What is the value of sin(-(7pi)/(12)) ?

The value of sin(-(7pi)/(12)) is -(sqrt(2)(sqrt(3)+1))/4

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sin(-(7pi)/(12))

Solution

sin (- \frac{7 π}{12})

Solution

- \frac{2 ( 3 + 1 )}{4}

Decimal Notation

- 0.96592 \dots

Solution steps

sin (- \frac{7 π}{12})

Use the following property:

sin (- x) = - sin (x)

sin (- \frac{7 π}{12}) = - sin (\frac{7 π}{12})

= - sin (\frac{7 π}{12})

Rewrite using trig identities:

sin (\frac{7 π}{12}) = \frac{2 ( 3 + 1 )}{4}

sin (\frac{7 π}{12})

Rewrite using trig identities:

sin (\frac{π}{3}) cos (\frac{π}{4}) + cos (\frac{π}{3}) sin (\frac{π}{4})

sin (\frac{7 π}{12})

Write

sin (\frac{7 π}{12})

sin (\frac{π}{3} + \frac{π}{4})

= sin (\frac{π}{3} + \frac{π}{4})

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= sin (\frac{π}{3}) cos (\frac{π}{4}) + cos (\frac{π}{3}) sin (\frac{π}{4})

= sin (\frac{π}{3}) cos (\frac{π}{4}) + cos (\frac{π}{3}) sin (\frac{π}{4})

Use the following trivial identity:

sin (\frac{π}{3}) = \frac{3}{2}

sin (\frac{π}{3})

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{3}{2}

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{2}{2}

cos (\frac{π}{4})

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

Use the following trivial identity:

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{π}{3})

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{1}{2}

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{2}{2}

sin (\frac{π}{4})

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

= \frac{3}{2} \cdot \frac{2}{2} + \frac{1}{2} \cdot \frac{2}{2}

Simplify

\frac{3}{2} \cdot \frac{2}{2} + \frac{1}{2} \cdot \frac{2}{2} : \frac{2 ( 3 + 1 )}{4}

\frac{3}{2} \cdot \frac{2}{2} + \frac{1}{2} \cdot \frac{2}{2}

Factor out common term

\frac{2}{2}

= \frac{2}{2} (\frac{3}{2} + \frac{1}{2})

\frac{3}{2} + \frac{1}{2} = \frac{3 + 1}{2}

\frac{3}{2} + \frac{1}{2}

Apply rule

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

= \frac{3 + 1}{2}

= \frac{2}{2} \cdot \frac{1 + 3}{2}

Multiply fractions:

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

= \frac{( 3 + 1 ) 2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 ( 1 + 3 )}{4}

= \frac{2 ( 3 + 1 )}{4}

= - \frac{2 ( 3 + 1 )}{4}

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Frequently Asked Questions (FAQ)

What is the value of sin(-(7pi)/(12)) ?
The value of sin(-(7pi)/(12)) is -(sqrt(2)(sqrt(3)+1))/4