I could rephrase it as 'Find the area between the graphs of sin x and arcsin x between x=-1 and x=0. Then find the area between the graphs of sin x and arcsin x between x=0 and x=1. What is the sum of these two areas?'

Reply 83

RDKGames

Study Forum Helper

Original post by Palette

You mean this?

Reply 84

Palette

Original post by RDKGames

You mean this?

Yes.

Reply 85

RDKGames

Study Forum Helper

Original post by Palette

Yes.

Just rephrase it to

\int^0_{-1} arcsin(x)\ dx + \int^1_0 sin(x)\ dx

I'm unsure if C4 integrates arcsine, but it's in the formula booklet anyway so it's doable.

Update: Yes it's doable. Basic inverse trig integration is covered in C3

(edited 7 years ago)

Reply 86

Zacken

Original post by RDKGames

I'm unsure if C4 integrates arcsine, but it's in the formula booklet anyway so it's doable.

You're getting confused with differentiation and integration. The way to integrate these is to use IBP with

u = \arcsin x

and

\mathrm{d}v = 1

Reply 87

RDKGames

Study Forum Helper

Original post by Zacken

You're getting confused with differentiation and integration. The way to integrate these is to use IBP with

u = \arcsin x

and

\mathrm{d}v = 1

Yeah I wasn't sure if it's integration or differentiation that is given to you in the formula booklet, still there so students can figure it out I'm sure.

Reply 88

Palette

Proving various trigonometric identities is by far my weakest point in C3.

How can I start proving that

\frac{\sec \theta -1}{\sec \theta +1}= \tan^2 \frac{\theta}{2}

Reply 89

Zacken

Original post by Palette

Proving various trigonometric identities is by far my weakest point in C3.

How can I start proving that

\frac{\sec \theta -1}{\sec \theta +1}= \tan^2 \frac{\theta}{2}

The obvious way; convert everything to sines and cosines first off:

\frac{1 - \cos \theta}{1 + \cos \theta}

, then well, everybody should be seeing that we want

\frac{\theta}{2}

so the only sensible plan of attack is to use

\cos \theta = 1 - 2\sin^2 \frac{\theta}{2}

and

\cos \theta = 2\cos^2 \frac{\theta}{2} - 1

in the numerator and denominator respectively, for two obvious reasons:

(i) It gets

\sin

in the numerator and

\cos

in the denominator; which is the form for

\tan

.

(ii) It gets rid of the pesky

1 \pm

so the division becomes straightforward.

Anywho, that gets you

\frac{\sin^2 \frac{\theta}{2}}{\cos^2 \frac{\theta}{2}}\equiv \cdots

Reply 90

Palette

Original post by Zacken

The obvious way; convert everything to sines and cosines first off:

\frac{1 - \cos \theta}{1 + \cos \theta}

, then well, everybody should be seeing that we want

\frac{\theta}{2}

so the only sensible plan of attack is to use

\cos \theta = 1 - 2\sin^2 \frac{\theta}{2}

and

\cos \theta = 2\cos^2 \frac{\theta}{2} - 1

in the numerator and denominator respectively, for two obvious reasons:

Everybody except me I suppose. I was thinking too much along the lines of the half angle formulae when I saw

\tan^2 \frac{\theta}{2}

(i) It gets

\sin

in the numerator and

\cos

in the denominator; which is the form for

\tan

.

(ii) It gets rid of the pesky

1 \pm

so the division becomes straightforward.

Anywho, that gets you

\frac{\sin^2 \frac{\theta}{2}}{\cos^2 \frac{\theta}{2}}\equiv \cdots

Thanks for the help!

Reply 91

Palette

M3 question:
Is there a way that I can visualise

\frac{dv}{dx}

in my head? I know it means 'the rate of change of velocity with respect to displacement' but it still feels a bit strange when all the questions in mechanics so far are to do with '_____ with respect to time'.

Reply 92

Zacken

Original post by Palette

M3 question:
Is there a way that I can visualise

\frac{dv}{dx}

Just write it as

\frac{\mathrm{d}v}{\mathrm{d}x} = \frac{\mathrm{d}v}{\mathrm{d}t} \cdot \frac{\mathrm{d}t}{\mathrm{d}x} = \frac{a}{v}

Reply 93

Palette

I am currently studying the mean value theorem, a piece of calculus that is taught in schools in the US but not in the UK.

I wanted to know some results that can be derived from the mean value theorem and one of them happens to be:

If

f(a,b)\rightarrow\mathbb(R)

is differentiable and

f'(x)=0

for all

x \in (a,b)

, then

f

is constant.

I am confused by what

f(a,b)

means.

f: (a,b) \rightarrow \mathbb{R}

(edited 7 years ago)

Reply 94

B_9710

Original post by Palette

f(a,b)\rightarrow \mathbb(R)

is differentiable and

f'(x)=0

for all

x \in (a,b)

, then

f

is constant.

I am confused by

f(a,b)

It's a function of 2 variables, a and b. (I believe).

Reply 95

Zacken

Original post by Palette

I am confused by what

f(a,b)

means.

Wow, where did you get that from? That's a horrible abuse of notation. It should be: the function

f: (a, b) \to \mathbb{R}

is differentiable, etc... i.e: the domain of

f

(a, b)

and the co-domain is

\mathbb{R}

Reply 96

Palette

Original post by Zacken

Wow, where did you get that from? That's a horrible abuse of notation. It should be: the function

f: (a, b) \to \mathbb{R}

is differentiable, etc... i.e: the domain of

f

(a, b)

and the co-domain is

\mathbb{R}

The TeX messed up when I was copying and pasting it. Was in a rush so I didn't have my TeX sheet with me. I can show you the source- it looks fine there:

http://math.stackexchange.com/questions/9749/applications-of-the-mean-value-theorem

It must have been TSR as I copied directly from the TeX source in the link above. Even had I spotted it, I wouldn't have known about the distinction between

f: (a,b)

and

f(a,b)

as I'd have thought that they were as interchangeable as

f(x)

and

f: x

.

Please forgive my numerous mathematical heresies over the past year.

(edited 7 years ago)

Reply 97

Zacken

Original post by Palette

I'd have thought that they were as interchangeable as

f(x)

and

f: x

Those aren't interchangeable.

Reply 98

Palette

Original post by Zacken

Those aren't interchangeable.

My C3 textbook states:

You can write functions in two different ways:

f(x)=

and

f:x \rightarrow

.

Is the textbook misleading or am I making some classic mistake in analysis?

P.S. I now understand why the TeX went wrong. I now remember that there were initially spacing issues with the TeX when I first copied it so I had to retype parts. Hence \mathbb{R} was accidentally typed as \mathbb(R) and the colon was left out in f: (a, b) as I was rushing it. I think that explains it!

(edited 7 years ago)

Reply 99

B_9710

Original post by Palette

My C3 textbook states:

You can write functions in two different ways:

f(x)=

and

f:x \rightarrow

You can write a function like this

\displaystyle f : \mathbb{R} \rightarrow \mathbb{R}

\displaystyle x\mapsto x+4

.
As you can see the function is defined completely, you have the domain, codomain and the mapping.
You won't see it at A-level but it can be quite important to have all these terms defined for a function.
For example if we have