STEP III

Question 1

If cos(B) = Sin(A) this implies A = (4n+1)pi/2 (+/-) B

Let B = A + x for some x in R

It follows the problem is now Cos(A+x) = Sin(A) *

Epanding using the addition formulae we see that * is transformed to;

Cos(A)Cos(x) - Sin(A)Sin(x) = Sin(A)

<->

Sin(A)(1+Sin(x))=CosA)Cos(x)

<->

tan(A) = \frac{Cos(x)}{1+Sin(x)}

Spoiler

We also note that multiples of pi can be added to the tangent functions argument without altering its value to obtain;

Tan(A) = Tan(\frac{\pi}{4} - \frac{x}{2} + n\pi) \; n \in \mathbb{Z}

<->

A = \pi(n+\frac{1}{4}) - \frac{x}{2}

and the result

A = \frac{(4n+1)\pi}{2} + B

follows - although interestingly we do not have

\pm B

wip is obviously needed to obtain this.

Bit of an odd method?

\sin{A} = \cos{B} \implies \cos{(\frac{\pi}{2} - A)} = \cos{B}

.

From looking at the cos graph, we can see that

\cos{x} = \cos{y} \implies x = y + 2k\pi

x = -y + 2k \pi

for some integer k, which gives

Unparseable latex formula:

\\ \frac{\pi}{2} - A = B + 2k \pi \implies A = (-4k+1)\frac{\pi}{2} - B \\[br]\\ \mbox{or} \\[br]\\ \frac{\pi}{2}- A = -B + 2k\pi \implies A = (-4k+1)\frac{\pi}{2} + B

which we can write as

A = (4n+1)\frac{\pi}{2} \pm B

for some integer n.

Reply 28

Dadeyemi

It might be a little neater just to do:

I = \displaystyle\int_0^1 x(x+1)^m \ \mathrm{d}x [br]\\= \int_0^1 (x+1-1)(x+1)^m \ \mathrm{d}x [br]\\= \int_0^1 (x+1)^{m+1}- \int_0^1 (x+1)^m \ \mathrm{d}x

then the results follow immediately from part (i)

Reply 29

nota bene

Adje

III/2 - with the proviso that I think I may have missed something somewhere.

Solution of Differential Equation

I don't agree with your constant after the integration

-\frac{1}{2} \ln (x^2 + a^2) + K

You say this equals

-\frac{1}{2} \ln [B(x^2 + a^2)]

with

B = - \frac{1}{2}\ln K

However, substituting that B into it gives

-\frac{1}{2}(\ln\ln(k^{-\frac{1}{2}})+\ln(x^2+a^2))

and -0.5ln(ln(1/sqrt(k))) =/ k.

I got:

Spoiler

Reply 30

Adjective

Yes, I mean K = -1/2 ln B, don't I? I was just shifting around the constant. Sorries.

Reply 31

nuodai

Indeed, but the

(2n-1)\pi

terms (and the others) correspond to

\sin 2t = 0

Reply 32

SimonM

STEP II, Question 5

Spoiler

Reply 33

SimonM

STEP III, Question 8

Spoiler

Reply 34

Oh I Really Don't Care

lulz

I just got negged for my solution to STEP III q.1.

Apparently my graphs are not up to the exacting standards that TSR requires.

II/10:

STEP I 2005 Question 10:

First part

Spoiler

ii)

Spoiler

Reply 37

∏ Squared Over 6

SimonM

STEP III, Question 3

Spoiler

In particular:

SimonM

(x^2+vx+w)^2-k = x^4+2vx^3+(v^2+2w)x^2+2vwx+w^2-k

This clearly satisfies our equation, so we're done.

I have just done this question and for the part above (in the same way as the first part) I got simultaneous equations and found the condition on a, b and c, which turned out to be the same as in the first part. I'm just wondering how you could tell that it "clearly satisfies our equation". What have I missed?

Thanks

Reply 38

SimonM