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

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

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-1-2arctan(-1)

- 1 - 2 arctan (- 1)

- 1 + \frac{π}{2}

Decimal Notation

32.7

Solution steps

- 1 - 2 arctan (- 1)

Rewrite using trig identities:

arctan (- 1) = - \frac{π}{4}

arctan (- 1)

Use the following property:

arctan (- x) = - arctan (x)

arctan (- 1) = - arctan (1)

= - arctan (1)

Use the following trivial identity:

arctan (1) = \frac{π}{4}

arctan (1)

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}{4}

= - \frac{π}{4}

= - 1 - 2 (- \frac{π}{4})

Simplify

- 1 - 2 (- \frac{π}{4}) : - 1 + \frac{π}{2}

- 1 - 2 (- \frac{π}{4})

Apply rule

- (- a) = a

= - 1 + 2 \cdot \frac{π}{4}

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}

= - 1 + \frac{π}{2}

= - 1 + \frac{π}{2}

\frac{8}{sin ( 3 0 ^{\circ} )}

tan (arcsin (\frac{- 3}{4}))

cos (3 0^{\circ}) + sin (3 0^{\circ})

arctan (32)

12 \cdot cos (4 0^{\circ})