What is the value of sqrt(1-cos^2(15)) ?

The value of sqrt(1-cos^2(15)) is (sqrt(-\sqrt{3)+2})/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sqrt(1-cos^2(15))

Solution

1 - cos^{2} (1 5^{\circ})

Solution

\frac{- 3 + 2}{2}

Decimal Notation

0.25881 \dots

Solution steps

1 - cos^{2} (1 5^{\circ})

Rewrite using trig identities:

cos^{2} (1 5^{\circ}) = \frac{1 + cos ( 3 0 ^{\circ} )}{2}

cos^{2} (1 5^{\circ})

Use the following identity:

cos^{2} (θ) = \frac{1 + cos ( 2 θ )}{2}

Use the Double Angle identity

cos (2 θ) = 2 cos^{2} (θ) - 1

Switch sides

2 cos^{2} (θ) - 1 = cos (2 θ)

Add

1

to both sides

2 sin^{2} (θ) = 1 + cos (2 θ)

Divide both sides by

2

cos^{2} (θ) = \frac{1 + cos ( 2 θ )}{2}

= \frac{1 + cos ( 2 \cdot 1 5 ^{\circ} )}{2}

Simplify

= \frac{1 + cos ( 3 0 ^{\circ} )}{2}

= 1 - \frac{1 + cos ( 3 0 ^{\circ} )}{2}

Simplify

= \frac{- cos ( 3 0 ^{\circ} ) + 1}{2}

Use the following trivial identity:

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

cos (3 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{3}{2}

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

Simplify

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

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

\frac{- \frac{3}{2} + 1}{2} = \frac{- 3 + 2}{4}

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

Join

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

- \frac{3}{2} + 1

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

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

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

Apply the fraction rule:

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

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

Multiply the numbers:

2 \cdot 2 = 4

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

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

Apply radical rule:

assuming

a \geq 0, b \geq 0

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

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

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

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

Popular Examples

5tan(65)

5 tan (6 5^{\circ})

6sin(pi/4)cos(pi/6)

6 sin (\frac{π}{4}) cos (\frac{π}{6})

arctan(tan(-(5pi)/3))

arctan (tan (- \frac{5 π}{3}))

2-(sin(2))/(cos(2))

2 - \frac{sin ( 2 )}{cos ( 2 )}

8/(tan(40))

\frac{8}{tan ( 4 0 ^{\circ} )}

Frequently Asked Questions (FAQ)

What is the value of sqrt(1-cos^2(15)) ?
The value of sqrt(1-cos^2(15)) is (sqrt(-\sqrt{3)+2})/2