What is the value of 1-2sin^2((5pi)/(12)) ?

The value of 1-2sin^2((5pi)/(12)) is 1-(2+sqrt(3))/2

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1-2sin^2((5pi)/(12))

Solution

1 - 2 sin^{2} (\frac{5 π}{12})

Solution

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

Decimal Notation

- 0.86602 \dots

Solution steps

1 - 2 sin^{2} (\frac{5 π}{12})

Rewrite using trig identities:

sin (\frac{5 π}{12}) = \frac{6 + 2}{4}

sin (\frac{5 π}{12})

Rewrite using trig identities:

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

sin (\frac{5 π}{12})

Write

sin (\frac{5 π}{12})

sin (\frac{π}{4} + \frac{π}{6})

= sin (\frac{π}{4} + \frac{π}{6})

Use the Angle Sum 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}

= 1 - 2 (\frac{6 + 2}{4})^{2}

Simplify

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

1 - 2 (\frac{6 + 2}{4})^{2}

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

2 (\frac{6 + 2}{4})^{2}

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

(\frac{6 + 2}{4})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( 6 + 2 ) ^{2}}{4 ^{2}}

(6 + 2)^{2} = 8 + 43

(6 + 2)^{2}

Apply Perfect Square Formula:

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

a = 6, b = 2

= (6)^{2} + 262 + (2)^{2}

Simplify

(6)^{2} + 262 + (2)^{2} : 8 + 43

(6)^{2} + 262 + (2)^{2}

(6)^{2} = 6

(6)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= (6^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 6^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 6

262 = 43

262

Factor integer

6 = 2 \cdot 3

= 2 2 \cdot 3 2

Apply radical rule:

2 \cdot 3 = 23

= 2232

Apply radical rule:

a a = a

22 = 2

= 2 \cdot 23

Multiply the numbers:

2 \cdot 2 = 4

= 43

(2)^{2} = 2

(2)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 2

= 6 + 43 + 2

Add the numbers:

6 + 2 = 8

= 8 + 43

= 8 + 43

= \frac{8 + 4 3}{4 ^{2}}

Factor

8 + 43 : 4 (2 + 3)

8 + 43

Rewrite as

= 4 \cdot 2 + 43

Factor out common term

4

= 4 (2 + 3)

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

Cancel the common factor:

4

= \frac{2 + 3}{4}

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

Multiply fractions:

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

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

Cancel the common factor:

2

= \frac{2 + 3}{2}

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

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

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Frequently Asked Questions (FAQ)

What is the value of 1-2sin^2((5pi)/(12)) ?
The value of 1-2sin^2((5pi)/(12)) is 1-(2+sqrt(3))/2