What is the value of arctan(((sqrt(3))/2)/(-1/2)) ?

The value of arctan(((sqrt(3))/2)/(-1/2)) is -pi/3

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arctan(((sqrt(3))/2)/(-1/2))

arctan (\frac{\frac{3}{2}}{- \frac{1}{2}})

- \frac{π}{3}

Decimal Notation

- 60

Solution steps

arctan (\frac{\frac{3}{2}}{- \frac{1}{2}})

Simplify:

\frac{\frac{3}{2}}{- \frac{1}{2}} = - 3

\frac{\frac{3}{2}}{- \frac{1}{2}}

Apply the fraction rule:

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

= - \frac{\frac{3}{2}}{\frac{1}{2}}

Divide fractions:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= - \frac{3 \cdot 2}{2 \cdot 1}

Refine

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

Cancel the common factor:

2

= - 3

= arctan (- 3)

Use the following property:

arctan (- x) = - arctan (x)

arctan (- 3) = - arctan (3)

= - arctan (3)

Use the following trivial identity:

arctan (3) = \frac{π}{3}

arctan (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}{3}

= - \frac{π}{3}

tan (\frac{π}{4} - 0.2) - tan (\frac{π}{4})

tan (arccos (\frac{20}{29}))

9.8 \cdot sin (6 0^{\circ})

cos (2 arcsin (\frac{3}{2}))

tan^{2} (4 5^{\circ} + 1 5^{\circ})

What is the value of arctan(((sqrt(3))/2)/(-1/2)) ?
The value of arctan(((sqrt(3))/2)/(-1/2)) is -pi/3