Is cos(90+30)=cos(120) ?

The answer to whether cos(90+30)=cos(120) is True

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prove cos(90+30)=cos(120)

Solution

prove

cos (9 0^{\circ} + 3 0^{\circ}) = cos (12 0^{\circ})

Solution

True

Solution steps

cos (9 0^{\circ} + 3 0^{\circ}) = cos (12 0^{\circ})

Manipulating left side

cos (9 0^{\circ} + 3 0^{\circ})

Rewrite using trig identities

cos (9 0^{\circ} + 3 0^{\circ})

Use the Angle Sum identity:

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

= cos (9 0^{\circ}) cos (3 0^{\circ}) - sin (9 0^{\circ}) sin (3 0^{\circ})

cos (9 0^{\circ}) cos (3 0^{\circ}) - sin (9 0^{\circ}) sin (3 0^{\circ}) = - \frac{1}{2}

cos (9 0^{\circ}) cos (3 0^{\circ}) - sin (9 0^{\circ}) sin (3 0^{\circ})

cos (9 0^{\circ}) cos (3 0^{\circ}) = 0

cos (9 0^{\circ}) cos (3 0^{\circ})

Simplify

cos (9 0^{\circ}) : 0

cos (9 0^{\circ})

Use the following trivial identity:

cos (9 0^{\circ}) = 0

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}

= 0

= 0 \cdot cos (3 0^{\circ})

Simplify

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

cos (3 0^{\circ})

Use the following trivial identity:

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

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

Apply rule

0 \cdot a = 0

= 0

sin (9 0^{\circ}) sin (3 0^{\circ}) = \frac{1}{2}

sin (9 0^{\circ}) sin (3 0^{\circ})

Simplify

sin (9 0^{\circ}) : 1

sin (9 0^{\circ})

Use the following trivial identity:

sin (9 0^{\circ}) = 1

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}

= 1

= 1 \cdot sin (3 0^{\circ})

Simplify

sin (3 0^{\circ}) : \frac{1}{2}

sin (3 0^{\circ})

Use the following trivial identity:

sin (3 0^{\circ}) = \frac{1}{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{1}{2}

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

Multiply:

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

= \frac{1}{2}

= 0 - \frac{1}{2}

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

= - \frac{1}{2}

= - \frac{1}{2}

= - \frac{1}{2}

Manipulating right side

cos (12 0^{\circ})

Simplify

= - \frac{1}{2}

We showed that the two sides could take the same form

\Rightarrow True

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Is cos(90+30)=cos(120) ?
The answer to whether cos(90+30)=cos(120) is True