Is tan(pi/3)sin((2pi)/3)=1+sin(pi/6) ?

The answer to whether tan(pi/3)sin((2pi)/3)=1+sin(pi/6) is True

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Popular >

prove tan(pi/3)sin((2pi)/3)=1+sin(pi/6)

Solution

prove

tan (\frac{π}{3}) sin (\frac{2 π}{3}) = 1 + sin (\frac{π}{6})

Solution

True

Solution steps

tan (\frac{π}{3}) sin (\frac{2 π}{3}) = 1 + sin (\frac{π}{6})

Manipulating left side

tan (\frac{π}{3}) sin (\frac{2 π}{3})

Simplify

sin (\frac{2 π}{3}) tan (\frac{π}{3}) : \frac{3}{2}

sin (\frac{2 π}{3}) tan (\frac{π}{3})

Simplify

sin (\frac{2 π}{3}) : \frac{3}{2}

sin (\frac{2 π}{3})

Use the following trivial identity:

sin (\frac{2 π}{3}) = \frac{3}{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{3}{2}

= \frac{3}{2} tan (\frac{π}{3})

Simplify

tan (\frac{π}{3}) : 3

tan (\frac{π}{3})

Use the following trivial identity:

tan (\frac{π}{3}) = 3

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 3

= 3 \frac{3}{2}

Simplify

\frac{3}{2} 3

Multiply fractions:

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

= \frac{3 3}{2}

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{3}{2}

= \frac{3}{2}

= \frac{3}{2}

Manipulating right side

1 + sin (\frac{π}{6})

Simplify

1 + sin (\frac{π}{6}) : \frac{3}{2}

1 + 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}

= 1 + \frac{1}{2}

Simplify

1 + \frac{1}{2}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

1 \cdot 2 + 1 = 3

1 \cdot 2 + 1

Multiply the numbers:

1 \cdot 2 = 2

= 2 + 1

Add the numbers:

2 + 1 = 3

= 3

= \frac{3}{2}

= \frac{3}{2}

= \frac{3}{2}

We showed that the two sides could take the same form

\Rightarrow True

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Is tan(pi/3)sin((2pi)/3)=1+sin(pi/6) ?
The answer to whether tan(pi/3)sin((2pi)/3)=1+sin(pi/6) is True