What is the value of 2(-1)^2+2arcsin(-1) ?

The value of 2(-1)^2+2arcsin(-1) is 2-pi

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2(-1)^2+2arcsin(-1)

2 (- 1)^{2} + 2 arcsin (- 1)

2 - π

Decimal Notation

- 65.41

Solution steps

2 (- 1)^{2} + 2 arcsin (- 1)

Rewrite using trig identities:

arcsin (- 1) = - \frac{π}{2}

arcsin (- 1)

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- 1) = - arcsin (1)

= - arcsin (1)

Use the following trivial identity:

arcsin (1) = \frac{π}{2}

arcsin (1)

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}{2}

= - \frac{π}{2}

= 2 (- 1)^{2} + 2 (- \frac{π}{2})

Simplify

2 (- 1)^{2} + 2 (- \frac{π}{2}) : 2 - π

2 (- 1)^{2} + 2 (- \frac{π}{2})

Remove parentheses:

(- a) = - a

= 2 (- 1)^{2} - 2 \cdot \frac{π}{2}

2 (- 1)^{2} = 2

2 (- 1)^{2}

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

= 2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 2

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

Multiply fractions:

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

= \frac{π 2}{2}

Cancel the common factor:

2

= π

= 2 - π

= 2 - π

arccos (\frac{13}{6 \cdot 86})

\frac{π}{3} cos (\frac{π}{3})

\frac{cos ^{2} ( \frac{π}{4} )}{2}

arctan (\frac{6}{17})

29 \cdot cos (- 5.9 1^{\circ})

What is the value of 2(-1)^2+2arcsin(-1) ?
The value of 2(-1)^2+2arcsin(-1) is 2-pi