Is tan((2pi)/3)=(sqrt(243/4))/(-9/2) ?

The answer to whether tan((2pi)/3)=(sqrt(243/4))/(-9/2) is True

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prove tan((2pi)/3)=(sqrt(243/4))/(-9/2)

Solution

prove

tan (\frac{2 π}{3}) = \frac{\frac{243}{4}}{- \frac{9}{2}}

Solution

True

Solution steps

tan (\frac{2 π}{3}) = \frac{\frac{243}{4}}{- \frac{9}{2}}

Manipulating left side

tan (\frac{2 π}{3})

Simplify

tan (\frac{2 π}{3}) : - 3

tan (\frac{2 π}{3})

Rewrite using trig identities:

\frac{sin ( \frac{2 π}{3} )}{cos ( \frac{2 π}{3} )}

tan (\frac{2 π}{3})

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{sin ( \frac{2 π}{3} )}{cos ( \frac{2 π}{3} )}

= \frac{sin ( \frac{2 π}{3} )}{cos ( \frac{2 π}{3} )}

Use the following trivial identity:

sin (\frac{2 π}{3}) = \frac{3}{2}

sin (\frac{2 π}{3})

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{3}{2}

Use the following trivial identity:

cos (\frac{2 π}{3}) = - \frac{1}{2}

cos (\frac{2 π}{3})

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - \frac{1}{2}

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

Simplify

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

\frac{\frac{3}{2}}{- \frac{1}{2}}

Apply the fraction rule:

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

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

Divide fractions:

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

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

Refine

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

Cancel the common factor:

2

= - 3

= - 3

= - 3

Manipulating right side

\frac{\frac{243}{4}}{- \frac{9}{2}}

Simplify

\frac{\frac{243}{4}}{- \frac{9}{2}} : - 3

\frac{\frac{243}{4}}{- \frac{9}{2}}

Apply the fraction rule:

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

= - \frac{\frac{243}{4}}{\frac{9}{2}}

Apply the fraction rule:

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

\frac{\frac{243}{4}}{\frac{9}{2}} = \frac{\frac{243}{4} \cdot 2}{9}

= - \frac{\frac{243}{4} \cdot 2}{9}

\frac{243}{4} = \frac{9 3}{2}

\frac{243}{4}

Apply radical rule:

n \frac{a}{b} = \frac{n a}{n b},

assuming

a \geq 0, b \geq 0

= \frac{243}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

= \frac{243}{2}

243 = 93

243

Prime factorization of

243 : 3^{5}

243

243

divides by

3243 = 81 \cdot 3

= 3 \cdot 81

81

divides by

381 = 27 \cdot 3

= 3 \cdot 3 \cdot 27

27

divides by

327 = 9 \cdot 3

= 3 \cdot 3 \cdot 3 \cdot 9

9

divides by

39 = 3 \cdot 3

= 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3

3

is a prime number, therefore no further factorization is possible

= 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3

= 3^{5}

= 3^{5}

Apply exponent rule:

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

= 3^{4} \cdot 3

Apply radical rule:

n ab = n a n b

= 3 3^{4}

Apply radical rule:

n a^{m} = a^{\frac{m}{n}}

3^{4} = 3^{\frac{4}{2}} = 3^{2}

= 3^{2} 3

Refine

= 93

= \frac{9 3}{2}

= - \frac{2 \cdot \frac{9 3}{2}}{9}

Multiply

\frac{9 3}{2} \cdot 2 : 93

\frac{9 3}{2} \cdot 2

Multiply fractions:

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

= \frac{9 3 \cdot 2}{2}

Cancel the common factor:

2

= 93

= - \frac{9 3}{9}

Divide the numbers:

\frac{9}{9} = 1

= - 3

= - 3

We showed that the two sides could take the same form

\Rightarrow True

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Is tan((2pi)/3)=(sqrt(243/4))/(-9/2) ?
The answer to whether tan((2pi)/3)=(sqrt(243/4))/(-9/2) is True