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.

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? $f(x)=\sin(2x)+2\sin(x)$

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 […]

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) […]

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}}} […]

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: $$\frac{\sin(u+\Delta u)-\sin(u)}{\Delta u}<\cos(u)<\frac{\sin(u-\Delta u)-\sin(u)}{-\Delta u}$$ […]

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

I am having trouble grasping some of the concepts regarding the unit circle. I think I have the basics down but I do not have an intuitive sense of what is going on. Is memorizing the radian measurements and their corresponding points the only way to master this? What are some ways one can memorize […]

Calculate: $$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$$ I don’t know how to use L’Hôpital’s Rule. I tried to make $\tan x =\frac{\sin x}{\cos x}$ for the term ${\sqrt{\tan x}}$.

The question is to find $x\in\left(0,\frac{\pi}{2}\right)$: $$\frac {\sqrt3 – 1}{\sin x} + \frac {\sqrt3 + 1}{\cos x} = 4\sqrt2 $$ What I did was to take the $\cos x$ fraction to the right and try to simplify ; But it looked very messy and trying to write $\sin x$ in terms of $\cos x$ didn’t […]

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