What is the value of arccos(sin(-(7pi)/6)) ?

The value of arccos(sin(-(7pi)/6)) is pi/3

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arccos(sin(-(7pi)/6))

arccos (sin (- \frac{7 π}{6}))

\frac{π}{3}

Decimal Notation

60

Solution steps

arccos (sin (- \frac{7 π}{6}))

Use the following property:

sin (- x) = - sin (x)

sin (- \frac{7 π}{6}) = - sin (\frac{7 π}{6})

= arccos (- sin (\frac{7 π}{6}))

Rewrite using trig identities:

sin (\frac{7 π}{6}) = - sin (\frac{5 π}{6})

sin (\frac{7 π}{6})

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{7 π}{6})

Simplify:

2 π - \frac{7 π}{6} = \frac{5 π}{6}

2 π - \frac{7 π}{6}

Convert element to fraction:

2 π = \frac{2 π 6}{6}

= \frac{2 π 6}{6} - \frac{7 π}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{2 π 6 - 7 π}{6}

2 π 6 - 7 π = 5 π

2 π 6 - 7 π

Multiply the numbers:

2 \cdot 6 = 12

= 12 π - 7 π

Add similar elements:

12 π - 7 π = 5 π

= 5 π

= \frac{5 π}{6}

= - sin (\frac{5 π}{6})

= arccos (- (- sin (\frac{5 π}{6})))

Simplify

= arccos (sin (\frac{5 π}{6}))

Rewrite using trig identities:

sin (\frac{5 π}{6}) = sin (\frac{π}{6})

sin (\frac{5 π}{6})

Use the basic trigonometric identity:

sin (x) = sin (π - x)

= sin (π - \frac{5 π}{6})

Simplify:

π - \frac{5 π}{6} = \frac{π}{6}

π - \frac{5 π}{6}

Convert element to fraction:

π = \frac{π 6}{6}

= \frac{π 6}{6} - \frac{5 π}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{π 6 - 5 π}{6}

Add similar elements:

6 π - 5 π = π

= \frac{π}{6}

= sin (\frac{π}{6})

= arccos (sin (\frac{π}{6}))

Rewrite using trig identities:

sin (\frac{π}{6}) = cos (\frac{π}{3})

sin (\frac{π}{6})

cos (\frac{π}{2} - x) = sin (x)

= cos (\frac{π}{2} - \frac{π}{6})

= cos (\frac{π}{3})

= arccos (cos (\frac{π}{3}))

Use the inverse trig property:

\frac{π}{3}

arccos (cos (\frac{π}{3}))

For

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

0 \leq \frac{π}{3} \leq π

= \frac{π}{3}

= \frac{π}{3}

arctan (\frac{2}{- 3})

sin (7 5^{\circ}) cos (1 5^{\circ}) + cos (7 5^{\circ}) sin (1 5^{\circ})

sin (arctan (- 5))

tan (\frac{2}{2})

arccos (- 2)

What is the value of arccos(sin(-(7pi)/6)) ?
The value of arccos(sin(-(7pi)/6)) is pi/3