Is cos(x-pi/6)-cos(x+pi/6)=sin(x) ?

The answer to whether cos(x-pi/6)-cos(x+pi/6)=sin(x) is True

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prove cos(x-pi/6)-cos(x+pi/6)=sin(x)

Solution

prove

cos (x - \frac{π}{6}) - cos (x + \frac{π}{6}) = sin (x)

Solution

True

Solution steps

cos (x - \frac{π}{6}) - cos (x + \frac{π}{6}) = sin (x)

Manipulating left side

cos (x - \frac{π}{6}) - cos (x + \frac{π}{6})

Rewrite using trig identities

cos (x - \frac{π}{6})

Use the Angle Difference identity:

cos (s - t) = cos (s) cos (t) + sin (s) sin (t)

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

Simplify

cos (x) cos (\frac{π}{6}) + sin (x) sin (\frac{π}{6}) : \frac{3}{2} cos (x) + \frac{1}{2} sin (x)

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

Simplify

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{3}{2} cos (x) + sin (\frac{π}{6}) sin (x)

Simplify

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{3}{2} cos (x) + \frac{1}{2} sin (x)

= \frac{3}{2} cos (x) + \frac{1}{2} sin (x)

= \frac{3}{2} cos (x) + \frac{1}{2} sin (x) - cos (x + \frac{π}{6})

Rewrite using trig identities

cos (x + \frac{π}{6})

Use the Angle Sum identity:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

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

Simplify

cos (x) cos (\frac{π}{6}) - sin (x) sin (\frac{π}{6}) : \frac{3}{2} cos (x) - \frac{1}{2} sin (x)

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

Simplify

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{3}{2} cos (x) - sin (\frac{π}{6}) sin (x)

Simplify

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{3}{2} cos (x) - \frac{1}{2} sin (x)

= \frac{3}{2} cos (x) - \frac{1}{2} sin (x)

= \frac{3}{2} cos (x) + \frac{1}{2} sin (x) - (\frac{3}{2} cos (x) - \frac{1}{2} sin (x))

Simplify

\frac{3}{2} cos (x) + \frac{1}{2} sin (x) - (\frac{3}{2} cos (x) - \frac{1}{2} sin (x)) : sin (x)

\frac{3}{2} cos (x) + \frac{1}{2} sin (x) - (\frac{3}{2} cos (x) - \frac{1}{2} sin (x))

- (\frac{3}{2} cos (x) - \frac{1}{2} sin (x)) : - \frac{3}{2} cos (x) + \frac{1}{2} sin (x)

- (\frac{3}{2} cos (x) - \frac{1}{2} sin (x))

Distribute parentheses

= - (\frac{3}{2} cos (x)) - (- \frac{1}{2} sin (x))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - \frac{3}{2} cos (x) + \frac{1}{2} sin (x)

= \frac{3}{2} cos (x) + \frac{1}{2} sin (x) - \frac{3}{2} cos (x) + \frac{1}{2} sin (x)

Simplify

\frac{3}{2} cos (x) + \frac{1}{2} sin (x) - \frac{3}{2} cos (x) + \frac{1}{2} sin (x) : sin (x)

\frac{3}{2} cos (x) + \frac{1}{2} sin (x) - \frac{3}{2} cos (x) + \frac{1}{2} sin (x)

Group like terms

= \frac{1}{2} sin (x) + \frac{1}{2} sin (x) + \frac{3}{2} cos (x) - \frac{3}{2} cos (x)

Add similar elements:

\frac{3}{2} cos (x) - \frac{3}{2} cos (x) = 0

\frac{3}{2} cos (x) - \frac{3}{2} cos (x)

Factor out common term

cos (x)

= cos (x) (\frac{3}{2} - \frac{3}{2})

\frac{3}{2} - \frac{3}{2} = 0

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

Since the denominators are equal, combine the fractions:

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

= \frac{3 - 3}{2}

Factor

3 - 3 : 0

3 - 3

Factor out common term

3

= 3 (1 - 1)

Refine

= 0

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

= 0

= \frac{1}{2} sin (x) + \frac{1}{2} sin (x)

Add similar elements:

\frac{1}{2} sin (x) + \frac{1}{2} sin (x) = sin (x)

\frac{1}{2} sin (x) + \frac{1}{2} sin (x)

Factor out common term

sin (x)

= sin (x) (\frac{1}{2} + \frac{1}{2})

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

\frac{1}{2} + \frac{1}{2}

Apply rule

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

= \frac{1 + 1}{2}

Refine

= 1

= sin (x)

= sin (x)

= sin (x)

= sin (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is cos(x-pi/6)-cos(x+pi/6)=sin(x) ?
The answer to whether cos(x-pi/6)-cos(x+pi/6)=sin(x) is True