What is the value of arccosh(5/4) ?

The value of arccosh(5/4) is ln(2)

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arccosh(5/4)

arccosh (\frac{5}{4})

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

arccosh (\frac{5}{4})

Use the Hyperbolic identity:

arccosh (x) = ln (x + x^{2} - 1)

= ln \frac{5}{4} + (\frac{5}{4})^{2} - 1

ln \frac{5}{4} + (\frac{5}{4})^{2} - 1 = ln (2)

ln \frac{5}{4} + (\frac{5}{4})^{2} - 1

(\frac{5}{4})^{2} - 1 = \frac{3}{4}

(\frac{5}{4})^{2} - 1

(\frac{5}{4})^{2} = \frac{25}{16}

(\frac{5}{4})^{2}

Apply exponent rule:

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

= \frac{5 ^{2}}{4 ^{2}}

5^{2} = 25

= \frac{25}{4 ^{2}}

4^{2} = 16

= \frac{25}{16}

= \frac{25}{16} - 1

Join

\frac{25}{16} - 1 : \frac{9}{16}

\frac{25}{16} - 1

Convert element to fraction:

1 = \frac{1 \cdot 16}{16}

= \frac{25}{16} - \frac{1 \cdot 16}{16}

Since the denominators are equal, combine the fractions:

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

= \frac{25 - 1 \cdot 16}{16}

25 - 1 \cdot 16 = 9

25 - 1 \cdot 16

Multiply the numbers:

1 \cdot 16 = 16

= 25 - 16

Subtract the numbers:

25 - 16 = 9

= 9

= \frac{9}{16}

= \frac{9}{16}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{9}{16}

16 = 4

16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

= \frac{9}{4}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{3}{4}

= ln (\frac{5}{4} + \frac{3}{4})

Join

\frac{5}{4} + \frac{3}{4} : 2

\frac{5}{4} + \frac{3}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{5 + 3}{4}

Add the numbers:

5 + 3 = 8

= \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= 2

= ln (2)

= ln (2)

- 8.4 \cdot cos (11 0^{\circ})

1 \cdot sin (4 5^{\circ})

5 \cdot arctan (12 0^{\circ})

arcsin (0.44125)

\frac{cos ^{4} ( \frac{π}{4} )}{4}