What is the value of 2*(3/(tan(18))*30)+240 ?

The value of 2*(3/(tan(18))*30)+240 is 180(sqrt(5)+2)sqrt(5-2\sqrt{5)}+240

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2(3/(tan(18))30)+240

Solution

2 \cdot (\frac{3}{tan ( 1 8 ^{\circ} )} \cdot 30) + 240

Solution

180 (5 + 2) 5 - 25 + 240

Decimal Notation

793.98303 \dots

Solution steps

2 (\frac{3}{tan ( 1 8 ^{\circ} )} \cdot 30) + 240

= 2 \cdot \frac{3}{tan ( 1 8 ^{\circ} )} \cdot 30 + 240

Rewrite using trig identities:

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

tan (1 8^{\circ})

Rewrite using trig identities:

\frac{1 - cos ( 3 6 ^{\circ} )}{1 + cos ( 3 6 ^{\circ} )}

tan (1 8^{\circ})

Write

tan (1 8^{\circ})

tan (\frac{3 6 ^{\circ}}{2})

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

Use the Half Angle identity:

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

Rewrite using trig identities:

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

Use the following identity

tan (θ) = \frac{sin ( θ )}{cos ( θ )}

Square both sides

tan^{2} (θ) = \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )}

Rewrite using trig identities:

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

Use the Double Angle identity

cos (2 θ) = 1 - 2 sin^{2} (θ)

Switch sides

2 sin^{2} (θ) - 1 = - cos (2 θ)

Add

1

to both sides

2 sin^{2} (θ) = 1 - cos (2 θ)

Divide both sides by

2

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

Rewrite using trig identities:

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}

tan^{2} (θ) = \frac{\frac{1 - c o s ( 2 θ )}{2}}{\frac{1 + c o s ( 2 θ )}{2}}

Simplify

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

Substitute

θ

with

\frac{θ}{2}

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

Simplify

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

Square root both sides

Choose the root sign according to the quadrant of

\frac{θ}{2}

range [0, 9 0^{\circ}] [9 0^{\circ}, 18 0^{\circ}] quadrant I II tan positive negative

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

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

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

Rewrite using trig identities:

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

Simplify

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

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

\frac{1 - \frac{5 + 1}{4}}{1 + \frac{5 + 1}{4}} = \frac{3 - 5}{5 + 5}

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

1 \cdot 4 + 5 + 1 = 5 + 5

1 \cdot 4 + 5 + 1

Multiply the numbers:

1 \cdot 4 = 4

= 4 + 5 + 1

Add the numbers:

4 + 1 = 5

= 5 + 5

= \frac{5 + 5}{4}

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

Join

1 - \frac{5 + 1}{4} : \frac{3 - 5}{4}

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply the numbers:

1 \cdot 4 = 4

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

Expand

4 - (5 + 1) : 3 - 5

4 - (5 + 1)

- (5 + 1) : - 5 - 1

- (5 + 1)

Distribute parentheses

= - (5) - (1)

Apply minus-plus rules

+ (- a) = - a

= - 5 - 1

= 4 - 5 - 1

Subtract the numbers:

4 - 1 = 3

= 3 - 5

= \frac{3 - 5}{4}

= \frac{\frac{3 - 5}{4}}{\frac{5 + 5}{4}}

Divide fractions:

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

= \frac{( 3 - 5 ) \cdot 4}{4 ( 5 + 5 )}

Cancel the common factor:

4

= \frac{3 - 5}{5 + 5}

= \frac{3 - 5}{5 + 5}

\frac{3 - 5}{5 + 5} = \frac{5 - 2 5}{5}

\frac{3 - 5}{5 + 5}

Multiply by the conjugate

\frac{5 - 5}{5 - 5}

= \frac{( 3 - 5 ) ( 5 - 5 )}{( 5 + 5 ) ( 5 - 5 )}

(3 - 5) (5 - 5) = 20 - 85

(3 - 5) (5 - 5)

Apply FOIL method:

(a + b) (c + d) = a c + a d + b c + b d

a = 3, b = - 5, c = 5, d = - 5

= 3 \cdot 5 + 3 (- 5) + (- 5) \cdot 5 + (- 5) (- 5)

Apply minus-plus rules

+ (- a) = - a, (- a) (- b) = ab

= 3 \cdot 5 - 35 - 55 + 55

Simplify

3 \cdot 5 - 35 - 55 + 55 : 20 - 85

3 \cdot 5 - 35 - 55 + 55

Add similar elements:

- 35 - 55 = - 85

= 3 \cdot 5 - 85 + 55

Multiply the numbers:

3 \cdot 5 = 15

= 15 - 85 + 55

Apply radical rule:

a a = a

55 = 5

= 15 - 85 + 5

Add the numbers:

15 + 5 = 20

= 20 - 85

= 20 - 85

(5 + 5) (5 - 5) = 20

(5 + 5) (5 - 5)

Apply Difference of Two Squares Formula:

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

a = 5, b = 5

= 5^{2} - (5)^{2}

Simplify

5^{2} - (5)^{2} : 20

5^{2} - (5)^{2}

5^{2} = 25

5^{2}

5^{2} = 25

= 25

(5)^{2} = 5

(5)^{2}

Apply radical rule:

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

= (5^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= 5^{\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

= 5

= 25 - 5

Subtract the numbers:

25 - 5 = 20

= 20

= 20

= \frac{20 - 8 5}{20}

Factor

20 - 85 : 4 (5 - 25)

20 - 85

Rewrite as

= 4 \cdot 5 - 4 \cdot 25

Factor out common term

4

= 4 (5 - 25)

= \frac{4 ( 5 - 2 5 )}{20}

Cancel the common factor:

4

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

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

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

= 2 \cdot \frac{3}{\frac{5 - 2 5}{5}} \cdot 30 + 240

Simplify

2 \cdot \frac{3}{\frac{5 - 2 5}{5}} \cdot 30 + 240 : 180 (5 + 2) 5 - 25 + 240

2 \cdot \frac{3}{\frac{5 - 2 5}{5}} \cdot 30 + 240

2 \cdot \frac{3}{\frac{5 - 2 5}{5}} \cdot 30 = \frac{180 5}{5 - 2 5}

2 \cdot \frac{3}{\frac{5 - 2 5}{5}} \cdot 30

\frac{3}{\frac{5 - 2 5}{5}} = \frac{3 5}{5 - 2 5}

\frac{3}{\frac{5 - 2 5}{5}}

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

\frac{5 - 2 5}{5}

Apply radical rule:

assuming

a \geq 0, b \geq 0

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

= \frac{3}{\frac{5 - 2 5}{5}}

Apply the fraction rule:

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

= \frac{3 5}{5 - 2 5}

= 2 \cdot 30 \cdot \frac{3 5}{5 - 2 5}

Multiply fractions:

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

= \frac{3 5 \cdot 2 \cdot 30}{5 - 2 5}

Multiply the numbers:

3 \cdot 2 \cdot 30 = 180

= \frac{180 5}{5 - 2 5}

= \frac{180 5}{5 - 2 5} + 240

\frac{180 5}{5 - 2 5} = 180 (5 + 2) 5 - 25

\frac{180 5}{5 - 2 5}

Multiply by the conjugate

\frac{5 - 2 5}{5 - 2 5}

= \frac{180 5 5 - 2 5}{5 - 2 5 5 - 2 5}

5 - 25 5 - 25 = 5 - 25

5 - 25 5 - 25

Apply radical rule:

a a = a

5 - 25 5 - 25 = 5 - 25

= 5 - 25

= \frac{180 5 5 - 2 5}{5 - 2 5}

Factor

5 - 25 : 5 (5 - 2)

5 - 25

5 = 55

= 55 - 25

Factor out common term

5

= 5 (5 - 2)

= \frac{180 5 5 - 2 5}{5 ( 5 - 2 )}

Cancel the common factor:

5

= \frac{180 5 - 2 5}{5 - 2}

Multiply by the conjugate

\frac{5 + 2}{5 + 2}

= \frac{180 5 - 2 5 ( 5 + 2 )}{( 5 - 2 ) ( 5 + 2 )}

(5 - 2) (5 + 2) = 1

(5 - 2) (5 + 2)

Apply Difference of Two Squares Formula:

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

a = 5, b = 2

= (5)^{2} - 2^{2}

Simplify

(5)^{2} - 2^{2} : 1

(5)^{2} - 2^{2}

(5)^{2} = 5

(5)^{2}

Apply radical rule:

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

= (5^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= 5^{\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

= 5

2^{2} = 4

2^{2}

2^{2} = 4

= 4

= 5 - 4

Subtract the numbers:

5 - 4 = 1

= 1

= 1

= \frac{180 ( 5 + 2 ) 5 - 2 5}{1}

Apply rule

\frac{a}{1} = a

= 180 (2 + 5) 5 - 25

= 180 (2 + 5) 5 - 25 + 240

= 180 (5 + 2) 5 - 25 + 240

Popular Examples

arcsin((14.5)/(49.5))

arcsin (\frac{14.5}{49.5})

sin(-(17pi)/6)

sin (- \frac{17 π}{6})

arccos(0.886)

arccos (0.886)

sec(arcsin(7/10))

sec (arcsin (\frac{7}{10}))

((1sin(90.25)))/(sin(37))

\frac{( 1 sin ( 90.2 5 ^{\circ} ) )}{sin ( 3 7 ^{\circ} )}

Frequently Asked Questions (FAQ)

What is the value of 2*(3/(tan(18))*30)+240 ?
The value of 2*(3/(tan(18))*30)+240 is 180(sqrt(5)+2)sqrt(5-2\sqrt{5)}+240