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Chapter 8: Problem 7

Prove the hyperbolic identity $$ \cosh 2 x=\sinh ^{2} x+\cosh ^{2} x $$

Short Answer

Expert verified
Question: Prove the hyperbolic identity: \(\cosh 2x = \sinh^2x + \cosh^2x\). Answer: Following the steps from the above solution, we rewrite the given identity using the definitions of hyperbolic sine and cosine functions and then simplify the expressions. After comparing the expressions, we find that \(\cosh 2x = \sinh^2x + \cosh^2x\), thus proving the hyperbolic identity.

Step by step solution

01

Write down the given identity

The given identity is: $$ \cosh 2x = \sinh^2 x + \cosh^2 x $$
02

Recall the definitions of hyperbolic functions

The hyperbolic sine and cosine functions are defined as: $$ \sinh x = \frac{e^x - e^{-x}}{2}, \quad \cosh x = \frac{e^x + e^{-x}}{2} $$
03

Apply the definition of hyperbolic cosine to 2x

We apply the definition of hyperbolic cosine to \(\cosh 2x\): $$ \cosh 2x = \frac{e^{2x} + e^{-2x}}{2} $$
04

Write the squares of hyperbolic sine and cosine functions

We write the squares of \(\sinh x\) and \(\cosh x\) and use their definitions: $$ \sinh^2x = \left(\frac{e^x - e^{-x}}{2}\right)^2 = \frac{e^{2x} - 2 + e^{-2x}}{4} $$ $$ \cosh^2x = \left(\frac{e^x + e^{-x}}{2}\right)^2 = \frac{e^{2x} + 2 + e^{-2x}}{4} $$
05

Add the squares of hyperbolic sine and cosine

Add the expressions for \(\sinh^2x\) and \(\cosh^2x\): $$ \sinh^2x + \cosh^2x = \frac{e^{2x} - 2 + e^{-2x}}{4} + \frac{e^{2x} + 2 + e^{-2x}}{4} = \frac{2e^{2x}+2e^{-2x}}{4} = \frac{e^{2x} + e^{-2x}}{2} $$
06

Compare the expressions

Recall that we found \(\cosh 2x=\frac{e^{2x}+e^{-2x}}{2}\) in step 3 and \(\sinh^2x+\cosh^2x=\frac{e^{2x}+e^{-2x}}{2}\) in step 5. Since both expressions are equal, the given hyperbolic identity is proven: $$ \cosh 2x = \sinh^2x + \cosh^2x $$

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Most popular questions from this chapter

Express the following using indices: (a) \(\log 35=1.5441\) (b) \(\log _{6} 1296=4\) (c) \(\ln 50=3.9120\) (d) \(\log _{9} 3=0.5\)

Simplify each expression as far as possible: (a) \(\mathrm{e}^{2 x} \mathrm{e}^{7 x}\) (b) \(\left(3 \mathrm{e}^{x}\right)\left(2 \mathrm{e}^{-x}\right)\) (c) \(\mathrm{e}^{2 x}\left(\mathrm{e}^{-2 x}+\mathrm{e}^{-x}+1\right)-\mathrm{e}^{x}\left(1+\mathrm{e}^{x}\right)\) (d) \(\frac{\mathrm{e}^{-3 x}}{2 \mathrm{e}^{-x}}\)

Solve (a) \(10^{\log x}=17\) (b) \(10^{2 \log x}=17\) (c) \(10^{x} 10^{2 x}=90\) (d) \(10^{2 x}=30\left(10^{2}\right)\)

Simplify to a single logarithmic expression: (a) \(\ln 4 y+\ln x\) (b) \(3 \ln t^{2}-2 \ln t\) (c) \(3 \log t-\log 3 t\) (d) \(\log 2 x+\log 5 x-1\)

Solve (a) \(\ln x=2.4050\) (b) \(\ln x=0.9611\) (c) \(\ln x=-0.9611\) (d) \(\ln x=-2.0000\)
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