sinh (log (3+sqrt{8}))= | vedclass

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Question

(B) We know that $\sinh(x) = \frac{e^x - e^{-x}}{2}$.Let $x = \log(3+\sqrt{8})$.Then $e^x = 3+\sqrt{8} = 3+2\sqrt{2} = (\sqrt{2}+1)^2$.Also,$e^{-x} =

Vedclass · Accepted Answer

$2^{3/2}$

Vedclass · Answer

(B) We know that $\sinh(x) = \frac{e^x - e^{-x}}{2}$.Let $x = \log(3+\sqrt{8})$.Then $e^x = 3+\sqrt{8} = 3+2\sqrt{2} = (\sqrt{2}+1)^2$.Also,$e^{-x} = \frac{1}{e^x} = \frac{1}{(\sqrt{2}+1)^2} = (\sqrt{2}-1)^2$.Substituting these into the formula:$\sinh(x) = \frac{(\sqrt{2}+1)^2 - (\sqrt{2}-1)^2}{2}$.Expanding the squares:$(\sqrt{2}+1)^2 = 2+1+2\sqrt{2} = 3+2\sqrt{2}$.$(\sqrt{2}-1)^2 = 2+1-2\sqrt{2} = 3-2\sqrt{2}$.$\sinh(x) = \frac{(3+2\sqrt{2}) - (3-2\sqrt{2})}{2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$.Since $2\sqrt{2} = 2^1 	imes 2^{1/2} = 2^{3/2}$,the correct option is $B$.

$\sinh (\log (3+\sqrt{8}))=$

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