Is sin(2pi-B)=-sin(B) ?

The answer to whether sin(2pi-B)=-sin(B) is True

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prove sin(2pi-B)=-sin(B)

prove

sin (2 π - B) = - sin (B)

True

Solution steps

sin (2 π - B) = - sin (B)

Manipulating left side

sin (2 π - B)

Rewrite using trig identities

sin (2 π - B)

Use the Angle Difference identity:

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

= sin (2 π) cos (B) - cos (2 π) sin (B)

Simplify

sin (2 π) cos (B) - cos (2 π) sin (B) : - sin (B)

sin (2 π) cos (B) - cos (2 π) sin (B)

sin (2 π) cos (B) = 0

sin (2 π) cos (B)

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 cos (B)

Apply rule

0 \cdot a = 0

= 0

cos (2 π) sin (B) = sin (B)

cos (2 π) sin (B)

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 sin (B)

Multiply:

1 \cdot sin (B) = sin (B)

= sin (B)

= 0 - sin (B)

0 - sin (B) = - sin (B)

= - sin (B)

= - sin (B)

= - sin (B)

We showed that the two sides could take the same form

\Rightarrow True

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