What is the general solution for sinh(y)=-2 ?

The general solution for sinh(y)=-2 is y=ln(-2+sqrt(5))

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sinh(y)=-2

Solution

sinh (y) = - 2

Solution

y = ln (- 2 + 5)

Degrees

y = - 82.71421 \dots^{\circ}

Solution steps

sinh (y) = - 2

Rewrite using trig identities

sinh (y) = - 2

Use the Hyperbolic identity:

sinh (x) = \frac{e ^{x} - e ^{- x}}{2}

\frac{e ^{y} - e ^{- y}}{2} = - 2

\frac{e ^{y} - e ^{- y}}{2} = - 2

\frac{e ^{y} - e ^{- y}}{2} = - 2 : y = ln (- 2 + 5)

\frac{e ^{y} - e ^{- y}}{2} = - 2

Multiply both sides by

2

\frac{e ^{y} - e ^{- y}}{2} \cdot 2 = - 2 \cdot 2

Simplify

e^{y} - e^{- y} = - 4

Apply exponent rules

e^{y} - e^{- y} = - 4

Apply exponent rule:

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

e^{- y} = (e^{y})^{- 1}

e^{y} - (e^{y})^{- 1} = - 4

e^{y} - (e^{y})^{- 1} = - 4

Rewrite the equation with

e^{y} = u

u - (u)^{- 1} = - 4

Solve

u - u^{- 1} = - 4 : u = - 2 + 5, u = - 2 - 5

u - u^{- 1} = - 4

Refine

u - \frac{1}{u} = - 4

Multiply both sides by

u

u - \frac{1}{u} = - 4

Multiply both sides by

u

uu - \frac{1}{u} u = - 4 u

Simplify

uu - \frac{1}{u} u = - 4 u

Simplify

uu : u^{2}

uu

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= u^{1 + 1}

Add the numbers:

1 + 1 = 2

= u^{2}

Simplify

- \frac{1}{u} u : - 1

- \frac{1}{u} u

Multiply fractions:

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

= - \frac{1 \cdot u}{u}

Cancel the common factor:

u

= - 1

u^{2} - 1 = - 4 u

u^{2} - 1 = - 4 u

u^{2} - 1 = - 4 u

Solve

u^{2} - 1 = - 4 u : u = - 2 + 5, u = - 2 - 5

u^{2} - 1 = - 4 u

Move

4 u

to the left side

u^{2} - 1 = - 4 u

Add

4 u

to both sides

u^{2} - 1 + 4 u = - 4 u + 4 u

Simplify

u^{2} - 1 + 4 u = 0

u^{2} - 1 + 4 u = 0

Write in the standard form

a x^{2} + b x + c = 0

u^{2} + 4 u - 1 = 0

Solve with the quadratic formula

u^{2} + 4 u - 1 = 0

Quadratic Equation Formula:

For

a = 1, b = 4, c = - 1

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 1) = 25

4^{2} - 4 \cdot 1 \cdot (- 1)

Apply rule

- (- a) = a

= 4^{2} + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4^{2} + 4

4^{2} = 16

= 16 + 4

Add the numbers:

16 + 4 = 20

= 20

Prime factorization of

20 : 2^{2} \cdot 5

20

20

divides by

220 = 10 \cdot 2

= 2 \cdot 10

10

divides by

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 5

= 2^{2} \cdot 5

= 2^{2} \cdot 5

Apply radical rule:

= 5 2^{2}

Apply radical rule:

2^{2} = 2

= 25

u_{1, 2} = \frac{- 4 \pm 2 5}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 4 + 2 5}{2 \cdot 1}, u_{2} = \frac{- 4 - 2 5}{2 \cdot 1}

u = \frac{- 4 + 2 5}{2 \cdot 1} : - 2 + 5

\frac{- 4 + 2 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4 + 2 5}{2}

Factor

- 4 + 25 : 2 (- 2 + 5)

- 4 + 25

Rewrite as

= - 2 \cdot 2 + 25

Factor out common term

2

= 2 (- 2 + 5)

= \frac{2 ( - 2 + 5 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 2 + 5

u = \frac{- 4 - 2 5}{2 \cdot 1} : - 2 - 5

\frac{- 4 - 2 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

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

Factor

- 4 - 25 : - 2 (2 + 5)

- 4 - 25

Rewrite as

= - 2 \cdot 2 - 25

Factor out common term

2

= - 2 (2 + 5)

= - \frac{2 ( 2 + 5 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - (2 + 5)

Negate

- (2 + 5) = - 2 - 5

= - 2 - 5

The solutions to the quadratic equation are:

u = - 2 + 5, u = - 2 - 5

u = - 2 + 5, u = - 2 - 5

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

u - u^{- 1}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - 2 + 5, u = - 2 - 5

u = - 2 + 5, u = - 2 - 5

Substitute back

u = e^{y},

solve for

y

Solve

e^{y} = - 2 + 5 : y = ln (- 2 + 5)

e^{y} = - 2 + 5

Apply exponent rules

e^{y} = - 2 + 5

f (x) = g (x)

, then

ln (f (x)) = ln (g (x))

ln (e^{y}) = ln (- 2 + 5)

Apply log rule:

ln (e^{a}) = a

ln (e^{y}) = y

y = ln (- 2 + 5)

y = ln (- 2 + 5)

Solve

e^{y} = - 2 - 5 :

No Solution for

y \in R

e^{y} = - 2 - 5

a^{f (y)}

cannot be zero or negative for

y \in R

No Solution for y \in R

y = ln (- 2 + 5)

y = ln (- 2 + 5)

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Frequently Asked Questions (FAQ)

What is the general solution for sinh(y)=-2 ?
The general solution for sinh(y)=-2 is y=ln(-2+sqrt(5))