What is the value of -ln|cos(pi/4)| ?

The value of -ln|cos(pi/4)| is 1/2 ln(2)

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-ln|cos(pi/4)|

- ln cos (\frac{π}{4})

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

Decimal Notation

0.34657 \dots

Solution steps

- ln cos (\frac{π}{4})

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{2}{2}

cos (\frac{π}{4})

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

= - ln \frac{2}{2}

Simplify

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

- ln \frac{2}{2}

Simplify

\frac{2}{2} : \frac{2}{2}

\frac{2}{2}

Apply absolute rule:

∣ a ∣ = a, a \geq 0

\frac{2}{2} = \frac{2}{2}

= \frac{2}{2}

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

Simplify

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

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

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

Apply rule

- (- a) = a

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

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

arccos (\frac{3}{33})

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\frac{sin ( 1 5 ^{\circ} ) \cdot 340}{sin ( 6 0 ^{\circ} )}

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\frac{sin ( 3 0 ^{\circ} )}{1}