What is the value of cos(2arccos(1/3)) ?

The value of cos(2arccos(1/3)) is -7/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

cos(2arccos(1/3))

cos (2 arccos (\frac{1}{3}))

- \frac{7}{9}

Decimal

- 0.77777 \dots

Solution steps

cos (2 arccos (\frac{1}{3}))

Rewrite using trig identities:

2 cos^{2} (arccos (\frac{1}{3})) - 1

cos (2 arccos (\frac{1}{3}))

Use the Double Angle identity:

cos (2 x) = 2 cos^{2} (x) - 1

= 2 cos^{2} (arccos (\frac{1}{3})) - 1

= 2 cos^{2} (arccos (\frac{1}{3})) - 1

Rewrite using trig identities:

cos (arccos (\frac{1}{3})) = \frac{1}{3}

Use the following identity:

cos (arccos (x)) = x

= \frac{1}{3}

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

Simplify

2 (\frac{1}{3})^{2} - 1 : - \frac{7}{9}

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

2 (\frac{1}{3})^{2} = \frac{2}{9}

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= 2 \cdot \frac{1}{3 ^{2}}

Multiply fractions:

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

= \frac{1 \cdot 2}{3 ^{2}}

Multiply the numbers:

1 \cdot 2 = 2

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

3^{2} = 9

= \frac{2}{9}

= \frac{2}{9} - 1

Convert element to fraction:

1 = \frac{1 \cdot 9}{9}

= - \frac{1 \cdot 9}{9} + \frac{2}{9}

Since the denominators are equal, combine the fractions:

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

= \frac{- 1 \cdot 9 + 2}{9}

- 1 \cdot 9 + 2 = - 7

- 1 \cdot 9 + 2

Multiply the numbers:

1 \cdot 9 = 9

= - 9 + 2

Add/Subtract the numbers:

- 9 + 2 = - 7

= - 7

= \frac{- 7}{9}

Apply the fraction rule:

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

= - \frac{7}{9}

= - \frac{7}{9}

arcsin (cos (\frac{3 π}{5}))

arcsin (\frac{1.52 \cdot sin ( 5 1 ^{\circ} )}{1.33})

\frac{20}{tan ( 1 5 ^{\circ} )}

cos (4 0^{\circ}) + cos (8 0^{\circ}) + cos (10 0^{\circ}) + cos (12 0^{\circ}) + cos (14 0^{\circ})

2 sin (5 0^{\circ}) cos (5 0^{\circ})