Is cos(4pi+x)=cos(x) ?

The answer to whether cos(4pi+x)=cos(x) is True

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prove cos(4pi+x)=cos(x)

prove

cos (4 π + x) = cos (x)

True

Solution steps

cos (4 π + x) = cos (x)

Manipulating left side

cos (4 π + x)

Rewrite using trig identities

cos (4 π + x)

Use the Angle Sum identity:

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

= cos (4 π) cos (x) - sin (4 π) sin (x)

Simplify

cos (4 π) cos (x) - sin (4 π) sin (x) : cos (x)

cos (4 π) cos (x) - sin (4 π) sin (x)

cos (4 π) cos (x) = cos (x)

cos (4 π) cos (x)

cos (4 π) = 1

cos (4 π)

cos (4 π) = cos (0)

cos (4 π)

Rewrite

4 π

2 π \cdot 2 + 0

= cos (2 π 2 + 0)

Apply the periodicity of

cos

cos (x + 2 π \cdot k) = cos (x)

cos (2 π \cdot 2 + 0) = cos (0)

= cos (0)

= cos (0)

Use the following trivial identity:

cos (0) = 1

cos (0)

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

= 1

= 1 \cdot cos (x)

Multiply:

1 \cdot cos (x) = cos (x)

= cos (x)

sin (4 π) sin (x) = 0

sin (4 π) sin (x)

sin (4 π) = 0

sin (4 π)

sin (4 π) = sin (0)

sin (4 π)

Rewrite

4 π

2 π \cdot 2 + 0

= sin (2 π 2 + 0)

Apply the periodicity of

sin

sin (x + 2 π \cdot k) = sin (x)

sin (2 π \cdot 2 + 0) = sin (0)

= sin (0)

= sin (0)

Use the following trivial identity:

sin (0) = 0

sin (0)

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

= 0

= 0 \cdot sin (x)

Apply rule

0 \cdot a = 0

= 0

= cos (x) - 0

cos (x) - 0 = cos (x)

= cos (x)

= cos (x)

= cos (x)

We showed that the two sides could take the same form

\Rightarrow True

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