C4, rcos(x+a), rsin(x+a) Why a rearrangement of a function finds a wrong angle?

Greetings,

I have re-edited this post to better explain what confuses me about the question seen below. Or more specifically one of the methods involved in this topic.

Considering the expression is in the form of "2cosx-sinx", I've decided to compare it with the expansion of "rcos(x+λ)=r(cosxcosλ-sinxsinλ)". It turned out that it relates to a wrong angle, with a tangent 0.5 instead of 2.

I have then rearranged the expression into "sinx-2cosx", compared it with "rsin(x-λ)=r(sinxcosλ-cosxsinλ)" and it worked. It relates to the correct angle, the tangent of which is 2.

So why is it that the cosine function finds one angle and the sine finds another.?

Thank you in advance for your time and assistance.

(edited 6 years ago)

Reply 1

OCR_B

4

Rcos(x-a). Hope this hint helps.

(edited 6 years ago)

Reply 2

Notnek

21

Original post by DeadManProp

Greetings,

I'm currently struggling with fully understanding part i) of Question 15 presented below. I have also attached photos of my failed attempt and a successful attempt at finding the tangent and the value of sine.

Considering the expression was in the form of "2cosx-sinx", I've tried to find the value of λ by comparing this expression to rcos(x+λ). I've ended up finding the wrong angle.

I have then rearranged the expression into "sinx-2cosx", compared it to rsin(x-λ) and it worked.

So why is it that the cosine function finds one angle and the sine finds another. And how can I predict in advance which rearrangement/expansion will be suitable for a given angle?

Thank you in advance for your time and assistance.

There is no need to use Rcos/sin to show that

\tan \lambda = 2

. You can just rearrange the equation

2\cos x - \sin x = 0

Reply 3

DeadManProp

OP

10

Original post by OCR_B

Rcos(x-a). Hope this hint helps.

rcos(x-a)=r(cosxcosa+sinxsina)=(rcosa)cosx+(rsina)sinx

=> When I compare it to 2cosx-sinx, I get:

rcosa=2, rsina=1 => tana=0.5 which is wrong (and there's a + instead of a - in the expansion.)

I don't understand the hint unfortunately :/

(edited 6 years ago)

OP

Any ideas please?

Original post by DeadManProp

Any ideas please?

Did you completely miss out Notnek’s reply above or...?

Reply 6

Notnek

21

Original post by DeadManProp

Any ideas please?

Did you read my post above? What you're doing is unnecessary for part i.

If you have an equation of the form acos(x) + bsin(y) = 0 then that can be solved easily by rearranging.

The Rcos/sin formulas are useful where you have an equation acos(x) + bsin(y) = a, where a is non-zero.

Reply 7

DeadManProp

OP

10

Sorry, I have not expressed myself correctly. Also a couple of attachments don't seem to work.

Let's forget about part i) for a moment. I have solved it by rearrangement as you have suggested. Thank you for that. :]

What confuses me specifically, about this exercise and method in general, is why the expansion of rcos(x+λ), which seems to match the expression "2cosx - sinx", relates to a wrong angle, whereas the rearrangement "sinx - 2cosx", along with the expansion of rsin(x-λ), relates to the correct angle.

I hope I'm making more sense now.

Thank you in advance as always! This is the critical part that I can't figure out.

(edited 6 years ago)

Reply 8

Notnek

21

Original post by DeadManProp

Sorry, I have not expressed myself correctly. Also a couple of attachments don't seem to work.

Let's forget about part i) for a moment. I have solved it by rearrangement as you have suggested. Thank you for that. :]

What confuses me specifically, about this exercise and method in general, is why the expansion of rcos(x+λ), which seems to match the expression "2cosx - sinx", relates to a wrong angle, whereas the rearrangement "sinx - 2cosx", along with the expansion of rsin(x-λ), relates to the correct angle.

I hope I'm making more sense now.

Thank you in advance as always! This is the critical part that I can't figure out.

So to be clear you are trying to solve 2cosx - sinx = 0 by expressing the left hand side in the form Rcos/sin. You are aware that this is not required but you still want to understand what's going wrong in this approach.

Have I got this right?

Your attachments aren't working for me.

Reply 9

Notnek

21

@DeadManProp uploading attachments on TSR is crap* and doesn't always work properly. If you're having trouble then try this.

@Lemur14 I posted a message to you here but Ignore it for now - I edited it out.

(edited 6 years ago)

Original post by DeadManProp

What confuses me specifically, about this exercise and method in general, is why the expansion of rcos(x+λ), which seems to match the expression "2cosx - sinx", relates to a wrong angle, whereas the rearrangement "sinx - 2cosx", along with the expansion of rsin(x-λ), relates to the correct angle.

Why are you choosing

\lambda

to go with

x

in the bracket for compound angle if we know that

\lambda

takes very specific values and so the relation doesn't actually hold, in all likelihood?

Compound angle approach isn't really an approach here, because

2\cos x - \sin x

is identical to both

\sqrt{5}\cos ( x + \arctan(\frac{1}{2}))

and

-\sqrt{5} \sin (x - \arctan(2))

and so really if the function is 0 at

x=\lambda

then all you can do is say that the two derived eqs. are 0 when

x = \lambda

. You can't combine them in any way to obtain

\tan \lambda

Reply 11

DeadManProp

OP

10

Ok, thank you all for that! My textbook is so terrible at explaining things sometimes that I don't even know what I'm supposed to be thinking about when solving questions. What a nightmare.

C4, rcos(x+a), rsin(x+a) Why a rearrangement of a function finds a wrong angle?

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