Recent content by Lith_30

\frac{b+h}{2x}\geq\frac{\sqrt{bh}}{x} assuming that x does not equal 0 \frac{b}{2x}+\frac{h}{2x}\geq\sqrt{\frac{b}{x}\frac{h}{x}} Now from the triangle diagram using trig we know \sin\left(\frac{\theta}{2}\right)=\frac{\frac{b}{2}}{x} and \cos\left(\frac{\theta}{2}\right)=\frac{h}{x}. Then we...

What day is this gonna happen? I might come if I'm free.

Blud was thrusting a different kind of sword in those goblins 💀

I will always be lurking

ok for part b) let's just ski[ the first few induction steps assume that the following statement is true (a_1+...+a_k)\left(\frac{1}{a_1}+...+\frac{1}{a_k}\right)\geq{}k^2 hence prove for n=k+1, that is (a_1+...+a_{k+1})\left(\frac{1}{a_1}+...+\frac{1}{a_{k+1}}\right)\geq{}(k+1)^2...

What did I say??? :confused2:

same ngl, back when maths was comprehensible 😢 also bruh for my solution just factor out e^{i\theta} for each of these (e^{2i\theta}+2e^{i\theta}\cos\left(\frac{\pi}{8}\right)+1). waaaay simpler

idk if there is an easier way but... the from part a you get the quadratic factors as \\z^8+1=(z^2+2z\cos\left(\frac{\pi}{8}\right)+1)(z^2-2z\cos\left(\frac{\pi}{8}\right)+1)(z^2+2z\cos\left(\frac{3\pi}{8}\right)+1)(z^2-2z\cos\left(\frac{3\pi}{8}\right)+1) now we use the substitution...

nah i'm a queen 💅

they said that z_2=\alpha{z_1} so then z_1^2+z_2^2=z_1^2+(\alpha{z_1})^2=z_1^2(1+\alpha^2)

The first equation \textbf{u}- \textbf{v}- \textbf{w}=0 means that all the vectors form a triangle, and from the given lengths we can say that this triangle is scalene. From there you can just use triangle rules to deduce what \sin(\theta) would be.

part ii) The question states that we should use the result from part i. Lets use the statement that n is divisible by 4, that is we suppose that n=4m, m\in\mathbb{Z}. Then we put that into the result from part i, 2(\sqrt{2})^{4m}\cos(m\pi)=2(\sqrt{2})^{4m}(-1)^{m} convert it back into terms of...

10/10 course combo :)👍