What is the value of ((320/3))/((sin(30)+cos(30))) ?

The value of ((320/3))/((sin(30)+cos(30))) is -(320(1-sqrt(3)))/3

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((320/3))/((sin(30)+cos(30)))

Solution

\frac{( \frac{320}{3} )}{( sin ( 3 0 ^{\circ} ) + cos ( 3 0 ^{\circ} ) )}

Solution

- \frac{320 ( 1 - 3 )}{3}

Decimal Notation

78.08541 \dots

Solution steps

\frac{( \frac{320}{3} )}{( sin ( 3 0 ^{\circ} ) + cos ( 3 0 ^{\circ} ) )}

= \frac{\frac{320}{3}}{sin ( 3 0 ^{\circ} ) + cos ( 3 0 ^{\circ} )}

Use the following trivial identity:

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

sin (3 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}

= \frac{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{320}{3}}{\frac{1}{2} + \frac{3}{2}}

Simplify

\frac{\frac{320}{3}}{\frac{1}{2} + \frac{3}{2}} : - \frac{320 ( 1 - 3 )}{3}

\frac{\frac{320}{3}}{\frac{1}{2} + \frac{3}{2}}

Combine the fractions

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

Apply rule

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

= \frac{1 + 3}{2}

= \frac{\frac{320}{3}}{\frac{1 + 3}{2}}

Divide fractions:

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

= \frac{320 \cdot 2}{3 ( 1 + 3 )}

Multiply the numbers:

320 \cdot 2 = 640

= \frac{640}{3 ( 1 + 3 )}

Rationalize

\frac{640}{3 ( 1 + 3 )} : - \frac{320 ( 1 - 3 )}{3}

\frac{640}{3 ( 1 + 3 )}

Multiply by the conjugate

\frac{1 - 3}{1 - 3}

= \frac{640 ( 1 - 3 )}{3 ( 1 + 3 ) ( 1 - 3 )}

3 (1 + 3) (1 - 3) = - 6

3 (1 + 3) (1 - 3)

Expand

(1 + 3) (1 - 3) : - 2

(1 + 3) (1 - 3)

Apply Difference of Two Squares Formula:

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

a = 1, b = 3

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

Simplify

1^{2} - (3)^{2} : - 2

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - (3)^{2}

(3)^{2} = 3

(3)^{2}

Apply radical rule:

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

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

Apply exponent rule:

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

= 3^{\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

= 3

= 1 - 3

Subtract the numbers:

1 - 3 = - 2

= - 2

= - 2

= 3 (- 2)

Expand

3 (- 2) : - 6

3 (- 2)

Distribute parentheses

= 3 (- 2)

Apply minus-plus rules

+ (- a) = - a

= - 3 \cdot 2

Multiply the numbers:

3 \cdot 2 = 6

= - 6

= - 6

= \frac{640 ( 1 - 3 )}{- 6}

Apply the fraction rule:

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

= - \frac{640 ( 1 - 3 )}{6}

Cancel the common factor:

2

= - \frac{320 ( 1 - 3 )}{3}

= - \frac{320 ( 1 - 3 )}{3}

= - \frac{320 ( 1 - 3 )}{3}

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Frequently Asked Questions (FAQ)

What is the value of ((320/3))/((sin(30)+cos(30))) ?
The value of ((320/3))/((sin(30)+cos(30))) is -(320(1-sqrt(3)))/3