Is sin(θ-pi/3)+cos(θ+(11pi)/6)=sin(θ) ?

The answer to whether sin(θ-pi/3)+cos(θ+(11pi)/6)=sin(θ) is True

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prove sin(θ-pi/3)+cos(θ+(11pi)/6)=sin(θ)

Solution

prove

sin (θ - \frac{π}{3}) + cos (θ + \frac{11 π}{6}) = sin (θ)

Solution

True

Solution steps

sin (θ - \frac{π}{3}) + cos (θ + \frac{11 π}{6}) = sin (θ)

Manipulating left side

sin (θ - \frac{π}{3}) + cos (θ + \frac{11 π}{6})

Rewrite using trig identities

sin (θ - \frac{π}{3})

Use the Angle Difference identity:

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

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

Simplify

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

sin (θ) cos (\frac{π}{3}) - cos (θ) sin (\frac{π}{3})

Simplify

cos (\frac{π}{3}) : \frac{1}{2}

cos (\frac{π}{3})

Use the following trivial identity:

cos (\frac{π}{3}) = \frac{1}{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{1}{2}

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

Simplify

sin (\frac{π}{3}) : \frac{3}{2}

sin (\frac{π}{3})

Use the following trivial identity:

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

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

= \frac{1}{2} sin (θ) - \frac{3}{2} cos (θ) + cos (θ + \frac{11 π}{6})

Rewrite using trig identities

cos (θ + \frac{11 π}{6})

Use the Angle Sum identity:

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

= cos (θ) cos (\frac{11 π}{6}) - sin (θ) sin (\frac{11 π}{6})

Simplify

cos (θ) cos (\frac{11 π}{6}) - sin (θ) sin (\frac{11 π}{6}) : \frac{3}{2} cos (θ) + \frac{1}{2} sin (θ)

cos (θ) cos (\frac{11 π}{6}) - sin (θ) sin (\frac{11 π}{6})

cos (\frac{11 π}{6}) = \frac{3}{2}

cos (\frac{11 π}{6})

Rewrite using trig identities:

cos (π) cos (\frac{5 π}{6}) - sin (π) sin (\frac{5 π}{6})

cos (\frac{11 π}{6})

Write

cos (\frac{11 π}{6})

cos (π + \frac{5 π}{6})

= cos (π + \frac{5 π}{6})

Use the Angle Sum identity:

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

= cos (π) cos (\frac{5 π}{6}) - sin (π) sin (\frac{5 π}{6})

= cos (π) cos (\frac{5 π}{6}) - sin (π) sin (\frac{5 π}{6})

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

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}

= (- 1)

Use the following trivial identity:

cos (\frac{5 π}{6}) = - \frac{3}{2}

cos (\frac{5 π}{6})

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}

Use the following trivial identity:

sin (π) = 0

sin (π)

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}

= 0

Use the following trivial identity:

sin (\frac{5 π}{6}) = \frac{1}{2}

sin (\frac{5 π}{6})

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}

= (- 1) (- \frac{3}{2}) - 0 \cdot \frac{1}{2}

Simplify

= \frac{3}{2}

= \frac{3}{2} cos (θ) - sin (\frac{11 π}{6}) sin (θ)

sin (\frac{11 π}{6}) = - \frac{1}{2}

sin (\frac{11 π}{6})

Rewrite using trig identities:

sin (π) cos (\frac{5 π}{6}) + cos (π) sin (\frac{5 π}{6})

sin (\frac{11 π}{6})

Write

sin (\frac{11 π}{6})

sin (π + \frac{5 π}{6})

= sin (π + \frac{5 π}{6})

Use the Angle Sum identity:

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

= sin (π) cos (\frac{5 π}{6}) + cos (π) sin (\frac{5 π}{6})

= sin (π) cos (\frac{5 π}{6}) + cos (π) sin (\frac{5 π}{6})

Use the following trivial identity:

sin (π) = 0

sin (π)

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}

= 0

Use the following trivial identity:

cos (\frac{5 π}{6}) = - \frac{3}{2}

cos (\frac{5 π}{6})

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}

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

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}

= (- 1)

Use the following trivial identity:

sin (\frac{5 π}{6}) = \frac{1}{2}

sin (\frac{5 π}{6})

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}

= 0 \cdot (- \frac{3}{2}) + (- 1) \frac{1}{2}

Simplify

= - \frac{1}{2}

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

Apply rule

- (- a) = a

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

Factor out common term

cos (θ)

= cos (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

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

Add similar elements:

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

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

Factor out common term

sin (θ)

= sin (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

= sin (θ)

= sin (θ)

= sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is sin(θ-pi/3)+cos(θ+(11pi)/6)=sin(θ) ?
The answer to whether sin(θ-pi/3)+cos(θ+(11pi)/6)=sin(θ) is True