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Mathematics 16 Online
OpenStudy (anonymous):

if sinh x = 15/8 find the values of the other hyperbolic functions at x. cosh x = tanh x = csch x = sech x = coth x = how do I do this ??

OpenStudy (anonymous):

try reading here

OpenStudy (anonymous):

read what ??

OpenStudy (anonymous):

cosh^2(x) - sinh^2(x) = 1

OpenStudy (anonymous):

subtitusi sinh = 15/8 to that equation

OpenStudy (anonymous):

does 1.301708279 = cosh ??

OpenStudy (anonymous):

can you do that?

OpenStudy (anonymous):

what do you mean by that ?

OpenStudy (anonymous):

subtitusi, sorry the network is bad

OpenStudy (anonymous):

cosh x = 17/8

OpenStudy (anonymous):

cosh^2(x)-(15/8)^2=1 cosh^2(x)=1+(15/8)^2 i am i on the right track ??

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

cos=\[\sqrt{1+(15/8)^2}\]

OpenStudy (anonymous):

yea it does

OpenStudy (anonymous):

ok so from here how do i find the rest ?

OpenStudy (anonymous):

this ??

OpenStudy (anonymous):

(15/8)^2 = 15^2/8^2

OpenStudy (anonymous):

what does that mean ?

OpenStudy (anonymous):

\[\sqrt{1+\ \left( \frac{ 15 }{ 8 } \right)^2}\]

OpenStudy (anonymous):

\[\sqrt{1+\frac{ 15^2 }{ 8^2 }}\]

OpenStudy (anonymous):

yea that's 17/8

OpenStudy (anonymous):

\[\sqrt{1+\frac{ 225 }{ 64 }}\]

OpenStudy (anonymous):

yes that right

OpenStudy (anonymous):

how do we find tanh (x) form there ?

OpenStudy (anonymous):

tanh x = sinh x / cosh x

OpenStudy (anonymous):

(15/8)/(17/8)= tanh(x)

OpenStudy (anonymous):

yes, simplify

OpenStudy (anonymous):

csch x = 1/sinh x sech x = 1/ cosh x coth x = 1 / tanh x

OpenStudy (anonymous):

So csch is 1/(15/8)

OpenStudy (anonymous):

yes, but please simplify become 8/15

OpenStudy (anonymous):

ok I understand this now than you soo much !!!!

OpenStudy (anonymous):

thank you soo much !

OpenStudy (anonymous):

oks

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