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Popular Trigonometry >

cosh(θ)= 8/3 \land θ<0,sinh(θ)

Solution

cosh (θ) = \frac{8}{3} and θ < 0, sinh (θ)

Solution

θ = ln (\frac{8 - 55}{3})

Decimal

θ = - 1.63680 \dots

Solution steps

cosh (θ) = \frac{8}{3} and θ < 0

cosh (θ) = \frac{8}{3} : θ = ln (\frac{8 + 55}{3}), θ = ln (\frac{8 - 55}{3})

cosh (θ) = \frac{8}{3}

Rewrite using trig identities

cosh (θ) = \frac{8}{3}

Use the Hyperbolic identity:

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

\frac{e ^{θ} + e ^{- θ}}{2} = \frac{8}{3}

\frac{e ^{θ} + e ^{- θ}}{2} = \frac{8}{3}

\frac{e ^{θ} + e ^{- θ}}{2} = \frac{8}{3} : θ = ln (\frac{8 + 55}{3}), θ = ln (\frac{8 - 55}{3})

\frac{e ^{θ} + e ^{- θ}}{2} = \frac{8}{3}

Apply fraction cross multiply: if

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

then

a \cdot d = b \cdot c

(e^{θ} + e^{- θ}) \cdot 3 = 2 \cdot 8

Simplify

(e^{θ} + e^{- θ}) \cdot 3 = 16

Apply exponent rules

(e^{θ} + e^{- θ}) \cdot 3 = 16

Apply exponent rule:

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

e^{- θ} = (e^{θ})^{- 1}

(e^{θ} + (e^{θ})^{- 1}) \cdot 3 = 16

(e^{θ} + (e^{θ})^{- 1}) \cdot 3 = 16

Rewrite the equation with

e^{θ} = u

(u + (u)^{- 1}) \cdot 3 = 16

Solve

(u + u^{- 1}) \cdot 3 = 16 : u = \frac{8 + 55}{3}, u = \frac{8 - 55}{3}

(u + u^{- 1}) \cdot 3 = 16

Refine

(u + \frac{1}{u}) \cdot 3 = 16

Simplify

(u + \frac{1}{u}) \cdot 3 : 3 (u + \frac{1}{u})

(u + \frac{1}{u}) \cdot 3

Apply the commutative law:

(u + \frac{1}{u}) \cdot 3 = 3 (u + \frac{1}{u})

3 (u + \frac{1}{u})

3 (u + \frac{1}{u}) = 16

Expand

3 (u + \frac{1}{u}) : 3 u + \frac{3}{u}

3 (u + \frac{1}{u})

Apply the distributive law:

a (b + c) = ab + a c

a = 3, b = u, c = \frac{1}{u}

= 3 u + 3 \cdot \frac{1}{u}

3 \cdot \frac{1}{u} = \frac{3}{u}

3 \cdot \frac{1}{u}

Multiply fractions:

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

= \frac{1 \cdot 3}{u}

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3}{u}

= 3 u + \frac{3}{u}

3 u + \frac{3}{u} = 16

Multiply both sides by

u

3 u + \frac{3}{u} = 16

Multiply both sides by

u

3 uu + \frac{3}{u} u = 16 u

Simplify

3 uu + \frac{3}{u} u = 16 u

Simplify

3 uu : 3 u^{2}

3 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= 3 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 3 u^{2}

Simplify

\frac{3}{u} u : 3

\frac{3}{u} u

Multiply fractions:

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

= \frac{3 u}{u}

Cancel the common factor:

u

= 3

3 u^{2} + 3 = 16 u

3 u^{2} + 3 = 16 u

3 u^{2} + 3 = 16 u

Solve

3 u^{2} + 3 = 16 u : u = \frac{8 + 55}{3}, u = \frac{8 - 55}{3}

3 u^{2} + 3 = 16 u

Move

16 u

to the left side

3 u^{2} + 3 = 16 u

Subtract

16 u

from both sides

3 u^{2} + 3 - 16 u = 16 u - 16 u

Simplify

3 u^{2} + 3 - 16 u = 0

3 u^{2} + 3 - 16 u = 0

Write in the standard form

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

3 u^{2} - 16 u + 3 = 0

Solve with the quadratic formula

3 u^{2} - 16 u + 3 = 0

Quadratic Equation Formula:

For

a = 3, b = - 16, c = 3

u_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

(- 16)^{2} - 4 \cdot 3 \cdot 3 = 255

(- 16)^{2} - 4 \cdot 3 \cdot 3

Apply exponent rule:

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

n

is even

(- 16)^{2} = 1 6^{2}

= 1 6^{2} - 4 \cdot 3 \cdot 3

Multiply the numbers:

4 \cdot 3 \cdot 3 = 36

= 1 6^{2} - 36

1 6^{2} = 256

= 256 - 36

Subtract the numbers:

256 - 36 = 220

= 220

Prime factorization of

220 : 2^{2} \cdot 5 \cdot 11

220

220

divides by

2220 = 110 \cdot 2

= 2 \cdot 110

110

divides by

2110 = 55 \cdot 2

= 2 \cdot 2 \cdot 55

55

divides by

555 = 11 \cdot 5

= 2 \cdot 2 \cdot 5 \cdot 11

2, 5, 11

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 5 \cdot 11

= 2^{2} \cdot 5 \cdot 11

= 2^{2} \cdot 5 \cdot 11

Apply radical rule:

= 2^{2} 5 \cdot 11

Apply radical rule:

2^{2} = 2

= 2 5 \cdot 11

Refine

= 255

u_{1, 2} = \frac{- ( - 16 ) \pm 2 55}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- ( - 16 ) + 2 55}{2 \cdot 3}, u_{2} = \frac{- ( - 16 ) - 2 55}{2 \cdot 3}

u = \frac{- ( - 16 ) + 2 55}{2 \cdot 3} : \frac{8 + 55}{3}

\frac{- ( - 16 ) + 2 55}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{16 + 2 55}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{16 + 2 55}{6}

Factor

16 + 255 : 2 (8 + 55)

16 + 255

Rewrite as

= 2 \cdot 8 + 255

Factor out common term

2

= 2 (8 + 55)

= \frac{2 ( 8 + 55 )}{6}

Cancel the common factor:

2

= \frac{8 + 55}{3}

u = \frac{- ( - 16 ) - 2 55}{2 \cdot 3} : \frac{8 - 55}{3}

\frac{- ( - 16 ) - 2 55}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{16 - 2 55}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{16 - 2 55}{6}

Factor

16 - 255 : 2 (8 - 55)

16 - 255

Rewrite as

= 2 \cdot 8 - 255

Factor out common term

2

= 2 (8 - 55)

= \frac{2 ( 8 - 55 )}{6}

Cancel the common factor:

2

= \frac{8 - 55}{3}

The solutions to the quadratic equation are:

u = \frac{8 + 55}{3}, u = \frac{8 - 55}{3}

u = \frac{8 + 55}{3}, u = \frac{8 - 55}{3}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

(u + u^{- 1}) 3

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = \frac{8 + 55}{3}, u = \frac{8 - 55}{3}

u = \frac{8 + 55}{3}, u = \frac{8 - 55}{3}

Substitute back

u = e^{θ},

solve for

θ

Solve

e^{θ} = \frac{8 + 55}{3} : θ = ln (\frac{8 + 55}{3})

e^{θ} = \frac{8 + 55}{3}

Apply exponent rules

e^{θ} = \frac{8 + 55}{3}

f (x) = g (x)

, then

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

ln (e^{θ}) = ln (\frac{8 + 55}{3})

Apply log rule:

ln (e^{a}) = a

ln (e^{θ}) = θ

θ = ln (\frac{8 + 55}{3})

θ = ln (\frac{8 + 55}{3})

Solve

e^{θ} = \frac{8 - 55}{3} : θ = ln (\frac{8 - 55}{3})

e^{θ} = \frac{8 - 55}{3}

Apply exponent rules

e^{θ} = \frac{8 - 55}{3}

f (x) = g (x)

, then

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

ln (e^{θ}) = ln (\frac{8 - 55}{3})

Apply log rule:

ln (e^{a}) = a

ln (e^{θ}) = θ

θ = ln (\frac{8 - 55}{3})

θ = ln (\frac{8 - 55}{3})

θ = ln (\frac{8 + 55}{3}), θ = ln (\frac{8 - 55}{3})

θ = ln (\frac{8 + 55}{3}), θ = ln (\frac{8 - 55}{3})

Combine the intervals

(θ = ln (\frac{8 - 55}{3}) or θ = ln (\frac{8 + 55}{3})) and θ < 0

Merge Overlapping Intervals

θ = ln (\frac{8 - 55}{3}) or θ = ln (\frac{8 + 55}{3}) and θ < 0

The intersection of two intervals is the set of numbers which are in both intervals

θ = ln (\frac{8 - 55}{3}) or θ = ln (\frac{8 + 55}{3})

and

θ < 0

θ = ln (\frac{8 - 55}{3})

θ = ln (\frac{8 - 55}{3})

Graph

Popular Examples

cos(θ)=(sqrt(3))/2 \land csc(θ)<0