dd1234
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This may seem like a rather simple question- but when you divide sinh^2 x by cosh^2 x would you get tanh^2 x or just tanh x?
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TomB1996
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Tan^2 happy maths! X


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dd1234
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(Original post by TomB1996)
Tan^2 happy maths! X


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Haha thought so! Feels like a really silly question, but I've been doing so much of FP2 I've started to forget some basic principles :rolleyes:
Thanks!
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Mr M
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(Original post by TomB1996)
Tan^2 happy maths! X


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Are you sure?!!
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SlowlorisIncognito
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Hi,

I've moved this thread to the maths study help sub- forum, as the forum you posted it in is for discussing maths university courses, not asking for help with maths questions.
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dd1234
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(Original post by Mr M)
Are you sure?!!
Is that not correct?
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Mr M
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(Original post by dd1234)
Is that not correct?
These are hyperbolic functions not trigonometric functions.
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dd1234
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(Original post by Mr M)
These are hyperbolic functions not trigonometric functions.
Oh right, I disregarded that mistake- thought you meant it wasn't tanh^2 x
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dd1234
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(Original post by Mr M)
These are hyperbolic functions not trigonometric functions.
Could I ask for help with this question- its the FP2 OCR paper January 2008 Qn 8 ii - I am not sure how to go forwards after substituting in the result found in part i ?

Thanks
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Mr M
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(Original post by dd1234)
Could I ask for help with this question- its the FP2 OCR paper January 2008 Qn 8 ii - I am not sure how to go forwards after substituting in the result found in part i ?

Thanks
Note that \sinh^2 x > 0 (you are not interested in x = 0)
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dd1234
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(Original post by Mr M)
Note that \sinh^2 x > 0 (you are not interested in x = 0)
Great, thank you, I have done the question correctly but have the same issue with qn 9 ii - how would you go about doing this question using the first part?

Thanks a lot for your help!
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Mr M
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(Original post by dd1234)
Great, thank you, I have done the question correctly but have the same issue with qn 9 ii - how would you go about doing this question using the first part?

Thanks a lot for your help!
So you should obtain \sinh x = \pm \frac{1}{2}. You now use the logarithmic form of the arsinh function from your formula book.
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dd1234
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(Original post by Mr M)
So you should obtain \sinh x = \pm \frac{1}{2}. You now use the logarithmic form of the arsinh function from your formula book.
I'm not sure how you get sinhx = 1/2
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Mr M
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(Original post by dd1234)
I'm not sure how you get sinhx = 1/2
Ask your teacher in the morning then.
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dd1234
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(Original post by Mr M)
Ask your teacher in the morning then.
I am unable to do so right now, is it possible for you to explain it?
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Mr M
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(Original post by dd1234)
I am unable to do so right now, is it possible for you to explain it?
Thanks
Show your working then.
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dd1234
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(Original post by Mr M)
Show your working then.
Integration by substitution, I figured it out, thanks
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HenryHiddler
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(Original post by dd1234)
This may seem like a rather simple question- but when you divide sinh^2 x by cosh^2 x would you get tanh^2 x or just tanh x?
tanh2 - just like normal trig, only remember Osborn's rule when working with hyperbolic trig functions.

You treat everything as normal trig, but if you have sin2 in the normal trig, that becomes -sinh2 in the hyperbolics.

IMPORTANT: But remember, replace ANY implied product of 2 sine terms by minus the product.
sinAsinB -> -sinhAsinhB
tan2A -> -tanh2A

Hope this helps
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dd1234
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(Original post by HenryHiddler)
tanh2 - just like normal trig, only remember Osborn's rule when working with hyperbolic trig functions.

You treat everything as normal trig, but if you have sin2 in the normal trig, that becomes -sinh2 in the hyperbolics.

IMPORTANT: But remember, replace ANY implied product of 2 sine terms by minus the product.
sinAsinB -> -sinhAsinhB
tan2A -> -tanh2A

Hope this helps
That is really helpful, thanks a lot
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