More P5 Hyperbolic identity help needed Watch

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jmzcherry
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#1
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#1
Can anyone do this? Is there a shorter way to do it than just writing all the e^x's out and then cancelling them etc?

sinh x= (2tanh 0.5x)/(1-tanh^2 0.5x)

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dvs
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RHS = (2tanh 0.5x)/(1-tanh^2 0.5x) = 2tanh(0.5x)[1/sech²(0.5x)] = 2sinh(0.5x)cosh(0.5x) = sinhx = LHS
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Syncman
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id be guessing. without doing it i wouldnt know. and its 11pm and ill be honest im in a TV mood not a maths one. But my guess is change the (1-tan^2 0.5x) to (sech^2 0.5x)

sinh x= (2tanh 0.5x)/(1/sech^2 0.5x)

sinh x= (2tanh 0.5x cosh^2 0.5x)

sinh x= (2sinh 0.5x cosh 0.5x) tan = sin over cos, then cancel a cos.
then u have 2sinAcosA = sin(2A) double angle formula.

sinh x = sinh x

dam ok so maybe i got a bit carried away and did it
i think this is right.
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