Is prove-2sin(2x+pi)=-2sin(2x-pi) ?

The answer to whether prove-2sin(2x+pi)=-2sin(2x-pi) is True

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prove-2sin(2x+pi)=-2sin(2x-pi)

Solution

prove

- 2 sin (2 x + π) = - 2 sin (2 x - π)

Solution

True

Solution steps

- 2 sin (2 x + π) = - 2 sin (2 x - π)

Manipulating left side

- 2 sin (2 x + π)

Rewrite using trig identities

sin (2 x + π)

Use the Angle Sum identity:

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

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

Simplify

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

sin (2 x) cos (π) + cos (2 x) sin (π)

sin (2 x) cos (π) = - sin (2 x)

sin (2 x) cos (π)

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) sin (2 x)

Refine

= - sin (2 x)

= - sin (2 x) + sin (π) cos (2 x)

cos (2 x) sin (π) = 0

cos (2 x) sin (π)

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 cos (2 x)

Apply rule

0 \cdot a = 0

= 0

= - sin (2 x) + 0

- sin (2 x) + 0 = - sin (2 x)

= - sin (2 x)

= - sin (2 x)

= - 2 (- sin (2 x))

Apply rule

- (- a) = a

= 2 sin (2 x)

Manipulating right side

- 2 sin (2 x - π)

Rewrite using trig identities

sin (2 x - π)

Use the Angle Difference identity:

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

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

Simplify

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

sin (2 x) cos (π) - cos (2 x) sin (π)

sin (2 x) cos (π) = - sin (2 x)

sin (2 x) cos (π)

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) sin (2 x)

Refine

= - sin (2 x)

= - sin (2 x) - sin (π) cos (2 x)

cos (2 x) sin (π) = 0

cos (2 x) sin (π)

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 cos (2 x)

Apply rule

0 \cdot a = 0

= 0

= - sin (2 x) - 0

- sin (2 x) - 0 = - sin (2 x)

= - sin (2 x)

= - sin (2 x)

= - 2 (- sin (2 x))

Apply rule

- (- a) = a

= 2 sin (2 x)

We showed that the two sides could take the same form

\Rightarrow True

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