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C3 Trig (sums and differences of sines/cosines)

Hi again.

This question asked me to express the first expression as a sum or difference of sines or cosines. What I don't understand is why the x and 7x terms switch sides from one side of the equation to the other in the answer given. It seems to mean that I got the wrong answer when doing it due to me keeping them the way round they originally were. I don't quite understand whether both ways would end up giving the same solution (the question does not ask or provide a way to solve) or whether I'm misunderstanding something else. Help!

7E 5c.JPG

Edit: Obviously the image given is not the final answer, but that is where the terms switch and the rest of the working follows as I would expect it to with the terms having been that way round in the first place.

Edit 2: Just to elaborate further, my working goes as follows:

-3/2(-2sinx sin7x)
= -3/2[cos(x+7x)-cos(x-7x)]
=-3/2[cos8x - cos(-6x)]

where the CD comes out with -3/2(cos8x-cos6x).

Edit 3: The only thing that occurs to me is that maybe it's because cos(-6x) is the same as cos(6x) so it doesn't matter, but then why do they bother switching the terms around?

(edited 11 years ago)

Reply 1

ghostwalker

Original post by Ronove

Edit 2: Just to elaborate further, my working goes as follows:

-3/2(-2sinx sin7x)
= -3/2[cos(x+7x)-cos(x-7x)]
=-3/2[cos8x - cos(-6x)]

where the CD comes out with -3/2(cos8x-cos6x).

You should know that cos(-x) = cos(x), so you can simplify your result.

Edit: I see you noticed that.

Edit2: It makes no difference switching the terms around, other than avoiding the -6x, when you convert. Which is only trivial once your realise cos(-6x) = cos(6x)

(edited 11 years ago)

Reply 2

Ronove

Original post by ghostwalker

You should know that cos(-x) = cos(x), so you can simplify your result.

Edit: I see you noticed that.

I don't understand why they just switch the terms, though? To me it seems like switching A and B could result in some errors, or is this not the case?

Reply 3

ghostwalker

Original post by Ronove

I don't understand why they just switch the terms, though? To me it seems like switching A and B could result in some errors, or is this not the case?

No. AB=BA.

But see the last edit in my previous post.

Reply 4

Ronove

Original post by ghostwalker

No. AB=BA.

But see the last edit in my previous post.

But what if it were a difference of sines? Oh... I just realised that when it's sines, the first expression has both sin and cos in so you should remember which term is A and which is B. I'm guessing this is what I was worrying about unnecessarily? ie. if you wrote something as 2sinBcosA and forgot they were switched and expressed it as sin(B+A) - sin(B-A) by accident... what with sin(-x) not being the same as sin(x). Confused!

Reply 5

ghostwalker

Original post by Ronove

Confused!

Hopefully, they won't call them A and B.

But if it helps, write down the formula you're using in terms of A and B, and then write A=..., B=..., and just substitute in.

Reply 6

Ronove

Original post by ghostwalker

Hopefully, they won't call them A and B.

But if it helps, write down the formula you're using in terms of A and B, and then write A=..., B=..., and just substitute in.

I'm not even sure if the scenario I described/am fearing is physically possible! But I'll certainly make sure to note down the terms if it seems like the hour is upon me. Thanks for making sense out of my nonsense. :tongue:

Reply 7

ghostwalker

Original post by Ronove

...

Glad it helped.

Reply 8

Ronove

I'm going to go ahead and show an example I tried to test a theory (to reassure myself) and I would appreciate it if someone could check I'm not making some obvious mistake. :smile:

I think I noticed that no matter how you write the expression 2 cosx sin7x you will get the same answer as long as you follow the correct method according to whether you've put the cos first or the sin first. (I wasn't talking about mixing the cos's term with the sin's term, rather switching the two around):

2 cosx sin 7x = sin(x+7x) - sin(x-7x) = sin8x - sin(-6x) = sin8x - -sin6x = sin8x + sin6x

and

2 sin7x cosx = sin (7x+x) + sin(7x-x) = sin8x + sin6x

Edit: Of course, I know that 2AB = 2BA but I wasn't sure how it would translate when you changed the expression. I'm going to stop rambling now...