Trig: r cos (x-alpha) question

Spitfir3 · 28 Dec 2011 at 20:29

Hey guys,

I'm having some trouble with this:

asinx+bcosx can be rewritten as rcos(x-alpha)

It's fair straight forward until I get to getting a value for alpha. In my notes you just use tan alpha = a/b (or b/a if it's acosx+bsinx) but apparently, it gives 2 possible values? I'm guessing the second is ascertained through use of symmetry? And then it says "only 1 value is right. Use to tell:

Sin alpha = a/r Cos alpha = b/r

It doesn't go any further than that :confused:

How would that help?

Oh I've seen some stuff through google where it's asinx+bcosx=1 and the methods used to get alpha, but all the questions in past papers have just been asinx+bcosx.

Cheers

Dr Delphi · 28 Dec 2011 at 22:15

OP I remember doing this a long long time ago, so I have done a little google and come up with this youtube proof (please check):

RomanNose · 28 Dec 2011 at 22:20

The second answer is from symmetry and they just want the smallest answer in the case of tan alpha=a/b, alpha should be between -90 and 90.

KingArthur · 28 Dec 2011 at 22:22

This is an easy question, I will guide you through it.

If asinx+bcosx can be rewritten as rcos(x-alpha), then it can also be written as (x-men) and asdsfsdkfjsd+bcosx, so if it can be rewritten in such a way, then that must mean that (x-alpha) is (x-gamma) radiation and it CANNOT pass through lead. So if that is the case then asinx is pie multiplied by rcos will give you bcosx*eifhweiofh82382391 squared and simplifying that will equal X=1701

Nitefly · 28 Dec 2011 at 22:23

KingArthur said:
This is an easy question, I will guide you through it.

It's good that you helped, but a teacher should never describe something as 'easy'.

Dr Delphi · 28 Dec 2011 at 22:24

KingArthur said:
This is an easy question, I will guide you through it.

If asinx+bcosx can be rewritten as rcos(x-alpha), then it can also be written as (x-men) and asdsfsdkfjsd+bcosx, so if it can be rewritten in such a way, then that must mean that (x-alpha) is (x-gamma) radiation and it CANNOT pass through lead. So if that is the case then asinx is pie multiplied by rcos will give you bcosx*eifhweiofh82382391 squared and simplifying that will equal X=1701

Yeah good answer - helping is fun?

OP I'm trying to think about this, will look at some A-Level books and get back to you.

Dunks · 28 Dec 2011 at 22:24

Nitefly said:
It's good that you helped, but a teacher should never describe something as 'easy'.

Did you actually read his post?

Nitefly · 28 Dec 2011 at 22:25

Dunks said:
Did you actually read his post?

I never read.

Dr Delphi · 28 Dec 2011 at 22:28

OP - are you forgetting to add that

R = SQRT(A^2 + B^2) ?

KingArthur · 28 Dec 2011 at 22:30

If asinx+bcosx can be rewritten as rcos(x-alpha), then it can also be written as (x-men) and asdsfsdkfjsd+bcosx, so if it can be rewritten in such a way, then that must mean that (x-alpha) is (x-gamma) radiation and it CANNOT pass through lead. So if that is the case then asinx is pie multiplied by rcos will give you bcosx*eifhweiofh82382391 squared and simplifying that will equal X=1701

Continued.

If bcosx*eifhweiofh82382391 squared and simplifying it equals X=1701 Remainder R= SQRT(A^2 + B^2), then it brings us to the other problem, because bcosx*eifhweiofh82382391 squared is not a Continuous gangrene number, it leaves us with the age old MatheLiterature problem of Kirk or picard. There is a simple way of remedying this problem, because Kirk + Picard will equal pi X radius squared epicness, so When we take away the Picard algorithm away, it leaves us with a mathematical superiority, which Proves Kirk is best Thus disproving string theory, with a remainder of the GPS location of The sword of Excalibur.

Simple Logic.

Dunks · 28 Dec 2011 at 22:39

Trying too hard now, 4/10.

TheCenturion · 28 Dec 2011 at 22:42

Spitfir3 said:
Hey guys,

I'm having some trouble with this:

asinx+bcosx can be rewritten as rcos(x-alpha)

It's fair straight forward until I get to getting a value for alpha. In my notes you just use tan alpha = a/b (or b/a if it's acosx+bsinx) but apparently, it gives 2 possible values? I'm guessing the second is ascertained through use of symmetry? And then it says "only 1 value is right. Use to tell:

Sin alpha = a/r Cos alpha = b/r

It doesn't go any further than that How would that help?

Oh I've seen some stuff through google where it's asinx+bcosx=1 and the methods used to get alpha, but all the questions in past papers have just been asinx+bcosx.

Cheers

Rsin alpha = a
RCos alpha = b

Rsin alpha/Rcos alpha = Tan alpha

Tan alpha = a/b

Spitfir3 · 29 Dec 2011 at 02:28

Hi guys,

Cheers for the replies so far. Basically, I have an understanding of the derivation, how you get r etc, the fact that alpha is the offset from 0 etc.

I basically don't understand the rule for understanding what value of alpha to use.

For example, I get -pi/3 (-60 degrees), when I graph this, it is incorrect but when I add pi to get 2pi/3 +60 degrees, it works.

I just don't thoroughly understand any of the solutions, cross referencing them with my notes makes it worse.

Here's an example:

There's others too, each offering different explanations to why alpha should be what it is.

Anyone shed any light on that?

Thanks

Spitfir3 · 29 Dec 2011 at 02:51

Doh, I just realised, checking if it's negative/positive with Cosx and Sinx shows you which quadrant it's in thus telling you if you need to chuck on a pi or leave it.

Edit,

Hang on I don't think that's right...

OH FOR GODS SAKE