evaluate $\cos 18^\circ$ without using the calculator

I only know $30^\circ$, $45^\circ$, $60^\circ$, $90^\circ$, $180^\circ$, $270^\circ$, and $360^\circ$ as standard angles but how can I prove that $$\cos 18^\circ=\frac{1}{4}\sqrt{10+2\sqrt{5}}$$

Verify the identity: $\tan^{-1} x +\tan^{-1} (1/x) = \pi /2$

Verify the identity: $\tan^{-1} x + \tan^{-1} (1/x) = \frac\pi 2, x > 0$ $$\alpha= \tan^{-1} x$$ $$\beta = \tan^{-1} (1/x)$$ $$\tan \alpha = x$$ $$\tan \beta = 1/x$$ $$\tan^{-1}[\tan(\alpha + \beta)]$$ $$\tan^{-1}\left [{\tan\alpha + \tan\beta\over 1 – \tan\alpha \tan\beta} \right]$$ $$\tan^{-1}\left[ {x + 1/x\over 1- x/x }\right]$$ $$\tan^{-1}\left[{x + (1/x)\over 0} \right]$$ I can’t […]

Calculating :$((\sqrt{3} + \tan (1^\circ)).((\sqrt{3} +\tan(2^\circ))…((\sqrt{3}+\tan(29^\circ))$

What is the easiest way to calculate : $$(\sqrt{3} + \tan (1^\circ)).((\sqrt{3} +\tan(2^\circ))…((\sqrt{3}+\tan(29^\circ)) $$

Calculation of $\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x} dx\;,$

Calculate the definite integral $$ I=\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\;dx $$ given that $a>b>0$ My Attempt: If we replace $x$ by $C$, then $$ I = \int_{0}^{\pi}\frac{\sin^2 C}{a^2+b^2-2ab\cos C}\;dC $$ Now we can use the Cosine Formula ($A+B+C=\pi$). Applying the formula gives $$ \begin{align} \cos C &= \frac{a^2+b^2-c^2}{2ab}\\ a^2+b^2-2ab\cos C &= c^2 \end{align} $$ From […]

Finite Series – reciprocals of sines

Find the sum of the finite series $$\sum _{k=1}^{k=89} \frac{1}{\sin(k^{\circ})\sin((k+1)^{\circ})}$$ This problem was asked in a test in my school. The answer seems to be $\dfrac{\cos1^{\circ}}{\sin^21^{\circ}}$ but I do not know how. I have tried reducing it using sum to product formulae and found out the actual value and it agrees well. Haven’t been successful […]