What is the value of arctan(-1/(sqrt(3)))+pi ?

The value of arctan(-1/(sqrt(3)))+pi is -pi/6+pi

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arctan(-1/(sqrt(3)))+pi

arctan (- \frac{1}{3}) + π

- \frac{π}{6} + π

Decimal Notation

150

Solution steps

arctan (- \frac{1}{3}) + π

Rewrite using trig identities:

arctan (- \frac{1}{3}) = - \frac{π}{6}

arctan (- \frac{1}{3})

arctan (- \frac{1}{3}) = arctan (- \frac{3}{3})

arctan (- \frac{1}{3})

\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}

= arctan (- \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}

= - \frac{π}{6} + π

arccos (0.26)

6 cos (2 π)

arccos (0.51)

cos (\frac{π}{7}) cos (\frac{π}{5}) - sin (\frac{π}{7}) sin (\frac{π}{5})

arccos (0.46)

What is the value of arctan(-1/(sqrt(3)))+pi ?
The value of arctan(-1/(sqrt(3)))+pi is -pi/6+pi