Year 12s: what will be the first way you research your future options? >>

FP3 Reduction formulae

This question is kinda annoying. I've managed to do both (a) and (b), but stuck on (c).

Any ideas on how to approach this?

Original post by P____P

This question is kinda annoying. I've managed to do both (a) and (b), but stuck on (c).

Any ideas on how to approach this?

Why not just use part a) to show they're all the same, and then I guess part b) for the second part.

(edited 8 years ago)

Reply 2

P____P

Original post by joostan

Why not just use part a)?

Yh that was my initial line of thought (on the pdf). But stuck after that point. :redface:

Reply 3

joostan

Original post by P____P

Yh that was my initial line of thought (on the pdf). But stuck after that point. :redface:

Looks good to me, what seems to be the problem?

Reply 4

P____P

Original post by joostan

Looks good to me, what seems to be the problem?

The problem seems to be that I am supposed to get a value and I have a function?

Reply 5

joostan

Original post by P____P

The problem seems to be that I am supposed to get a value and I have a function?

Sorry, the line above that with the question mark is correct, and of course

\sin(n \pi)=0

Reply 6

P____P

Original post by joostan

Sorry, the line above that with the question mark is correct, and of course

\sin(n \pi)=0

So, that means that the first integral is equivalent to the second integral, but how do I simplify that?

Reply 7

joostan

Original post by P____P

So, that means that the first integral is equivalent to the second integral, but how do I simplify that?

Well you know for each

n \in \mathbb{N} , \ I_n-I_{n-1}=0

which surely implies the result. . .

(edited 8 years ago)

Reply 8

Zacken

Original post by P____P

This question is kinda annoying. I've managed to do both (a) and (b), but stuck on (c).

Any ideas on how to approach this?

This means

\displaystyle I_n = I_{n-1} = I_{n-2} = \cdots = I_{0} = \int_0^{\pi/2} \frac{\sin x}{\sin x} \, \mathrm{d}x

- you can easily do the integral.

Reply 9

P____P

Original post by joostan

Well you know for each

n \in \mathbb{N} , \ I_n-I_{n-1}=0

which surely implies the result. . .

Original post by Zacken

This means

\displaystyle I_n = I_{n-1} = I_{n-2} = \cdots = I_{0} = \int_0^{\pi/2} \frac{\sin x}{\sin x} \, \mathrm{d}x

- you can easily do the integral.

Thanks for the help.

Much appreciated :smile:

Quick Reply

Latest

Articles for you

How to use past exam papers to revise effectively

What’s happening with GCSE, A-level and Btec exams in 2023?

What is an LLM and how could it benefit your career?

Powerful questions business students forget to ask their career advisers

FP3 Reduction formulae

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you