Is sin(120)=sin(180-60) ?

The answer to whether sin(120)=sin(180-60) is True

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prove sin(120)=sin(180-60)

Solution

prove

sin (12 0^{\circ}) = sin (18 0^{\circ} - 6 0^{\circ})

Solution

True

Solution steps

sin (12 0^{\circ}) = sin (18 0^{\circ} - 6 0^{\circ})

Manipulating left side

sin (12 0^{\circ})

Simplify

= \frac{3}{2}

Manipulating right side

sin (18 0^{\circ} - 6 0^{\circ})

Rewrite using trig identities

sin (18 0^{\circ} - 6 0^{\circ})

Use the Angle Difference identity:

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

= sin (18 0^{\circ}) cos (6 0^{\circ}) - cos (18 0^{\circ}) sin (6 0^{\circ})

sin (18 0^{\circ}) cos (6 0^{\circ}) - cos (18 0^{\circ}) sin (6 0^{\circ}) = \frac{3}{2}

sin (18 0^{\circ}) cos (6 0^{\circ}) - cos (18 0^{\circ}) sin (6 0^{\circ})

sin (18 0^{\circ}) cos (6 0^{\circ}) = 0

sin (18 0^{\circ}) cos (6 0^{\circ})

Simplify

sin (18 0^{\circ}) : 0

sin (18 0^{\circ})

Use the following trivial identity:

sin (18 0^{\circ}) = 0

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}

= 0

= 0 \cdot cos (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}

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

Apply rule

0 \cdot a = 0

= 0

cos (18 0^{\circ}) sin (6 0^{\circ}) = - \frac{3}{2}

cos (18 0^{\circ}) sin (6 0^{\circ})

Simplify

cos (18 0^{\circ}) : - 1

cos (18 0^{\circ})

Use the following trivial identity:

cos (18 0^{\circ}) = (- 1)

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}

= - 1

= - 1 \cdot sin (6 0^{\circ})

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}

= - 1 \cdot \frac{3}{2}

Multiply:

1 \cdot \frac{3}{2} = \frac{3}{2}

= - \frac{3}{2}

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

Simplify

0 - (- \frac{3}{2})

Apply rule

- (- a) = a

= 0 + \frac{3}{2}

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

= \frac{3}{2}

= \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 sin(120)=sin(180-60) ?
The answer to whether sin(120)=sin(180-60) is True