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What's the fastest way to simplify \left ( 1+\cos{\frac{\pi}{8}}+i\sin{\frac{\pi}{8}}\right )^4+\left ( 1+\cos{\frac{\pi}{8}}-i\sin{\frac{\pi}{8}}\right )^4 without expanding everything/using exact ratios? Thanks!

\\$(i)if $y=\frac{1}{1+\tan{x}}$ express $\frac{dy}{dx}$ in terms of $\sin{2x}\\$(ii)By substituting $x=\frac{\pi}{4}-t$ show that $\int \frac{dx}{1+\sin{x}}=\frac{1}{2}\tan\left ( x-\frac{\pi}{4} \right )+C\\$(iii)Using the results of (i) and (ii) prove the identity...

So this was the last question of a recent topic test which I'm still stuck with. Where on earth is alpha and beta anyway? :( Thanks!

$By dividing the numerator and denominator by $\cos^2x$ prove that $\\\\\int_{0}^{\frac{\pi}{4}}\frac{dx}{9\cos^2x-\sin^2x}=\frac{1}{6}\ln2 I've tried rearranging it and using various trig identities, but I kept going around in circles and not getting it into any recognisable form. Wolfram...

I have trouble with this question \\$If $y=x^2+3x^2\ln{x}+x^3,$ find the constants $a, b$ and $c$ if\\\\ $x^2\frac{d^2y}{dx^2}+ay=bx^3+cx\frac{dy}{dx} Any help is appreciated, thanks!

I'm stuck on a problem in Cambridge Year 11 Chapter 4G Question 16 k) Prove \frac{(\sin ^{2}\alpha -\cos ^2\alpha )(1-\sin \alpha \cos \alpha )}{\cos \alpha (\sec \alpha -\csc\alpha)(\sin ^3\alpha +\cos^3\alpha )}=\sin \alpha My answer ended up being $LHS$=\sin(1-\sin\alpha\cos\alpha)...

Just found a cool article from the UBS website written the Senior Global Economist, Paul Donovan. It combines my favourite school subject AND favourite book series/movie The Hunger Games :) It's quite interesting, what do you guys think of it? Read more in this PDF here...