Is arccos(-x)=arccos(x) ?

The answer to whether arccos(-x)=arccos(x) is False

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prove arccos(-x)=arccos(x)

prove

arccos (- x) = arccos (x)

False

Solution steps

arccos (- x) = arccos (x)

Show that the two sides are not equal

For

arccos (- x) = arccos (x)

plug in

x = 1

arccos (- 1) = π (Decimal : Degrees : 3.14159 \dots 18 0^{\circ})

arccos (- 1)

Use the following trivial identity:

arccos (- 1) = π

arccos (- 1)

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}

= π

= π

arccos (1) = 0 (Degrees : 0^{\circ})

arccos (1)

Use the following trivial identity:

arccos (1) = 0

arccos (1)

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}

= 0

= 0

The two sides are not equal

\Rightarrow False

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