What is the value of (cos(ln(1^2)))+(107)/(5(1)) ?

The value of (cos(ln(1^2)))+(107)/(5(1)) is 112/5

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(cos(ln(1^2)))+(107)/(5(1))

(cos (ln (1^{2}))) + \frac{107}{5 ( 1 )}

\frac{112}{5}

Decimal Notation

22.4

Solution steps

(cos (ln (1^{2}))) + \frac{107}{5 ( 1 )}

= cos (ln (1^{2})) + \frac{107}{5 \cdot 1}

Simplify:

ln (1^{2}) = 0

ln (1^{2})

Apply rule

1^{a} = 1

= ln (1)

Apply log rule:

lo g_{a} (1) = 0

= 0

cos (0) + \frac{107}{5 \cdot 1} = \frac{5 cos ( 0 ) + 107}{5}

cos (0) + \frac{107}{5 \cdot 1}

Multiply the numbers:

5 \cdot 1 = 5

= cos (0) + \frac{107}{5}

Convert element to fraction:

cos (0) = \frac{cos ( 0 ) 5}{5}

= \frac{cos ( 0 ) \cdot 5}{5} + \frac{107}{5}

Since the denominators are equal, combine the fractions:

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

= \frac{cos ( 0 ) \cdot 5 + 107}{5}

= \frac{5 cos ( 0 ) + 107}{5}

Use the following trivial identity:

cos (0) = 1

cos (0)

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}

= 1

= \frac{5 \cdot 1 + 107}{5}

Simplify

= \frac{112}{5}

tan (arcsec (2))

6 sin (15 0^{\circ})

50 tan (5 3^{\circ})

tan (\frac{5 π}{14})

\frac{84 sin ( 11 2 ^{\circ} )}{sin ( 9 8 ^{\circ} )}

What is the value of (cos(ln(1^2)))+(107)/(5(1)) ?
The value of (cos(ln(1^2)))+(107)/(5(1)) is 112/5