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Show that the roots of z6 + z3 + 1 = 0 are among the roots of z9 - 1 = 0. Hence find the roots of z6 + z3 + 1 = 0 in modulus/argument form

Alternative method: Let f(x) = 1 / (1 + sin x) f(-x) = 1 / (1 - sin x) [note: sin -x = -sin x] Therefore the integral would equal to: / pi/4 | | [1 / (1 + sin x)] + [1 / (1 - sin x)] dx | / 0 = / pi/4 | 2 | ------------ dx | 1 - sin2 x / 0 = / pi/4 | 2 | ------------ dx | cos2 x...

Correct, but reasoning/working out?

I did it the Year 11 way - ie. finding the perpendicular distance (radius) and equating it to the other radii in terms of x and y. From there, find the centre by solving the two equations simultaneously and find x and y to get a value for the radius...

I had the idea but I manipulated the y = 4/3 x incorrectly And then I ended up getting x1 as 3 and y1 as -6 T_T

Where'd you get the question? And as if post a question you don't know the solution to here...

conics2008 is the answer to the triangle and circle question pi?

Edit: Wrong solution my bad, it looked like the diamond equation P(x) is a monic polynomial of degree 4 with interger coefficients and constant term 4. One zero is root 2, another zero is rational and the sum of the zeros is positive. Factorise P(x) over real numbers.

Where did you get this question from specifically? If that method and the protractor method don't work Then I don't know

Well if you must show working Then the method that undalay suggested would be a good idea Construct LM and LK so that it is a triangle Measure their lengths If LM2 + LK2 = MK2 Then the triangle must be right angled since Pythagoras theorem applies

How can you just jump from the second last line to the last without working?

Actually I take back what I said about coming up with the proof I thought I did with co-interior angles But after some thinking It is impossible to prove algebraically without any given conditions As L can be moved around and the angle would vary Consider this Since there's no restriction on L...

Mm I looked at your profile and it says you're only 14 years old? Is this question really suitable to be in the 4U forum?

While we wait: If In = Int [tann x] dx where n >= 0, show that In = [tann-1x / (n - 1)] - In-2

Haha honestly, I don't even know myself since I didn't bring up the question =p

Yea it's a permutation question but it's a lot more complicated than just 7P4. You'd need to consider some cases because right now, you're not including solutions that have repetitions eg. AAAA, NONO, APAN

Alternatively, let everything else be 0 Anyway guys, let's stay on topic :| I pasted all the questions to the first post so it's more accessible for new posters =)

Is this even a question?

z = ±root(-4) = ±2i Lol at De Moirve's theorem method

Irrelevant