What is the value of arccos(-(sqrt(2))/2)+arctan((sqrt(3))/3) ?

The value of arccos(-(sqrt(2))/2)+arctan((sqrt(3))/3) is (11pi)/(12)

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arccos(-(sqrt(2))/2)+arctan((sqrt(3))/3)

Solution

arccos (- \frac{2}{2}) + arctan (\frac{3}{3})

Solution

\frac{11 π}{12}

Decimal Notation

165

Solution steps

arccos (- \frac{2}{2}) + arctan (\frac{3}{3})

Use the following trivial identity:

arccos (- \frac{2}{2}) = \frac{3 π}{4}

arccos (- \frac{2}{2})

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{3 π}{4}

Use the following trivial identity:

arctan (\frac{3}{3}) = \frac{π}{6}

arctan (\frac{3}{3})

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{6}

= \frac{3 π}{4} + \frac{π}{6}

Simplify

\frac{3 π}{4} + \frac{π}{6} : \frac{11 π}{12}

\frac{3 π}{4} + \frac{π}{6}

Least Common Multiplier of

4, 6 : 12

4, 6

Least Common Multiplier (LCM)

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Prime factorization of

6 : 2 \cdot 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

Multiply each factor the greatest number of times it occurs in either

4

6

= 2 \cdot 2 \cdot 3

Multiply the numbers:

2 \cdot 2 \cdot 3 = 12

= 12

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

12

For

\frac{3 π}{4} :

multiply the denominator and numerator by

3

\frac{3 π}{4} = \frac{3 π 3}{4 \cdot 3} = \frac{9 π}{12}

For

\frac{π}{6} :

multiply the denominator and numerator by

2

\frac{π}{6} = \frac{π 2}{6 \cdot 2} = \frac{π 2}{12}

= \frac{9 π}{12} + \frac{π 2}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{9 π + π 2}{12}

Add similar elements:

9 π + 2 π = 11 π

= \frac{11 π}{12}

= \frac{11 π}{12}

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Frequently Asked Questions (FAQ)

What is the value of arccos(-(sqrt(2))/2)+arctan((sqrt(3))/3) ?
The value of arccos(-(sqrt(2))/2)+arctan((sqrt(3))/3) is (11pi)/(12)