Recent content by Nav123

\frac{\pi}{2}<2<\pi (1) 0<\pi-2<\frac{\pi}{2} so its in the first quadrant or using your logic: \cos(\pi-2)=-\cos(2) but \cos(2)<0 from 1 \cos(\pi-2)>0

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z=w(|z|-1) |z|=|w|||z|-1| |z|^2=|w|^2||z|-1|^2 |z|^2=|w|^2(|z|^2-2|z|+1) |z|^2(1-|w|^2)+2|w|^2|z|-|w|^2=0 then just solve the quadratic for |z| Drogonskis method is better and less algebra but this method requires less thinking

I=\int_{0}^{\frac{\pi}{8}}\frac{\left(\sin x-\cos x\right)\ln\left(\tan2x\right)}{\left(\sin x+\cos x\right)\sqrt{\cos4x}}dx I=\int_{0}^{\frac{\pi}{8}}\frac{\left(\sin x-\cos x\right)^2\ln\left(\tan2x\right)}{\left(\sin x+\cos x\right)\left(\sin x-\cos x\right)\sqrt{\cos4x}}dx...

Yes I guess I should have clarified that this same argument can be repeated for different orders of 0,z_1,z_2,z_3 on the circle, but I assumed by the symmetry it is quite easy to see how.

For part 3: Unfortunately circle geometry has been removed from the syllabus but here is a circle geometry proof anyways: We essentially have to prove: arg\left(\frac{1}{z_2}-\frac{1}{z_1}\right)=arg\left(\frac{1}{z_3}-\frac{1}{z_2}\right) As the two vectors must have the same direction...

Fair enough, and I agree nearer to your exam you should definitely do exams in timed conditions. I was mainly talking about when you first start off past papers (atleast for me), if I got the question myself without any outside help I improved much better than looking at the solutions, because I...

Yes and you also have to shift your perspective during an exam. Whilst in practice your sole purpose is to answer all the questions not caring too much about time, in an exam you have to be realistic in how you spend your time (you probably already know this). But many people advise you practice...

Theres a difference between you personally getting a question out yourself and the tutor doing it for you, or even guiding you. Basically try hard questions all by yourself trying every method or thought that comes to mind until you get the answer. Then think to yourself why that path worked...

Yeah standard textbook question really. Should have been either vectors or proofs.

I would expect atleast some probability as it is one of the topics that changed a lot with the new syllabus. So expect conditional probability, venn diagrams etc.

Definitely tricky I think it has something to do with squeeze theorem but not exactly sure.

Not quite sure where I missed the half, but I might not have checked it properly.

Not sure if this is still active but I'll give this a go for fun (let me know if I've made any mistakes or there is an easier way). The x-ints are x= \pm \frac{1}{2\sqrt{2}} Using the corrected identity (Thanks to vernburn for showing that there is a factor of a half): \int_{-a}^{a}...

Oh yep thats perfectly fine, unless the question specifically states use the substitution u=... in which case you have to use that substitution.