Is cos(60+θ)-cos(60-θ)=-sqrt(3)sin(θ) ?

The answer to whether cos(60+θ)-cos(60-θ)=-sqrt(3)sin(θ) is True

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prove cos(60+θ)-cos(60-θ)=-sqrt(3)sin(θ)

Solution

prove

cos (6 0^{\circ} + θ) - cos (6 0^{\circ} - θ) = - 3 sin (θ)

Solution

True

Solution steps

cos (6 0^{\circ} + θ) - cos (6 0^{\circ} - θ) = - 3 sin (θ)

Manipulating left side

cos (6 0^{\circ} + θ) - cos (6 0^{\circ} - θ)

Rewrite using trig identities

cos (6 0^{\circ} - θ)

Use the Angle Difference identity:

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

= cos (6 0^{\circ}) cos (θ) + sin (6 0^{\circ}) sin (θ)

Simplify

cos (6 0^{\circ}) cos (θ) + sin (6 0^{\circ}) sin (θ) : \frac{1}{2} cos (θ) + \frac{3}{2} sin (θ)

cos (6 0^{\circ}) cos (θ) + sin (6 0^{\circ}) sin (θ)

Simplify

cos (6 0^{\circ}) : \frac{1}{2}

cos (6 0^{\circ})

Use the following trivial identity:

cos (6 0^{\circ}) = \frac{1}{2}

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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{1}{2} cos (θ) + sin (6 0^{\circ}) sin (θ)

Simplify

sin (6 0^{\circ}) : \frac{3}{2}

sin (6 0^{\circ})

Use the following trivial identity:

sin (6 0^{\circ}) = \frac{3}{2}

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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{1}{2} cos (θ) + \frac{3}{2} sin (θ)

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

= cos (6 0^{\circ} + θ) - (\frac{1}{2} cos (θ) + \frac{3}{2} sin (θ))

Rewrite using trig identities

cos (6 0^{\circ} + θ)

Use the Angle Sum identity:

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

= cos (6 0^{\circ}) cos (θ) - sin (6 0^{\circ}) sin (θ)

Simplify

cos (6 0^{\circ}) cos (θ) - sin (6 0^{\circ}) sin (θ) : \frac{1}{2} cos (θ) - \frac{3}{2} sin (θ)

cos (6 0^{\circ}) cos (θ) - sin (6 0^{\circ}) sin (θ)

Simplify

cos (6 0^{\circ}) : \frac{1}{2}

cos (6 0^{\circ})

Use the following trivial identity:

cos (6 0^{\circ}) = \frac{1}{2}

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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{1}{2} cos (θ) - sin (6 0^{\circ}) sin (θ)

Simplify

sin (6 0^{\circ}) : \frac{3}{2}

sin (6 0^{\circ})

Use the following trivial identity:

sin (6 0^{\circ}) = \frac{3}{2}

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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{1}{2} cos (θ) - \frac{3}{2} sin (θ)

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

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

Simplify

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

+ (- a) = - a

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

\frac{1}{2} cos (θ) - \frac{1}{2} cos (θ) = 0

\frac{1}{2} cos (θ) - \frac{1}{2} cos (θ)

Factor out common term

cos (θ)

= cos (θ) (\frac{1}{2} - \frac{1}{2})

\frac{1}{2} - \frac{1}{2} = 0

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

Refine

= 0

= 0

= - \frac{3}{2} sin (θ) - \frac{3}{2} sin (θ)

Add similar elements:

- \frac{3}{2} sin (θ) - \frac{3}{2} sin (θ) = - 3 sin (θ)

- \frac{3}{2} sin (θ) - \frac{3}{2} sin (θ)

Factor out common term

sin (θ)

= sin (θ) (- \frac{3}{2} - \frac{3}{2})

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

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

Apply rule

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

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

Factor

- 3 - 3 : - 23

- 3 - 3

Factor out common term

3

= - 3 (1 + 1)

Refine

= - 23

= - \frac{2 3}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 3

= - 3 sin (θ)

= - 3 sin (θ)

= - 3 sin (θ)

= - 3 sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is cos(60+θ)-cos(60-θ)=-sqrt(3)sin(θ) ?
The answer to whether cos(60+θ)-cos(60-θ)=-sqrt(3)sin(θ) is True