Is tan(pi/6)= 1/(sqrt(3)) ?

The answer to whether tan(pi/6)= 1/(sqrt(3)) is True

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prove tan(pi/6)= 1/(sqrt(3))

prove

tan (\frac{π}{6}) = \frac{1}{3}

True

Solution steps

tan (\frac{π}{6}) = \frac{1}{3}

Manipulating left side

tan (\frac{π}{6})

Express with sin, cos

tan (\frac{π}{6})

Use the basic trigonometric identity:

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

= \frac{sin ( \frac{π}{6} )}{cos ( \frac{π}{6} )}

\frac{sin ( \frac{π}{6} )}{cos ( \frac{π}{6} )} = \frac{1}{3}

\frac{sin ( \frac{π}{6} )}{cos ( \frac{π}{6} )}

sin (\frac{π}{6}) = \frac{1}{2}

sin (\frac{π}{6})

Use the following trivial identity:

sin (\frac{π}{6}) = \frac{1}{2}

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{1}{2}

= \frac{\frac{1}{2}}{cos ( \frac{π}{6} )}

cos (\frac{π}{6}) = \frac{3}{2}

cos (\frac{π}{6})

Use the following trivial identity:

cos (\frac{π}{6}) = \frac{3}{2}

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{3}{2}

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

Simplify

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

Divide fractions:

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

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

Refine

= \frac{2}{2 3}

Cancel the common factor:

2

= \frac{1}{3}

= \frac{1}{3}

= \frac{1}{3}

= \frac{1}{3}

We showed that the two sides could take the same form

\Rightarrow True

cos (4 a) = 1 - 8 sin^{2} (a) cos^{2} (a)

(cos (x)) (tan (x) + cot (x)) = csc (x)

cos (2 0^{\circ}) = cos (θ) + cos (θ)

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

1 - cos (4 x) = 2 sin^{2} (2 x)

Is tan(pi/6)= 1/(sqrt(3)) ?
The answer to whether tan(pi/6)= 1/(sqrt(3)) is True