What is the general solution for sinh(a)= 1/2 ?

The general solution for sinh(a)= 1/2 is a=ln((1+sqrt(5))/2)

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Popular >

sinh(a)= 1/2

Solution

sinh (a) = \frac{1}{2}

Solution

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

Degrees

a = 27.57140 \dots^{\circ}

Solution steps

sinh (a) = \frac{1}{2}

Rewrite using trig identities

sinh (a) = \frac{1}{2}

Use the Hyperbolic identity:

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

\frac{e ^{a} - e ^{- a}}{2} = \frac{1}{2}

\frac{e ^{a} - e ^{- a}}{2} = \frac{1}{2}

\frac{e ^{a} - e ^{- a}}{2} = \frac{1}{2} : a = ln (\frac{1 + 5}{2})

\frac{e ^{a} - e ^{- a}}{2} = \frac{1}{2}

Multiply both sides by

2

\frac{e ^{a} - e ^{- a}}{2} \cdot 2 = \frac{1}{2} \cdot 2

Simplify

e^{a} - e^{- a} = 1

Apply exponent rules

e^{a} - e^{- a} = 1

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{a} = u

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

Solve

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

u - u^{- 1} = 1

Refine

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

Multiply both sides by

u

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

Multiply both sides by

u

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

Simplify

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

Simplify

1 \cdot u : u

1 \cdot u

Multiply:

1 \cdot u = u

= u

u^{2} - 1 = u

u^{2} - 1 = u

u^{2} - 1 = u

Solve

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

u^{2} - 1 = u

Move

u

to the left side

u^{2} - 1 = u

Subtract

u

from both sides

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

Simplify

u^{2} - 1 - u = 0

u^{2} - 1 - u = 0

Write in the standard form

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

u^{2} - u - 1 = 0

Solve with the quadratic formula

u^{2} - u - 1 = 0

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

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

Separate the solutions

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

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

\frac{- ( - 1 ) + 5}{2 \cdot 1}

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 + 5}{2}

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

\frac{- ( - 1 ) - 5}{2 \cdot 1}

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 - 5}{2}

The solutions to the quadratic equation are:

u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

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 = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

Substitute back

u = e^{a},

solve for

a

Solve

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

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

Apply exponent rules

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

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{a}) = a

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

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

Solve

e^{a} = \frac{1 - 5}{2} :

No Solution for

a \in R

e^{a} = \frac{1 - 5}{2}

a^{f (a)}

cannot be zero or negative for

a \in R

No Solution for a \in R

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

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

Popular Examples

sin(x)=(-1}{\frac{sqrt(5))/2}

sin (x) = \frac{- 1}{\frac{5}{2}}

2csc^2(θ)-3cot(θ)+3=0

2 csc^{2} (θ) - 3 cot (θ) + 3 = 0

(2sin(x)+1)/(sqrt(cos(x)))=0

\frac{2 sin ( x ) + 1}{cos ( x )} = 0

sin(x)=0.03725

sin (x) = 0.03725

tan(x)=2.747

tan (x) = 2.747

Frequently Asked Questions (FAQ)

What is the general solution for sinh(a)= 1/2 ?
The general solution for sinh(a)= 1/2 is a=ln((1+sqrt(5))/2)