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Hyperbolic Trig proof

Prove 4cosh2t+sinh22t=4cosh4t 4cosh^2 t + sinh^2 2t = 4cosh^4 t. I'd rather not go into exponential, as there's a 2t. I've tried a couple of identities, none of which seem to make it easier to solve :/
(edited 10 years ago)
Original post by TSR561
Prove 4cosh2t+sinh22t=4cosh4t 4cosh^2 t + sinh^2 2t = 4cosh^4 t. I'd rather not go into exponential, as there's a 2t. I've tried a couple of identities, none of which seem to make it easier to solve :/

sinh⁡2t=2sinh⁡tcosh⁡t\sinh 2t = 2\sinh{t}\cosh{t}
(edited 10 years ago)
Original post by TSR561
Prove 4cosh2t+sinh22t=4cosh4t 4cosh^2 t + sinh^2 2t = 4cosh^4 t. I'd rather not go into exponential, as there's a 2t. I've tried a couple of identities, none of which seem to make it easier to solve :/


Working from the RHS:

4cosh⁡4t=4cosh⁡2tcosh⁡2t=4cosh⁡2t(1+sinh⁡2t)\begin{aligned} 4\cosh^4 t & = 4 \cosh^2 t \cosh^2 t \\ & = 4 \cosh^2 t (1 + \sinh^2 t ) \end{aligned}
Reply 3
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TSR561
OP
12
Original post by Felix Felicis
sinh⁡22t=2sinh⁡tcosh⁡t\sinh^2 2t = 2\sinh{t}\cosh{t}


You mean sinh⁡2t=2sinh⁡2tcosh⁡2t\sinh2t = 2\sinh{2t}\cosh{2t} ? In which case sinh⁡22t=4sinh⁡22tcosh⁡22t\sinh^2 2t = 4\sinh^2{2t}\cosh^2{2t}, right? Now whether that simplifies down to what you originally said...
Original post by TSR561
You mean sinh⁡2t=2sinh⁡2tcosh⁡2t\sinh2t = 2\sinh{2t}\cosh{2t} ? In which case sinh⁡22t=4sinh⁡22tcosh⁡22t\sinh^2 2t = 4\sinh^2{2t}\cosh^2{2t}, right? Now whether that simplifies down to what you originally said...

No, sorry, typo, I meant sinh⁡2t=2sinh⁡tcosh⁡t  ⟹  sinh⁡22t=4sinh⁡2tcosh⁡2t\sinh{2t} = 2\sinh{t}\cosh{t} \implies \sinh^2 2t = 4\sinh^2 {t}\cosh^2 {t} and now with some factorisation, it is easy to show LHS = RHS.
(edited 10 years ago)
Original post by Felix Felicis
No, sorry, typo, I meant sinh⁡2t=2sinh⁡tcosh⁡t  ⟹  sinh⁡22t=4sinh⁡tcosh⁡t\sinh{2t} = 2\sinh{t}\cosh{t} \implies \sinh^2 2t = 4\sinh{t}\cosh{t} and now with some factorisation, it is easy to show LHS = RHS.


∗ sinh⁡2t=2sinh⁡tcosh⁡t  ⟹  sinh⁡22t=(2sinh⁡tcosh⁡t)2=4sinh⁡2tcosh⁡2t* \ \sinh{2t} = 2\sinh{t}\cosh{t} \implies \sinh^2 2t = \left( 2\sinh t \cosh t \right)^2 = 4\sinh^2{t}\cosh^2{t}
(edited 10 years ago)
Original post by Khallil
∗ sinh⁡2t=2sinh⁡tcosh⁡t  ⟹  sinh⁡22t=4sinh⁡2tcosh⁡2t* \ \sinh{2t} = 2\sinh{t}\cosh{t} \implies \sinh^2 2t = 4\sinh^2{t}\cosh^2{t}

Not my day today at all. :sigh:

Thanks.
Reply 7
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TSR561
OP
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Original post by Felix Felicis
No, sorry, typo, I meant sinh⁡2t=2sinh⁡tcosh⁡t  ⟹  sinh⁡22t=4sinh⁡2tcosh⁡2t\sinh{2t} = 2\sinh{t}\cosh{t} \implies \sinh^2 2t = 4\sinh^2 {t}\cosh^2 {t} and now with some factorisation, it is easy to show LHS = RHS.



Original post by Khallil
∗ sinh⁡2t=2sinh⁡tcosh⁡t  ⟹  sinh⁡22t=(2sinh⁡tcosh⁡t)2=4sinh⁡2tcosh⁡2t* \ \sinh{2t} = 2\sinh{t}\cosh{t} \implies \sinh^2 2t = \left( 2\sinh t \cosh t \right)^2 = 4\sinh^2{t}\cosh^2{t}


Got it.Thanks a lot, guys!
Original post by Felix Felicis
Not my day today at all. :sigh:

Thanks.
Original post by TSR561
Got it.Thanks a lot, guys!


No prob :smile:

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