Prove that sinhx, i feel like im going round in circles
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Original post by Phichi
#FeelingIgnored
sorry dude, i appriceate your help, trying to work it all out as well as reply to you both is a heavy task when ive just done a 10 till 10 shift at work too
Original post by Ramjam
Yeh i am, trig has never been a strong subject of mine 
from this line multiply top and bottom of the fraction by cosh2(x/2)
this will give you
2sinh(x/2)cosh(x/2) on the top
cosh2(x/2) -sinh2(x/2) at the bottom.
can you see this?
Original post by TeeEm
from this line multiply top and bottom of the fraction by cosh2(x/2)
this will give you
2sinh(x/2)cosh(x/2) on the top
cosh2(x/2) -sinh2(x/2) at the bottom.
can you see this?
this will give you
2sinh(x/2)cosh(x/2) on the top
cosh2(x/2) -sinh2(x/2) at the bottom.
can you see this?
yep i can see that as when you multiply on the top it cancels out the bottem
Original post by Ramjam
yep i can see that as when you multiply on the top it cancels out the bottem
it cancels some bits completely and other not
read my last post. Do you have what I claim you should have?
Original post by TeeEm
it cancels some bits completely and other not
read my last post. Do you have what I claim you should have?
read my last post. Do you have what I claim you should have?
I believe so, so just to clarify you have times top and bottom by cosh2x/2
Original post by Ramjam
I believe so, so just to clarify you have times top and bottom by cosh2x/2
cosh2(x/2)
"cosh squared of x over 2"
Original post by TeeEm
cosh2(x/2)
"cosh squared of x over 2"
"cosh squared of x over 2"
opps sorry forgot the superscript, yeh i have the same
Original post by Ramjam
opps sorry forgot the superscript, yeh i have the same
then the bottom is a well known hyperbolic identity
do you recognise it?
Original post by Ramjam
Thanks everyone for you help, i understand how to do this now and will work on my trig, you all have been very helpfull!
no problem
Original post by Ramjam
Thanks everyone for you help, i understand how to do this now and will work on my trig, you all have been very helpfull!
my pleasure
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