What is the value of cos(1290) ?

The value of cos(1290) is -(sqrt(3))/2

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cos(1290)

cos (129 0^{\circ})

- \frac{3}{2}

Decimal Notation

- 0.86602 \dots

Solution steps

cos (129 0^{\circ})

cos (129 0^{\circ}) = cos (21 0^{\circ})

cos (129 0^{\circ})

Rewrite

129 0^{\circ}

36 0^{\circ} \cdot 3 + 21 0^{\circ}

= cos (36 0^{\circ} 3 + 21 0^{\circ})

Apply the periodicity of

cos

cos (x + 36 0^{\circ} \cdot k) = cos (x)

cos (36 0^{\circ} \cdot 3 + 21 0^{\circ}) = cos (21 0^{\circ})

= cos (21 0^{\circ})

= cos (21 0^{\circ})

Rewrite using trig identities:

cos (18 0^{\circ}) cos (3 0^{\circ}) - sin (18 0^{\circ}) sin (3 0^{\circ})

cos (21 0^{\circ})

Write

cos (21 0^{\circ})

cos (18 0^{\circ} + 3 0^{\circ})

= cos (18 0^{\circ} + 3 0^{\circ})

Use the Angle Sum identity:

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

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

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

Use the following trivial identity:

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

cos (18 0^{\circ})

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)

Use the following trivial identity:

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

cos (3 0^{\circ})

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}

Use the following trivial identity:

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

sin (18 0^{\circ})

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

Use the following trivial identity:

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

sin (3 0^{\circ})

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

Simplify

= - \frac{3}{2}

tan (arcsin (\frac{4}{5}) - arccos (1))

tan (arcsin (\frac{2}{3}) - arccos (\frac{1}{3}))

cos (30 0^{\circ}) cos (1 5^{\circ}) - sin (30 0^{\circ}) sin (1 5^{\circ})

2 cos (2 π)

\frac{0.5}{sin ( 0.5 )}