For maths-based tomfoolery enter here!

[D.P.]

Hmm that's iffy, on the IB you'd get no marks for that

If it's got a "3 bar equals sign" I've always been taught to derive one side by manipulating the other rather than mess around with the original identity.

equivalence is not quite the same as straight equals AFAIK.

 

[Royality]

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.

 

[Psiko]

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.

 

[sid]

The reason you can't get the other solutions is because you've divided through presumably.

Thats why you can't do it, you loose a lot of other allowed solutions.

sid

 

[Royality]

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.

 

[Psiko]

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.

 

[Dave]

The reason you can't get the other solutions is because you've divided through presumably.

Thats why you can't do it, you loose a lot of other allowed solutions.

sid

Finally someone gets what I was (ineloquently) trying to say!

I've put up a quick answer here but to do these you just need to get comfortable with manipulation using trig identities. If you unsure on how a certain step was done just ask and I'm sure someone here'll help

 

[Royality]

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.

 

[Dave]

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

Yeah makes sense. I just edited my above post when you posted this, but I solved it slightly differently, went from the other side. As long as you have the original identity used on the formula sheet then you should be fine.