Articles of trigonometry

To show inverse of tan x

It quite confuses me. Where do I start? Please help.

Find the coordinates in an isosceles triangle

Given: $A = (0,0)$ $B = (0,-10)$ $AB = AC$ Using the angle between $AB$ and $AC$, how are the coordinates at C calculated?

Find the minimum of $\displaystyle \frac{1}{\sin^2(\angle A)} + \frac{1}{\sin^2(\angle B)} + \frac{1}{\sin^2(\angle C)}$

Is it possible to find the minimum value of $E$ where $$E = \frac{1}{\sin^2(\angle A)} + \frac{1}{\sin^2(\angle B)} + \frac{1}{\sin^2(\angle C)}$$for any $\triangle ABC$. I’ve got the feeling that $\min(E) = 4$ and that the critical value occurs when $ABC$ is equilateral.

How do you find the height of a triangle given $3$ angles and the base side? Image given.

This question has me absolutely stumped. This is the image of the question, how can I work out $x$? I’ve been doing a variety of attempts but I just cant get it.

Maximum of a trigonometric function without derivatives

I know that I can find the maximum of this function by using derivatives but is there an other way of finding the maximum that does not involve derivatives? Maybe use a well-known inequality or identity? $f(x)=\sin(2x)+2\sin(x)$

Limit $\lim_{x\to 0} \frac{\tan ^3 x – \sin ^3 x}{x^5}$ without l'Hôpital's rule.

I need to solve $$\lim_{x\to 0} \dfrac{\tan ^3 x – \sin ^3 x}{x^5}$$ I did like this: $\lim \limits_{x\to 0} \dfrac{\tan ^3 x – \sin ^3 x}{x^5} = \lim \limits_{x\to 0} \dfrac{\tan ^3 x}{x^5} – \dfrac{\sin ^3 x}{x^5}$ $=\dfrac 1{x^2} – \dfrac 1{x^2} =0$ But it’s wrong. Where I have gone wrong and how to […]

Why is this Definite Integral wrong?

This is the main problem $$\int_{0}^{\pi} \frac{\sin{x}}{1+\cos^{2}{x}}\,dx$$ and if I let $u=\cos{x}$ then $$-\int_{1}^{-1}\frac{du}{1+u^{2}}$$ which equals $-\tan^{-1}{u}$ from $1$ to $-1$ and where I make the mistake that I don’t understand is $$-\tan^{-1}(-1) = -\frac{3\pi}{4}$$ but this gives me the wrong answer, In order to get the correct one I’m supposed to get that $$-\tan^{-1}(-1) […]

What is the 90th derivative of $\cos(x^5)$ where x = 0?

Trying to figure out how to calculate the 90th derivative of $\cos(x^5)$ evaluated at 0. This is what I tried, but I guess I must have done something wrong or am not understanding something fundamental: $\cos(x) = \displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n)!} {(x)}^{2n}$ $\cos(x^5) = \displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n)!} {(x^5)}^{2n}=$ $1-\dfrac{{x^{5\cdot2}}}{2!}+\dfrac{{x^{5\cdot4}}}{4!}-\dfrac{{x^{5\cdot6}}}{6!}+…-\dfrac{{x^{5\cdot18}}}{18!}+…-\dfrac{{x^{5\cdot90}}}{90!}+…+\dfrac{{x^{5\cdot190}}}{180!}+…$ $f(x) = \displaystyle\sum_{n=0}^\infty \dfrac{f^{(n)}(0)}{n!} \cdot {x^{n}} $ $\dfrac{f^{90}(0)}{90!}\cdot {{x^{90}}} […]

Make this visual derivative of sine more rigorous

Is this the correct way to make this visualization of the derivative of sine more… rigorous? At least, for $u\in(0,\pi/2)$. Borrowed from Proofs without words. To try to make this rigorous, I argued that when $u\pm\Delta u$ is in the first quadrant, that we have the following geometrically obtained bounds: $$\frac{\sin(u+\Delta u)-\sin(u)}{\Delta u}<\cos(u)<\frac{\sin(u-\Delta u)-\sin(u)}{-\Delta u}$$ […]

How many solutions does $\cos x + \cos (2x) + \cos (3x) +\cos (4x) = -\frac{1}{2}$ have in $$?

I simplify it to $\cos x(\cos (2x)+\cos (3x))=-\frac{1}{4}$ but don’t know how to go further.