Is cos(pi+x)+cos(2pi-x)=0 ?

The answer to whether cos(pi+x)+cos(2pi-x)=0 is True

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prove cos(pi+x)+cos(2pi-x)=0

Solution

prove

cos (π + x) + cos (2 π - x) = 0

Solution

True

Solution steps

cos (π + x) + cos (2 π - x) = 0

Manipulating left side

cos (π + x) + cos (2 π - x)

Rewrite using trig identities

cos (2 π - x)

Use the Angle Difference identity:

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

= cos (2 π) cos (x) + sin (2 π) sin (x)

Simplify

cos (2 π) cos (x) + sin (2 π) sin (x) : cos (x)

cos (2 π) cos (x) + sin (2 π) sin (x)

cos (2 π) cos (x) = cos (x)

cos (2 π) cos (x)

cos (2 π) = 1

cos (2 π)

cos (2 π) = cos (0)

cos (2 π)

Rewrite

2 π

2 π + 0

= cos (2 π + 0)

Apply the periodicity of

cos

cos (x + 2 π) = cos (x)

cos (2 π + 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 (2 π) sin (x) = 0

sin (2 π) sin (x)

sin (2 π) = 0

sin (2 π)

sin (2 π) = sin (0)

sin (2 π)

Rewrite

2 π

2 π + 0

= sin (2 π + 0)

Apply the periodicity of

sin

sin (x + 2 π) = sin (x)

sin (2 π + 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 (π + x) + cos (x)

Rewrite using trig identities

cos (π + x)

Use the Angle Sum identity:

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

= cos (π) cos (x) - sin (π) sin (x)

Simplify

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

cos (π) cos (x) - sin (π) sin (x)

cos (π) cos (x) = - cos (x)

cos (π) cos (x)

Simplify

cos (π) : - 1

cos (π)

Use the following trivial identity:

cos (π) = (- 1)

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 \cdot cos (x)

Multiply:

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

= - cos (x)

= - cos (x) - sin (π) sin (x)

sin (π) sin (x) = 0

sin (π) sin (x)

Simplify

sin (π) : 0

sin (π)

Use the following trivial identity:

sin (π) = 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 \cdot sin (x)

Apply rule

0 \cdot a = 0

= 0

= - cos (x) - 0

- cos (x) - 0 = - cos (x)

= - cos (x)

= - cos (x)

= - cos (x) + cos (x)

Add similar elements:

- cos (x) + cos (x) = 0

= 0

We showed that the two sides could take the same form

\Rightarrow True

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Is cos(pi+x)+cos(2pi-x)=0 ?
The answer to whether cos(pi+x)+cos(2pi-x)=0 is True