What is the value of (140sin(63))/(139) ?

The value of (140sin(63))/(139) is (35sqrt(2)sqrt(\sqrt{2)sqrt(5-\sqrt{5)}+4})/(139)

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Popular >

(140sin(63))/(139)

Solution

\frac{140 sin ( 6 3 ^{\circ} )}{139}

Solution

\frac{35 2 2 5 - 5 + 4}{139}

Decimal Notation

0.89741 \dots

Solution steps

\frac{140 sin ( 6 3 ^{\circ} )}{139}

Rewrite using trig identities:

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

sin (6 3^{\circ})

Rewrite using trig identities:

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

sin (6 3^{\circ})

Write

sin (6 3^{\circ})

sin (\frac{12 6 ^{\circ}}{2})

= sin (\frac{12 6 ^{\circ}}{2})

Use the Half Angle identity:

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

Use the Double Angle identity

cos (2 θ) = 1 - 2 sin^{2} (θ)

Substitute

θ

with

\frac{θ}{2}

cos (θ) = 1 - 2 sin^{2} (\frac{θ}{2})

Switch sides

2 sin^{2} (\frac{θ}{2}) = 1 - cos (θ)

Divide both sides by

2

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

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}] [18 0^{\circ}, 27 0^{\circ}] [27 0^{\circ}, 36 0^{\circ}] quadrant I II III I V sin positive positive negative negative cos positive negative negative positive

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

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

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

Rewrite using trig identities:

cos (12 6^{\circ}) = - \frac{2 5 - 5}{4}

cos (12 6^{\circ})

Rewrite using trig identities:

- sin (3 6^{\circ})

cos (12 6^{\circ})

Use the following identity:

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

= sin (9 0^{\circ} - 12 6^{\circ})

Simplify

= sin (- 3 6^{\circ})

Use the following property:

sin (- x) = - sin (x)

sin (- 3 6^{\circ}) = - sin (3 6^{\circ})

= - sin (3 6^{\circ})

= - sin (3 6^{\circ})

Rewrite using trig identities:

sin (3 6^{\circ}) = \frac{2 5 - 5}{4}

sin (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}

Square both sides

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

Use the following identity:

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

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

Substitute

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

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

Refine

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

Take the square root of both sides

sin (3 6^{\circ}) = \pm \frac{5 - 5}{8}

sin (3 6^{\circ})

cannot be negative

sin (3 6^{\circ}) = \frac{5 - 5}{8}

Refine

sin (3 6^{\circ}) = \frac{\frac{5 - 5}{2}}{2}

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

Simplify

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

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

= \frac{1 - ( - \frac{2 5 - 5}{4} )}{2}

Simplify

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

\frac{1 - ( - \frac{2 5 - 5}{4} )}{2}

Apply rule

- (- a) = a

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

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

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

Join

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

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

Convert element to fraction:

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

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

Multiply the numbers:

1 \cdot 4 = 4

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

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

Apply the fraction rule:

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

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

Multiply the numbers:

4 \cdot 2 = 8

= \frac{4 + 2 5 - 5}{8}

= \frac{4 + 2 5 - 5}{8}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{4 + 2 5 - 5}{8}

8 = 22

8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

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

Rationalize

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

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

Multiply by the conjugate

\frac{2}{2}

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

222 = 4

222

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

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

Add similar elements:

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

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

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

2 \cdot \frac{1}{2}

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 2^{1 + 1}

Add the numbers:

1 + 1 = 2

= 2^{2}

2^{2} = 4

= 4

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

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

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

= \frac{140 \cdot \frac{2 4 + 2 5 - 5}{4}}{139}

Simplify

\frac{140 \cdot \frac{2 4 + 2 5 - 5}{4}}{139} : \frac{35 2 2 5 - 5 + 4}{139}

\frac{140 \cdot \frac{2 4 + 2 5 - 5}{4}}{139}

Multiply

140 \cdot \frac{2 4 + 2 5 - 5}{4} : 352 2 5 - 5 + 4

140 \cdot \frac{2 4 + 2 5 - 5}{4}

Multiply fractions:

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

= \frac{2 4 + 2 5 - 5 \cdot 140}{4}

Divide the numbers:

\frac{140}{4} = 35

= 352 2 5 - 5 + 4

= \frac{35 2 2 5 - 5 + 4}{139}

= \frac{35 2 2 5 - 5 + 4}{139}

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Frequently Asked Questions (FAQ)

What is the value of (140sin(63))/(139) ?
The value of (140sin(63))/(139) is (35sqrt(2)sqrt(\sqrt{2)sqrt(5-\sqrt{5)}+4})/(139)