https://maths.org/step/sites/maths.org.step/files/s2s3/Trig_topic_notes.pdf

In the commentary bit at the end of the symmetry section on the first page it says sin(-a) = sin(a) etc. I assume it means sin(-a) = -sin(a).

Can somebody give the final answer for question ii from the warmup section?

I've got an answer but it doesn't look quite like what I would've expected.

Hi all,

All the STEP questions in the assignments have the question number and paper year visible.

For Foundation assignments 1-25 this is on the webpage of the module. For the STEP II and STEP III modules the question numbers are given on the assignment pdfs.

If anyone finds an incorrect or missing one please let me know!

The STEP dates for 2018 have been released, see http://www.admissionstestingservice.org/for-test-takers/step/dates-and-c....

Basically:

Feb 26th: Entries Extranet opens
Apr 27th: Standard entry closes
May 11th: Late entry closes
Jun12th: STEP I (am)
Jun18th: STEP II (am)
Jun21st: STEP III (am)

I'm on part 3 but i'm confused what i am meant to do now. I have got 8x^3 -6x-root2=0 after subbing in y as 2x but i'm not sure how I can relate this to the previous question because in ii the equation involves cos3theta but in this equation we have root 2.
Any help would be appreciated.

iii) You will not be able to separate the players if $$r>n-r+1,$$ $$2r>n+1,$$ $$r>\frac{1}{2} (n+1)$$ ? On the solutions it has $r>\frac{1}{2}(n-1)$ as the case instead so I was a bit confused, is this an error in the solutions or have I missed something?
Thanks a lot

I don't understand how you can show that $e^xe^{-x} = 1$? I understand that because of the differentiation it must be a constant, and that when $x=0 , e^x=1$, but how does this show that $e^xe^{-x} = 1$? I am possibly just being thick, or perhaps something is assumed which I thought wasn't meant to be?

I'm not sure I get some parts of the modulo function. I'll run through what I did in the first part and then go through what I don't get about the second.

Okay, so for the first part I put that xmoda = (x+k)moda, where k is any multiple of a. Later, we get N1 + N2 = a(m1 + m2) + n1 + n2.
Therefore, if we put N1 + N2 into the modulo function under a, we get the same function as if we put n1 + n2. This worked for N1N2 as well.

I'm working through Assignment 3 again and I came across the floor function. This reminded me of something I started thinking about last year: Floor (x) is often defined informally as "the integer part of" and that's fine for positive numbers and zero, but is this definition true for negative numbers? As I understand it Floor (-2.5) = -3 whereas, the way I see it at least, "the integer part of" -2.5 is -2.

So, are Floor (x) and "the integer part of" not quite the same?

Hi all,

The Mixed STEP I and II Mechanics questions and Mixed STEP I and II Statistics and Probability questions now have hints and solutions documents available.

Please let us know if you spot any errors!