What is the value of ln(sin(pi/3))-ln(sin(pi/4)) ?

The value of ln(sin(pi/3))-ln(sin(pi/4)) is 1/2 ln(3)-1/2 ln(2)

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ln(sin(pi/3))-ln(sin(pi/4))

Solution

ln (sin (\frac{π}{3})) - ln (sin (\frac{π}{4}))

Solution

\frac{1}{2} ln (3) - \frac{1}{2} ln (2)

Decimal Notation

0.20273 \dots

Solution steps

ln (sin (\frac{π}{3})) - ln (sin (\frac{π}{4}))

Use the following trivial identity:

sin (\frac{π}{3}) = \frac{3}{2}

sin (\frac{π}{3})

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{3}{2}

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{2}{2}

sin (\frac{π}{4})

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

= ln (\frac{3}{2}) - ln (\frac{2}{2})

Simplify

ln (\frac{3}{2}) - ln (\frac{2}{2}) : \frac{1}{2} ln (3) - \frac{1}{2} ln (2)

ln (\frac{3}{2}) - ln (\frac{2}{2})

ln (\frac{3}{2}) = \frac{1}{2} ln (3) - ln (2)

ln (\frac{3}{2})

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

ln (\frac{3}{2}) = ln (3) - ln (2)

= ln (3) - ln (2)

Simplify

ln (3) : \frac{1}{2} ln (3)

ln (3)

Rewrite as

= ln (3^{\frac{1}{2}})

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

= \frac{1}{2} ln (3)

= \frac{1}{2} ln (3) - ln (2)

ln (\frac{2}{2}) = - \frac{1}{2} ln (2)

ln (\frac{2}{2})

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

ln (\frac{2}{2}) = ln (2) - ln (2)

= ln (2) - ln (2)

Simplify

ln (2) : \frac{1}{2} ln (2)

ln (2)

Rewrite as

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

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

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

Factor out common term

ln (2)

= ln (2) (\frac{1}{2} - 1)

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

\frac{1}{2} - 1

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

1 - 1 \cdot 2 = - 1

1 - 1 \cdot 2

Multiply the numbers:

1 \cdot 2 = 2

= 1 - 2

Subtract the numbers:

1 - 2 = - 1

= - 1

= \frac{- 1}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{2}

= - \frac{1}{2} ln (2)

= \frac{1}{2} ln (3) - ln (2) - (- \frac{1}{2} ln (2))

Apply rule

- (- a) = a

= \frac{1}{2} ln (3) - ln (2) + \frac{1}{2} ln (2)

Add similar elements:

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

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

Factor out common term

ln (2)

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

- 1 + \frac{1}{2} = - \frac{1}{2}

- 1 + \frac{1}{2}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

- 1 \cdot 2 + 1 = - 1

- 1 \cdot 2 + 1

Multiply the numbers:

1 \cdot 2 = 2

= - 2 + 1

Add/Subtract the numbers:

- 2 + 1 = - 1

= - 1

= \frac{- 1}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{2}

= - \frac{1}{2} ln (2)

= \frac{1}{2} ln (3) - \frac{1}{2} ln (2)

= \frac{1}{2} ln (3) - \frac{1}{2} ln (2)

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Frequently Asked Questions (FAQ)

What is the value of ln(sin(pi/3))-ln(sin(pi/4)) ?
The value of ln(sin(pi/3))-ln(sin(pi/4)) is 1/2 ln(3)-1/2 ln(2)