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Prove whether the following are identities 2tanh 1/2x / 1−tanh^2 1/2 x = sinh x​

Sagot :

LammettHash

Recall that

[tex]\cosh^2(x) - \sinh^2(x) = 1[/tex]

Dividing both sides by cosh²(x) gives

[tex]1 - \tanh^2(x) = \mathrm{sech}^2(x)[/tex]

Also, recall the identity

[tex]\sinh(2x) = 2\sinh(x)\cosh(x)[/tex]

Then

[tex]\dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = \dfrac{2\tanh\left(\frac x2\right)}{\mathrm{sech}^2\left(\frac x2\right)} \\\\ \dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = 2\tanh\left(\dfrac x2\right)\cosh^2\left(\dfrac x2\right) \\\\ \dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = 2\sinh\left(\dfrac x2\right)\cosh\left(\dfrac x2\right) \\\\\dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = \sinh(x)[/tex]

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