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Re: Hyperbolic Functions
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It aint cubed on the assignment sheet but do u reckon it might me wrong.
I got to this sin3θ = 3sinθ - 4sin3θ substituting jA for θ gives sin3jA = 3sinjA - 4(sin3jA) since sinjA = jsinhA jsinh3A = 3jsinhA - 4(jsinh3A) jsinh3A = 3jsinhA - 4jsinh3A just that negative sign!!! |
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Don't sweat over it anymore |
well cheers for the help. Can you go that with the 4sin3jA and put the 4 like 4(sin3jA) so it goes to 4jsinh3A
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then divide thru by j sinh3A = 3sinhA - 4sinh3A but like you say that's not what is on your sheet |
Ok well cheers for the help matey :thumbsup:
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The proof is in the text book, which have just dug out and blown the cobwebs off :D
http://upload5.postimage.org/241045/Image1.jpg |
Minter, well I now know they dont make up the qu themselves.
What book is that? Cheers |
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what are you studying? :study: |
The answer is
42 |
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