Is tan((5pi)/3)=-sqrt(3) ?

The answer to whether tan((5pi)/3)=-sqrt(3) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove tan((5pi)/3)=-sqrt(3)

prove

tan (\frac{5 π}{3}) = - 3

True

Solution steps

tan (\frac{5 π}{3}) = - 3

Manipulating left side

tan (\frac{5 π}{3})

Simplify

tan (\frac{5 π}{3}) : - 3

tan (\frac{5 π}{3})

tan (\frac{5 π}{3}) = tan (\frac{2 π}{3})

tan (\frac{5 π}{3})

Rewrite

\frac{5 π}{3}

π + \frac{2 π}{3}

= tan (π + \frac{2 π}{3})

Apply the periodicity of

tan

tan (x + π) = tan (x)

tan (π + \frac{2 π}{3}) = tan (\frac{2 π}{3})

= tan (\frac{2 π}{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

We showed that the two sides could take the same form

\Rightarrow True

sin (A) = \frac{3}{5}

cos (2 x) = cos (- 2 x)

tan (x) \cdot csc (x) = sec (x)

tan (θ) sin (θ) = 2 sin^{2} (θ)

3 csc (x) cot (x) + 3 csc (x) = 0