What is the value of cos(18)sin(-72) ?

The value of cos(18)sin(-72) is (-5-sqrt(5))/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(18)sin(-72)

Solution

cos (1 8^{\circ}) sin (- 7 2^{\circ})

Solution

\frac{- 5 - 5}{8}

Decimal Notation

- 0.90450 \dots

Solution steps

cos (1 8^{\circ}) sin (- 7 2^{\circ})

Rewrite using trig identities:

\frac{- sin ( 5 4 ^{\circ} ) - sin ( 9 0 ^{\circ} )}{2}

cos (1 8^{\circ}) sin (- 7 2^{\circ})

Use the Product to Sum identity:

sin (s) cos (t) = \frac{1}{2} (sin (s + t) + sin (s - t))

= \frac{1}{2} (sin (- 7 2^{\circ} + 1 8^{\circ}) + sin (- 7 2^{\circ} - 1 8^{\circ}))

Simplify

= \frac{sin ( - 5 4 ^{\circ} ) + sin ( - 9 0 ^{\circ} )}{2}

Use the following property:

sin (- x) = - sin (x)

sin (- 9 0^{\circ}) = - sin (9 0^{\circ})

= \frac{sin ( - 5 4 ^{\circ} ) - sin ( 9 0 ^{\circ} )}{2}

Use the following property:

sin (- x) = - sin (x)

sin (- 5 4^{\circ}) = - sin (5 4^{\circ})

= \frac{- sin ( 5 4 ^{\circ} ) - sin ( 9 0 ^{\circ} )}{2}

= \frac{- sin ( 5 4 ^{\circ} ) - sin ( 9 0 ^{\circ} )}{2}

Rewrite using trig identities:

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

sin (5 4^{\circ})

Rewrite using trig identities:

cos (3 6^{\circ})

sin (5 4^{\circ})

Use the following identity:

sin (x) = cos (9 0^{\circ} - x)

= cos (9 0^{\circ} - 5 4^{\circ})

Simplify

= cos (3 6^{\circ})

= cos (3 6^{\circ})

Rewrite using trig identities:

\frac{5 + 1}{4}

cos (3 6^{\circ})

Show that:

cos (3 6^{\circ}) - sin (1 8^{\circ}) = \frac{1}{2}

Use the following product to sum identity:

2 sin (x) cos (y) = sin (x + y) - sin (x - y)

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = sin (5 4^{\circ}) - sin (1 8^{\circ})

Show that:

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

sin (7 2^{\circ}) = 2 sin (3 6^{\circ}) cos (3 6^{\circ})

sin (7 2^{\circ}) sin (3 6^{\circ}) = 4 sin (3 6^{\circ}) sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

sin (3 6^{\circ})

sin (7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Use the following identity:

sin (x) = cos (9 0^{\circ} - x)

sin (7 2^{\circ}) = cos (9 0^{\circ} - 7 2^{\circ})

cos (9 0^{\circ} - 7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

cos (1 8^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

cos (1 8^{\circ})

1 = 4 sin (1 8^{\circ}) cos (3 6^{\circ})

Divide both sides by

2

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

Substitute

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

\frac{1}{2} = sin (5 4^{\circ}) - sin (1 8^{\circ})

sin (5 4^{\circ}) = cos (9 0^{\circ} - 5 4^{\circ})

\frac{1}{2} = cos (9 0^{\circ} - 5 4^{\circ}) - sin (1 8^{\circ})

\frac{1}{2} = cos (3 6^{\circ}) - sin (1 8^{\circ})

Show that:

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{4}

Use the factorization rule:

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

a = cos (3 6^{\circ}) + sin (1 8^{\circ})

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = ((cos (3 6^{\circ}) + sin (1 8^{\circ})) + (cos (3 6^{\circ}) - sin (1 8^{\circ}))) ((cos (3 6^{\circ}) + sin (1 8^{\circ})) - (cos (3 6^{\circ}) - sin (1 8^{\circ})))

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = 2 (2 cos (3 6^{\circ}) sin (1 8^{\circ}))

Show that:

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

sin (7 2^{\circ}) = 2 sin (3 6^{\circ}) cos (3 6^{\circ})

sin (7 2^{\circ}) sin (3 6^{\circ}) = 4 sin (3 6^{\circ}) sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

sin (3 6^{\circ})

sin (7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Use the following identity:

sin (x) = cos (9 0^{\circ} - x)

sin (7 2^{\circ}) = cos (9 0^{\circ} - 7 2^{\circ})

cos (9 0^{\circ} - 7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

cos (1 8^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

cos (1 8^{\circ})

1 = 4 sin (1 8^{\circ}) cos (3 6^{\circ})

Divide both sides by

2

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

Substitute

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = 1

Substitute

cos (3 6^{\circ}) - sin (1 8^{\circ}) = \frac{1}{2}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (\frac{1}{2})^{2} = 1

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - \frac{1}{4} = 1

Add

\frac{1}{4}

to both sides

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{4}

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} = \frac{5}{4}

Take the square root of both sides

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \pm \frac{5}{4}

cos (3 6^{\circ})

cannot be negative

sin (1 8^{\circ})

cannot be negative

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{4}

Add the following equations

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{2}

((cos (3 6^{\circ}) + sin (1 8^{\circ})) + (cos (3 6^{\circ}) - sin (1 8^{\circ}))) = (\frac{5}{2} + \frac{1}{2})

Refine

cos (3 6^{\circ}) = \frac{5 + 1}{4}

= \frac{5 + 1}{4}

= \frac{5 + 1}{4}

Use the following trivial identity:

sin (9 0^{\circ}) = 1

sin (9 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}

= 1

= \frac{- \frac{5 + 1}{4} - 1}{2}

Simplify

\frac{- \frac{5 + 1}{4} - 1}{2} : \frac{- 5 - 5}{8}

\frac{- \frac{5 + 1}{4} - 1}{2}

Join

- \frac{5 + 1}{4} - 1 : \frac{- 5 - 5}{4}

- \frac{5 + 1}{4} - 1

Convert element to fraction:

1 = \frac{1 \cdot 4}{4}

= - \frac{5 + 1}{4} - \frac{1 \cdot 4}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{- ( 5 + 1 ) - 1 \cdot 4}{4}

Multiply the numbers:

1 \cdot 4 = 4

= \frac{- ( 1 + 5 ) - 4}{4}

Expand

- (5 + 1) - 4 : - 5 - 5

- (5 + 1) - 4

- (5 + 1) : - 5 - 1

- (5 + 1)

Distribute parentheses

= - (5) - (1)

Apply minus-plus rules

+ (- a) = - a

= - 5 - 1

= - 5 - 1 - 4

Subtract the numbers:

- 1 - 4 = - 5

= - 5 - 5

= \frac{- 5 - 5}{4}

= \frac{\frac{- 5 - 5}{4}}{2}

Apply the fraction rule:

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

= \frac{- 5 - 5}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{- 5 - 5}{8}

= \frac{- 5 - 5}{8}

Popular Examples

arccos(5/10)

arccos (\frac{5}{10})

sin((3pi)/(10))-sin(pi/(10))

sin (\frac{3 π}{10}) - sin (\frac{π}{10})

sin(6/8)

sin (\frac{6}{8})

arccos(7)

arccos (7)

3sin((5pi)/6)

3 sin (\frac{5 π}{6})

Frequently Asked Questions (FAQ)

What is the value of cos(18)sin(-72) ?
The value of cos(18)sin(-72) is (-5-sqrt(5))/8