Is cos(θ-60)+cos(θ+60)=cos(θ) ?

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

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prove cos(θ-60)+cos(θ+60)=cos(θ)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

cos (θ - 6 0^{\circ})

Use the Angle Difference identity:

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

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

Simplify

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

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

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) + cos (θ + 6 0^{\circ})

Rewrite using trig identities

cos (θ + 6 0^{\circ})

Use the Angle Sum identity:

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

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

Simplify

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

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

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) : cos (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) = cos (θ)

\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} = 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

= cos (θ)

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

Add similar elements:

\frac{3}{2} sin (θ) - \frac{3}{2} sin (θ) = 0

\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} = 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

= cos (θ)

= cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

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