Prove-that-tanh-log-3-1-2-

Question Number 51887 by Tawa1 last updated on 31/Dec/18

Prove that; \tanh (\log \sqrt{3}) = \frac{1}{2}

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jan/19

e^{i x} = c o s x + i s i n x

e^{- i x} = c o s x - i s i n x

x = i θ

e^{- θ} = c o s (i θ) + i s i n (i θ)

= c o s h (θ) - s i n h (θ)

e^{θ} = c o s h (θ) + s i n h (θ)

t a n h (θ) = \frac{e^{θ} - e^{- θ}}{e^{θ} + e^{- θ}}

t a n h (l n \sqrt{3}) = \frac{e^{l n \sqrt{3}} - e^{- l n \sqrt{3}}}{e^{l n \sqrt{3}} + e^{- l n \sqrt{3}}} = \frac{e^{\frac{1}{2} l n 3} - e^{- \frac{1}{2} l n 3}}{e^{\frac{1}{2} l n 3} + e^{- \frac{1}{2} l n 3}} .

= \frac{{(3)}^{\frac{1}{2}} - {(3)}^{\frac{- 1}{2}}}{{(3)}^{\frac{1}{2}} + {(3)}^{\frac{- 1}{2}}} = \frac{3^{\frac{1}{2} + \frac{1}{2}} - 1}{3^{\frac{1}{2} + \frac{1}{2}} + 1} . = \frac{3 - 1}{3 + 1} = \frac{1}{2}

a^{x} = e^{x l n a}

Commented by Tawa1 last updated on 31/Dec/18

God bless you sir

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