What is the value of arcsinh(8/15) ?

The value of arcsinh(8/15) is ln(5/3)

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arcsinh(8/15)

arcsinh (\frac{8}{15})

ln (\frac{5}{3})

Decimal Notation

0.51082 \dots

Solution steps

arcsinh (\frac{8}{15})

Use the Hyperbolic identity:

arcsinh (x) = ln (x + x^{2} + 1)

= ln \frac{8}{15} + (\frac{8}{15})^{2} + 1

ln \frac{8}{15} + (\frac{8}{15})^{2} + 1 = ln (\frac{5}{3})

ln \frac{8}{15} + (\frac{8}{15})^{2} + 1

(\frac{8}{15})^{2} + 1 = \frac{17}{15}

(\frac{8}{15})^{2} + 1

(\frac{8}{15})^{2} = \frac{64}{225}

(\frac{8}{15})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{8 ^{2}}{1 5 ^{2}}

8^{2} = 64

= \frac{64}{1 5 ^{2}}

1 5^{2} = 225

= \frac{64}{225}

= \frac{64}{225} + 1

Join

\frac{64}{225} + 1 : \frac{289}{225}

\frac{64}{225} + 1

Convert element to fraction:

1 = \frac{1 \cdot 225}{225}

= \frac{64}{225} + \frac{1 \cdot 225}{225}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{64 + 1 \cdot 225}{225}

64 + 1 \cdot 225 = 289

64 + 1 \cdot 225

Multiply the numbers:

1 \cdot 225 = 225

= 64 + 225

Add the numbers:

64 + 225 = 289

= 289

= \frac{289}{225}

= \frac{289}{225}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{289}{225}

225 = 15

225

Factor the number:

225 = 1 5^{2}

= 1 5^{2}

Apply radical rule:

1 5^{2} = 15

= 15

= \frac{289}{15}

289 = 17

289

Factor the number:

289 = 1 7^{2}

= 1 7^{2}

Apply radical rule:

1 7^{2} = 17

= 17

= \frac{17}{15}

= ln (\frac{8}{15} + \frac{17}{15})

Join

\frac{8}{15} + \frac{17}{15} : \frac{5}{3}

\frac{8}{15} + \frac{17}{15}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{8 + 17}{15}

Add the numbers:

8 + 17 = 25

= \frac{25}{15}

Cancel the common factor:

5

= \frac{5}{3}

= ln (\frac{5}{3})

= ln (\frac{5}{3})

\frac{( sin ( 56. 2 ^{\circ} ) \cdot 19 ) ^{2}}{( 2 \cdot 9.8 )}

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200 (e^{- 2 π} cos (2))