What is the value of 1+cos(360) ?

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1+cos(360)

1 + cos (36 0^{\circ})

2

Solution steps

1 + cos (36 0^{\circ})

Rewrite using trig identities:

cos (36 0^{\circ}) = 1

cos (36 0^{\circ})

cos (36 0^{\circ}) = cos (0)

cos (36 0^{\circ})

Rewrite

36 0^{\circ}

36 0^{\circ} + 0

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

Apply the periodicity of

cos

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

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

= cos (0)

= cos (0)

Use the following trivial identity:

cos (0) = 1

cos (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}

= 1

= 1

= 1 + 1

Simplify

= 2

\frac{92}{tan ( 1 2 ^{\circ} )}

- 2 cos (- \frac{π}{4})

\frac{cos ( 5 0 ^{\circ} )}{cos ( 3 0 ^{\circ} )}

\frac{1 + cos ( 2 0 ^{\circ} ) - sin ( 2 0 ^{\circ} )}{1 - cos ( 2 0 ^{\circ} ) - sin ( 2 0 ^{\circ} )}

arccos (- 1) + arctan (3)