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Royality · 20 May 2007 at 20:35

Who here likes A-level Identities. That's right - everyone!

Please can you help me.

With the parameters of 0<x<2pi what are the solutions to the riddle below:

2cotx + 2cosec^2x = 5

I've deduced one of the answers to be 1.11 (which is correct). However the other answers of 2.82, 4.25 and 5.96 are eluding me.

I've an idea of what's going wrong but my fiddling is too time consuming. I'm sure I'll understand more when the dam is broken etc.

Could anyone explain to me where I'm going wrong (and possibly stay in this thread for the rest of the evening.

)

Royality · 20 May 2007 at 20:50

marc_howarth said:
from sin^2 + cos^2 = 1 you can get

1 + cot^2 = cosec^2 by dividing through bin sin^2

back in to the original equation...

2cotx + 2 (1 + cot^2x) = 5

expand the brackets and make it equal to 0

2cot^2x + 2cotx - 3 = 0

solve like a normal quadratic

Sorry! Annoyingly I got the equation wrong. Real equation is:

5cotx + 2cosec^2x = 5

Oh and the rearranging I can do, it's the solving of the resulting equations to find solutions I can't do (i.e to get cotx = 1/2 and -3 to give values of 2.82, 4.25, 5.96).

Royality · 20 May 2007 at 20:59

Shamrock said:
The way I remember doing it was using a CAST diagram.

Which I am not qualified to operate (I can't understand how it works).

Royality · 20 May 2007 at 21:07

marc_howarth said:
ok well never mind i did think that quadratic was a bit of a *****

so if cotx = 1/2 and -3

tan x = 2 and -1/3

bring out the calculator here and do arctan of those values to get

tan x = 2
x = 1.11

tanx = -1/3
x = -0.35

but now you have to remember that tan repeats over and over again with a gap of 180 degrees between each one. this question is in radians however so a gap of Pi between each tan repetition

simply add Pi to those values of x untill you're out of range to get the solutions

Okay, that helped a lot.

Hehe, I have more problems though...

Prove that cosec A + cot A = 1/(cosec A - cot A)

I decided to divide left from right to equal 1.

Can you help with this, i'd greatly appreciate it; why is it that maths books never include blooming solutions?

Royality · 20 May 2007 at 21:21

marc_howarth said:
have a feeling this could be a long night....

\gets working

Wow I actually did the one above.

Apparently if you divide a fraction by a fraction the answer = (numerator numerator) x (denominator denominator) / (numerator denominator) x (denominator numerator)

i.e (5/4)/(16/12) = (5x12)/(4x16) = 60/64 = 15/16

How weird is it that I never realised that...

Royality · 20 May 2007 at 22:12

Dave said:
If you're proving that LHS = RHS you have to manipulate one side to match the other, you can't divide one side by the other. Start from the left or right and use identities until it matches the other.

I'd give it a bash for you but I'm in an incredibly lazy mood I'm afraid

If you divide one side from another the answer will equal 1 when two identities are the same (like dividing 2 by 2). Hence you can use that method... that's what my book says at least!

I just can't get my head round these bloody identities, it's a miracle I got 56 in my C3 exam first attempt in January; unfortunately I need around 85 to get an A now.

Royality · 21 May 2007 at 20:39

Yawn! Another night, another struggle with identities...

Can anyone help with this bum of a question:

Prove the following identity - cot1/2x - tan1/2x = 2cotx

I really can't deal with the 1/2x's which aren't coefficients or indices.

Royality · 21 May 2007 at 21:42

Psiko said:
Well, let's start with the RHS. Try writing cotx = cosx/sinx, and then expand cosx and sinx in terms of cos(x/2) and sin(x/2) (using your normal double angle identities, but replacing x with x/2.)

I hope that helps. See if you can work it out with the hint, rather than just being given the answer.

I don't get what you mean when you say expand in terms of cos(X/2) etc. etc. Sigh. This is really frustrating. :mad:

Royality · 21 May 2007 at 22:59

Psiko said:
Do you know the double angle formulae?

sin2x = 2cosx sinx
cos2x = cos^2 x - sin^2 x

Now replace x with x/2 in these, and what do you get? Now put those into RHS = 2cotx = 2cosx/sinx. Then notice bits cancel and leave you with the LHS.

I knew of the double angle formula just not that you could do that!

Does this make sense:

2cotA = 2.cos^2(1/2x) - 2.sin^2(1/2x) / 2.cos1/2x.sin1/2x
= cos^2(1/2x) - sin^2(1/2x) / cos1/2x.sin1/2x
= (cos1/2x.cos1/2x) - (sin1/2x.sin1/2x) / cos1/2x.sin1/2x
= cot1/2x - tan1/2x

I think i dun gone did it.

Thanks for the help.

It's really useful than sin x = cos^2(1/2x) - sin^2(1/2x) etc.