What is the value of sqrt(1-cos^2(120)-cos^2(60)) ?

The value of sqrt(1-cos^2(120)-cos^2(60)) is sqrt(1/2)

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Popular >

sqrt(1-cos^2(120)-cos^2(60))

Solution

1 - cos^{2} (12 0^{\circ}) - cos^{2} (6 0^{\circ})

Solution

\frac{1}{2}

Decimal Notation

0.70710 \dots

Solution steps

1 - cos^{2} (12 0^{\circ}) - cos^{2} (6 0^{\circ})

Use the following trivial identity:

cos (12 0^{\circ}) = - \frac{1}{2}

cos (12 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:

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}

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

Simplify

1 - (- \frac{1}{2})^{2} - (\frac{1}{2})^{2} : \frac{1}{2}

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

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

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Combine the fractions

- \frac{1}{2 ^{2}} - \frac{1}{2 ^{2}} : - \frac{1}{2}

Apply rule

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

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

Subtract the numbers:

- 1 - 1 = - 2

= \frac{- 2}{2 ^{2}}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{2 ^{2}}

Cancel the common factor:

2

= - \frac{1}{2}

= 1 - \frac{1}{2}

Join

1 - \frac{1}{2} : \frac{1}{2}

1 - \frac{1}{2}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

1 \cdot 2 - 1 = 1

1 \cdot 2 - 1

Multiply the numbers:

1 \cdot 2 = 2

= 2 - 1

Subtract the numbers:

2 - 1 = 1

= 1

= \frac{1}{2}

= \frac{1}{2}

= \frac{1}{2}

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Frequently Asked Questions (FAQ)

What is the value of sqrt(1-cos^2(120)-cos^2(60)) ?
The value of sqrt(1-cos^2(120)-cos^2(60)) is sqrt(1/2)