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

The value of arctan((-2)/(2sqrt(3))) is -pi/6

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

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

- \frac{π}{6}

Decimal Notation

- 30

Solution steps

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

Simplify:

\frac{- 2}{2 3} = - \frac{3}{3}

\frac{- 2}{2 3}

Apply the fraction rule:

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

= - \frac{2}{2 3}

Divide the numbers:

\frac{2}{2} = 1

= - \frac{1}{3}

Rationalize

- \frac{1}{3} : - \frac{3}{3}

- \frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

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

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= - \frac{3}{3}

= - \frac{3}{3}

= arctan (- \frac{3}{3})

Use the following property:

arctan (- x) = - arctan (x)

arctan (- \frac{3}{3}) = - arctan (\frac{3}{3})

= - arctan (\frac{3}{3})

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{π}{6}

cos (e)

cos (- \frac{11 π}{2})

4 \cdot cos (3 0^{\circ})

\frac{tan ( 1 0 ^{\circ} ) + tan ( 5 0 ^{\circ} )}{1 - tan ( 1 0 ^{\circ} ) tan ( 5 0 ^{\circ} )}

sinh (2 π)

What is the value of arctan((-2)/(2sqrt(3))) ?
The value of arctan((-2)/(2sqrt(3))) is -pi/6