Hyperbolic functions question

Original post by TomB1996

Tan^2

happy maths! X

Posted from TSR Mobile

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!

Reply 3

Original post by TomB1996

Tan^2

happy maths! X

Posted from TSR Mobile

Are you sure?!!

Reply 4

SlowlorisIncognito

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.

Reply 5

Original post by Mr M

Are you sure?!!

Is that not correct?

Reply 6

Original post by dd1234

Is that not correct?

These are hyperbolic functions not trigonometric functions.

Reply 7

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

Reply 8

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 :smile:

Reply 9

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 :smile:

Note that

\sinh^2 x > 0

(you are not interested in x = 0)

Reply 10

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! :smile:

Reply 11

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! :smile:

So you should obtain

\sinh x = \pm \frac{1}{2}

. You now use the logarithmic form of the arsinh function from your formula book.

Reply 12

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 :s-smilie:

Reply 13

Original post by dd1234

I'm not sure how you get sinhx = 1/2 :s-smilie:

Ask your teacher in the morning then.

Reply 14

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?
Thanks

Reply 15

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.

Reply 16

Original post by Mr M

Show your working then.

Integration by substitution, I figured it out, thanks

Reply 17

HenryHiddler

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 -sinh²in the hyperbolics.

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

Hope this helps :smile:

(edited 9 years ago)

Reply 18