What is the value of arccos(sin(-(4pi)/3)) ?

The value of arccos(sin(-(4pi)/3)) is pi/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

arccos(sin(-(4pi)/3))

arccos (sin (- \frac{4 π}{3}))

\frac{π}{6}

Decimal Notation

30

Solution steps

arccos (sin (- \frac{4 π}{3}))

Use the following property:

sin (- x) = - sin (x)

sin (- \frac{4 π}{3}) = - sin (\frac{4 π}{3})

= arccos (- sin (\frac{4 π}{3}))

Rewrite using trig identities:

sin (\frac{4 π}{3}) = - sin (\frac{2 π}{3})

sin (\frac{4 π}{3})

Use the following identity:

sin (x) = - sin (2 π - x)

sin (x)

Use the following property:

sin (θ) = - sin (- θ)

sin (x) = - sin (- x)

= - sin (- x)

Apply the periodicity of

sin

sin (2 π + θ) = sin (θ)

- sin (- x) = - sin (2 π - x)

= - sin (2 π - x)

= - sin (2 π - \frac{4 π}{3})

Simplify:

2 π - \frac{4 π}{3} = \frac{2 π}{3}

2 π - \frac{4 π}{3}

Convert element to fraction:

2 π = \frac{2 π 3}{3}

= \frac{2 π 3}{3} - \frac{4 π}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{2 π 3 - 4 π}{3}

2 π 3 - 4 π = 2 π

2 π 3 - 4 π

Multiply the numbers:

2 \cdot 3 = 6

= 6 π - 4 π

Add similar elements:

6 π - 4 π = 2 π

= 2 π

= \frac{2 π}{3}

= - sin (\frac{2 π}{3})

= arccos (- (- sin (\frac{2 π}{3})))

Simplify

= arccos (sin (\frac{2 π}{3}))

Rewrite using trig identities:

sin (\frac{2 π}{3}) = sin (\frac{π}{3})

sin (\frac{2 π}{3})

Use the basic trigonometric identity:

sin (x) = sin (π - x)

= sin (π - \frac{2 π}{3})

Simplify:

π - \frac{2 π}{3} = \frac{π}{3}

π - \frac{2 π}{3}

Convert element to fraction:

π = \frac{π 3}{3}

= \frac{π 3}{3} - \frac{2 π}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{π 3 - 2 π}{3}

Add similar elements:

3 π - 2 π = π

= \frac{π}{3}

= sin (\frac{π}{3})

= arccos (sin (\frac{π}{3}))

Rewrite using trig identities:

sin (\frac{π}{3}) = cos (\frac{π}{6})

sin (\frac{π}{3})

cos (\frac{π}{2} - x) = sin (x)

= cos (\frac{π}{2} - \frac{π}{3})

= cos (\frac{π}{6})

= arccos (cos (\frac{π}{6}))

Use the inverse trig property:

\frac{π}{6}

arccos (cos (\frac{π}{6}))

For

0 \leq x \leq π, a rccos (cos (x)) = x

0 \leq \frac{π}{6} \leq π

= \frac{π}{6}

= \frac{π}{6}

sin (2 arctan (\frac{4}{3}))

sin (21 5^{\circ})

9 sin (\frac{2 π}{3})

6 cos (π)

sin (2 arcsin (\frac{7}{25}))

What is the value of arccos(sin(-(4pi)/3)) ?
The value of arccos(sin(-(4pi)/3)) is pi/6