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# Can anyone one help with this fiddly maths? watch

1. Can anyone help me out here, I think I'm nearly there......

Show [sinh (x + iy)]^2 = 1/2 [cosh 2x - cos 2y]

Previously the hint was to express sinh (x + iy) in u + iv form. I've done that, and several bits have cancelled out to leave the left hand side as:

2 i sinh x cosh x sin y cos y, which is also equal to

1/2 i sinh 2x sin 2y, although i'm not sure how that relates to the RHS.

Another question in the same book:
If u + iv = ln [(x + iy + a) / (x + iy - a)], prove that

1) x^2 + y^2 - 2 a x coth u + a^2 = 0

2) x = a sinh u / (cosh u - cos v)

3) (x + iy)^2 = a^2 (cosh u + cos v) / (cosh u - cos v)

2. Lost me there mate.
3. bump..... hopefully there r sum ppl doing maths who can help with this.
4. (Original post by 4Ed)
bump..... hopefully there r sum ppl doing maths who can help with this.
hmm for the first one can you expand it using siny(x+iy) into separate parts using the identity and then consider the real and imaginary parts separately. Let me know if this doesn't work, I havn't looked at it properly.
5. i did that and u get (sinh x cos y) ^ 2 - (cosh x sin y) ^ 2 + 2 i sinh x cos y cosh x sin y, the former two terms cancel out, and ur left with

2 i sinh x cos y cosh x sin y,

which i'm not sure how to make equal to the RHS. its probly quite a simple step, but i cant see it...
6. (Original post by 4Ed)
i did that and u get (sinh x cos y) ^ 2 - (cosh x sin y) ^ 2 + 2 i sinh x cos y cosh x sin y, the former two terms cancel out, and ur left with

2 i sinh x cos y cosh x sin y,

which i'm not sure how to make equal to the RHS. its probly quite a simple step, but i cant see it...
Are you certain the terms cancel, I'm not convinced, I get some terms left over.
7. eeek maybe i should have been more awake when i did the problem. i expanded both out in terms of e, and i thought they cancelled...... lemme go find that paper with the working on it.
8. (Original post by 4Ed)
eeek maybe i should have been more awake when i did the problem. i expanded both out in terms of e, and i thought they cancelled...... lemme go find that paper with the working on it.
I don't think you need e's here:

(sinh x cos y) ^ 2 - (cosh x sin y) ^ 2 = sinh^2 x cos^2 y - cosh^2 x sin^2 y = sinh^2 x - sinh^2 x sin^2 y - cosh^2 x sin^2 y = sinh^2 x + sin^2 y?
9. im not convinced they cancel now, but sinh^2 + cosh^2 does not equal to 1........... so im not sure that term is right either.
10. (Original post by 4Ed)
im not convinced they cancel now, but sinh^2 + cosh^2 does not equal to 1........... so im not sure that term is right either.
Indeed, my mistake... sinh^2x+cosh^2x = cosh2x?
11. I'll have a go if you remind me of what sinh x and cosh x is in terms of e^x
12. The thing about maths is that you forget everything you learned after the exam!!
13. (Original post by 4Ed)
bump..... hopefully there r sum ppl doing maths who can help with this.
Bumping posts is just not cool.
14. i think so...... but does that help achieve the RHS?
15. (Original post by 4Ed)
i think so...... but does that help achieve the RHS?
Well it looks like a step forward, although I don't know what to do with the i yet.
16. (Original post by Tek)
Bumping posts is just not cool.

sorry, but i started this thread, and i would like to know the answers to this question.

I'll have a go if you remind me of what sinh x and cosh x is in terms of e^x
sinh x = 1/2 (e^x - e^-x) cosh x is 1/2 (e^x + e^-x)

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