What is the value of 1-2sin^2(105) ?

The value of 1-2sin^2(105) is 1-(2+sqrt(3))/2

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1-2sin^2(105)

Solution

1 - 2 sin^{2} (10 5^{\circ})

Solution

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

Decimal Notation

- 0.86602 \dots

Solution steps

1 - 2 sin^{2} (10 5^{\circ})

Rewrite using trig identities:

sin (10 5^{\circ}) = \frac{2 ( 3 + 1 )}{4}

sin (10 5^{\circ})

Rewrite using trig identities:

sin (6 0^{\circ}) cos (4 5^{\circ}) + cos (6 0^{\circ}) sin (4 5^{\circ})

sin (10 5^{\circ})

Write

sin (10 5^{\circ})

sin (6 0^{\circ} + 4 5^{\circ})

= sin (6 0^{\circ} + 4 5^{\circ})

Use the Angle Sum identity:

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

= sin (6 0^{\circ}) cos (4 5^{\circ}) + cos (6 0^{\circ}) sin (4 5^{\circ})

= sin (6 0^{\circ}) cos (4 5^{\circ}) + cos (6 0^{\circ}) sin (4 5^{\circ})

Use the following trivial identity:

sin (6 0^{\circ}) = \frac{3}{2}

sin (6 0^{\circ})

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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 (4 5^{\circ}) = \frac{2}{2}

cos (4 5^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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 (6 0^{\circ}) = \frac{1}{2}

cos (6 0^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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 (4 5^{\circ}) = \frac{2}{2}

sin (4 5^{\circ})

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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}

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

Simplify

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

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

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

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

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

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

Apply exponent rule:

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

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

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

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

Factor

4^{2} : 2^{4}

Factor

4 = 2^{2}

= (2^{2})^{2}

Simplify

(2^{2})^{2} : 2^{4}

(2^{2})^{2}

Apply exponent rule:

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

= 2^{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= 2^{4}

= 2^{4}

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

Cancel the common factor:

2

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

(3 + 1)^{2} = 4 + 23

(3 + 1)^{2}

Apply Perfect Square Formula:

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

a = 3, b = 1

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

Simplify

(3)^{2} + 23 \cdot 1 + 1^{2} : 4 + 23

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

Apply rule

1^{a} = 1

1^{2} = 1

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

(3)^{2} = 3

(3)^{2}

Apply radical rule:

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

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

Apply exponent rule:

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

= 3^{\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

= 3

23 \cdot 1 = 23

23 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 23

= 3 + 23 + 1

Add the numbers:

3 + 1 = 4

= 4 + 23

= 4 + 23

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

Factor

4 + 23 : 2 (2 + 3)

4 + 23

Rewrite as

= 2 \cdot 2 + 23

Factor out common term

2

= 2 (2 + 3)

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

Cancel the common factor:

2

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

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

Multiply fractions:

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

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

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(105) ?
The value of 1-2sin^2(105) is 1-(2+sqrt(3))/2