Is prove-1/2 =sin(-pi/(12))sqrt(2+\sqrt{3)} ?

The answer to whether prove-1/2 =sin(-pi/(12))sqrt(2+\sqrt{3)} is True

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Popular Trigonometry >

prove-1/2 =sin(-pi/(12))sqrt(2+\sqrt{3)}

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

sin (- \frac{π}{12}) 2 + 3

Use the negative angle identity:

sin (- x) = - sin (x)

= 2 + 3 (- sin (\frac{π}{12}))

Expand

(- sin (\frac{π}{12})) 2 + 3 : - \frac{1}{2}

(- sin (\frac{π}{12})) 2 + 3

Remove parentheses:

(- a) = - a

= - sin (\frac{π}{12}) 2 + 3

sin (\frac{π}{12}) = \frac{6 - 2}{4}

sin (\frac{π}{12})

Rewrite using trig identities:

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

sin (\frac{π}{12})

Write

sin (\frac{π}{12})

sin (\frac{π}{4} - \frac{π}{6})

= sin (\frac{π}{4} - \frac{π}{6})

Use the Angle Difference identity:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

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

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

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}

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{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:

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} \cdot \frac{1}{2}

Simplify

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

\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{6 - 2}{4} 2 + 3

Simplify

- \frac{6 - 2}{4} 2 + 3

Multiply fractions:

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

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

Expand

(6 - 2) 2 + 3 : 2

(6 - 2) 2 + 3

Apply the distributive law:

a (b - c) = ab - a c

a = 2 + 3, b = 6, c = 2

= 2 + 3 6 - 2 + 3 2

= 6 2 + 3 - 2 2 + 3

Simplify

6 2 + 3 - 2 2 + 3 : 2

6 2 + 3 - 2 2 + 3

6 2 + 3 = 3 + 3

6 2 + 3

Apply radical rule:

a b = a \cdot b

6 2 + 3 = 6 (2 + 3)

= 6 (2 + 3)

Expand

6 (2 + 3) : 12 + 63

6 (2 + 3)

Apply the distributive law:

a (b + c) = ab + a c

a = 6, b = 2, c = 3

= 6 \cdot 2 + 63

Multiply the numbers:

6 \cdot 2 = 12

= 12 + 63

= 12 + 63

12 + 63 = 3 + 3

12 + 63

= 3 + 63 + 9

= (3)^{2} + 63 + (9)^{2}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= (3)^{2} + 63 + 3^{2}

23 \cdot 3 = 63

23 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 63

= (3)^{2} + 23 \cdot 3 + 3^{2}

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

(3)^{2} + 23 \cdot 3 + 3^{2} = (3 + 3)^{2}

= (3 + 3)^{2}

Apply radical rule:

(3 + 3)^{2} = 3 + 3

= 3 + 3

= 3 + 3

2 2 + 3 = 3 + 1

2 2 + 3

Apply radical rule:

a b = a \cdot b

2 2 + 3 = 2 (2 + 3)

= 2 (2 + 3)

Expand

2 (2 + 3) : 4 + 23

2 (2 + 3)

Apply the distributive law:

a (b + c) = ab + a c

a = 2, b = 2, c = 3

= 2 \cdot 2 + 23

Multiply the numbers:

2 \cdot 2 = 4

= 4 + 23

= 4 + 23

4 + 23 = 3 + 1

4 + 23

= 3 + 23 + 1

= (3)^{2} + 23 + (1)^{2}

1 = 1

1

Apply rule

1 = 1

= 1

= (3)^{2} + 23 + 1^{2}

23 \cdot 1 = 23

23 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 23

= (3)^{2} + 23 \cdot 1 + 1^{2}

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

(3)^{2} + 23 \cdot 1 + 1^{2} = (3 + 1)^{2}

= (3 + 1)^{2}

Apply radical rule:

(3 + 1)^{2} = 3 + 1

= 3 + 1

= 3 + 1

= 3 + 3 - (1 + 3)

- (3 + 1) : - 3 - 1

- (3 + 1)

Distribute parentheses

= - (3) - (1)

Apply minus-plus rules

+ (- a) = - a

= - 3 - 1

= 3 + 3 - 3 - 1

Simplify

3 + 3 - 3 - 1 : 2

3 + 3 - 3 - 1

Add similar elements:

3 - 3 = 0

= 3 - 1

Subtract the numbers:

3 - 1 = 2

= 2

= 2

= 2

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

= - \frac{1}{2}

= - \frac{1}{2}

= - \frac{1}{2}

We showed that the two sides could take the same form

\Rightarrow True

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Is prove-1/2 =sin(-pi/(12))sqrt(2+\sqrt{3)} ?
The answer to whether prove-1/2 =sin(-pi/(12))sqrt(2+\sqrt{3)} is True