What is the value of arccsch(2) ?

The value of arccsch(2) is ln((1+sqrt(5))/2)

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arccsch(2)

arccsch (2)

ln (\frac{1 + 5}{2})

Decimal Notation

0.48121 \dots

Solution steps

arccsch (2)

Use the Hyperbolic identity:

arccsch (x) = ln (\frac{1}{x} + \frac{1}{x ^{2}} + 1)

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

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

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

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

\frac{1}{2 ^{2}} + 1

2^{2} = 4

= \frac{1}{4} + 1

Join

\frac{1}{4} + 1 : \frac{5}{4}

\frac{1}{4} + 1

Convert element to fraction:

1 = \frac{1 \cdot 4}{4}

= \frac{1}{4} + \frac{1 \cdot 4}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{1 + 1 \cdot 4}{4}

1 + 1 \cdot 4 = 5

1 + 1 \cdot 4

Multiply the numbers:

1 \cdot 4 = 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

= \frac{5}{4}

= \frac{5}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{5}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{5}{2}

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

Join

\frac{1}{2} + \frac{5}{2} : \frac{1 + 5}{2}

\frac{1}{2} + \frac{5}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{1 + 5}{2}

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

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

10 \cdot tan (4 0^{\circ})

sin (\frac{16 π}{1.4})

csc (1.5)

cos^{6} (\frac{π}{2})

sin (\frac{7 π}{16})