Is sin((11pi}{12})=sin(\frac{3pi)/4+pi/6) ?

The answer to whether sin((11pi}{12})=sin(\frac{3pi)/4+pi/6) is True

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prove sin((11pi}{12})=sin(\frac{3pi)/4+pi/6)

Solution

prove

sin (\frac{11 π}{12}) = sin (\frac{3 π}{4} + \frac{π}{6})

Solution

True

Solution steps

sin (\frac{11 π}{12}) = sin (\frac{3 π}{4} + \frac{π}{6})

Manipulating left side

sin (\frac{11 π}{12})

Simplify

sin (\frac{11 π}{12}) : \frac{6 - 2}{4}

sin (\frac{11 π}{12})

Rewrite using trig identities:

sin (\frac{3 π}{4}) cos (\frac{π}{6}) + cos (\frac{3 π}{4}) sin (\frac{π}{6})

sin (\frac{11 π}{12})

Write

sin (\frac{11 π}{12})

sin (\frac{3 π}{4} + \frac{π}{6})

= sin (\frac{3 π}{4} + \frac{π}{6})

Use the Angle Sum identity:

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

= sin (\frac{3 π}{4}) cos (\frac{π}{6}) + cos (\frac{3 π}{4}) sin (\frac{π}{6})

= sin (\frac{3 π}{4}) cos (\frac{π}{6}) + cos (\frac{3 π}{4}) sin (\frac{π}{6})

Use the following trivial identity:

sin (\frac{3 π}{4}) = \frac{2}{2}

sin (\frac{3 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}

Use the following trivial identity:

cos (\frac{π}{6}) = \frac{3}{2}

cos (\frac{π}{6})

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{3}{2}

Use the following trivial identity:

cos (\frac{3 π}{4}) = - \frac{2}{2}

cos (\frac{3 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:

sin (\frac{π}{6}) = \frac{1}{2}

sin (\frac{π}{6})

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{1}{2}

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

Simplify

\frac{2}{2} \cdot \frac{3}{2} + (- \frac{2}{2}) \frac{1}{2} : \frac{6 - 2}{4}

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

Remove parentheses:

(- a) = - a

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

\frac{2}{2} \cdot \frac{3}{2} = \frac{6}{4}

\frac{2}{2} \cdot \frac{3}{2}

Multiply fractions:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 3}{4}

Simplify

23 : 6

23

Apply radical rule:

a b = a \cdot b

23 = 2 \cdot 3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

= \frac{6}{4}

\frac{2}{2} \cdot \frac{1}{2} = \frac{2}{4}

\frac{2}{2} \cdot \frac{1}{2}

Multiply fractions:

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

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

Multiply:

2 \cdot 1 = 2

= \frac{2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

= \frac{6}{4} - \frac{2}{4}

Apply rule

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

= \frac{6 - 2}{4}

= \frac{6 - 2}{4}

= \frac{- 2 + 6}{4}

Manipulating right side

sin (\frac{3 π}{4} + \frac{π}{6})

Rewrite using trig identities

sin (\frac{3 π}{4} + \frac{π}{6})

Use the Angle Sum identity:

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

= sin (\frac{3 π}{4}) cos (\frac{π}{6}) + cos (\frac{3 π}{4}) sin (\frac{π}{6})

sin (\frac{3 π}{4}) cos (\frac{π}{6}) + cos (\frac{3 π}{4}) sin (\frac{π}{6}) = \frac{6 - 2}{4}

sin (\frac{3 π}{4}) cos (\frac{π}{6}) + cos (\frac{3 π}{4}) sin (\frac{π}{6})

sin (\frac{3 π}{4}) cos (\frac{π}{6}) = \frac{6}{4}

sin (\frac{3 π}{4}) cos (\frac{π}{6})

Simplify

sin (\frac{3 π}{4}) : \frac{2}{2}

sin (\frac{3 π}{4})

Use the following trivial identity:

sin (\frac{3 π}{4}) = \frac{2}{2}

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{2}{2} cos (\frac{π}{6})

Simplify

cos (\frac{π}{6}) : \frac{3}{2}

cos (\frac{π}{6})

Use the following trivial identity:

cos (\frac{π}{6}) = \frac{3}{2}

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{3}{2}

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

Multiply fractions:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 3}{4}

Simplify

23 : 6

23

Apply radical rule:

a b = a \cdot b

23 = 2 \cdot 3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

= \frac{6}{4}

cos (\frac{3 π}{4}) sin (\frac{π}{6}) = - \frac{2}{4}

cos (\frac{3 π}{4}) sin (\frac{π}{6})

Simplify

cos (\frac{3 π}{4}) : - \frac{2}{2}

cos (\frac{3 π}{4})

Use the following trivial identity:

cos (\frac{3 π}{4}) = - \frac{2}{2}

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}

= - \frac{2}{2} sin (\frac{π}{6})

Simplify

sin (\frac{π}{6}) : \frac{1}{2}

sin (\frac{π}{6})

Use the following trivial identity:

sin (\frac{π}{6}) = \frac{1}{2}

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{1}{2}

= - \frac{1}{2} \cdot \frac{2}{2}

Multiply fractions:

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

= - \frac{2 \cdot 1}{2 \cdot 2}

Multiply:

2 \cdot 1 = 2

= - \frac{2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= - \frac{2}{4}

= \frac{6}{4} - \frac{2}{4}

Apply rule

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

= \frac{6 - 2}{4}

= \frac{6 - 2}{4}

= \frac{6 - 2}{4}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is sin((11pi}{12})=sin(\frac{3pi)/4+pi/6) ?
The answer to whether sin((11pi}{12})=sin(\frac{3pi)/4+pi/6) is True