What is the value of -2arcsin(0/2+(sqrt(2))/2) ?

The value of -2arcsin(0/2+(sqrt(2))/2) is -pi/2

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-2arcsin(0/2+(sqrt(2))/2)

- 2 arcsin (\frac{0}{2} + \frac{2}{2})

- \frac{π}{2}

Decimal Notation

- 90

Solution steps

- 2 arcsin (\frac{0}{2} + \frac{2}{2})

Use the following trivial identity:

arcsin (\frac{0}{2} + \frac{2}{2}) = \frac{π}{4}

arcsin (\frac{0}{2} + \frac{2}{2})

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{4}

= - 2 \cdot \frac{π}{4}

Simplify

- 2 \cdot \frac{π}{4} : - \frac{π}{2}

- 2 \cdot \frac{π}{4}

Multiply fractions:

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

= - \frac{π 2}{4}

Cancel the common factor:

2

= - \frac{π}{2}

= - \frac{π}{2}

\frac{0.05 \cdot ( 3.1 \cdot cos ( 1 4 ^{\circ} ) ) ( 2 \cdot cos ( 1 4 ^{\circ} ) )}{30}

cos (60 0^{\circ}) - i sin (60 0^{\circ})

4 (i sin (6 0^{\circ}))

- cos (13 5^{\circ})

cos (13 9^{\circ})

What is the value of -2arcsin(0/2+(sqrt(2))/2) ?
The value of -2arcsin(0/2+(sqrt(2))/2) is -pi/2