Is 2sin(2x)sin(2x-pi/2)=-cos(4x-pi/2) ?

The answer to whether 2sin(2x)sin(2x-pi/2)=-cos(4x-pi/2) is True

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Popular >

prove 2sin(2x)sin(2x-pi/2)=-cos(4x-pi/2)

Solution

prove

2 sin (2 x) sin (2 x - \frac{π}{2}) = - cos (4 x - \frac{π}{2})

Solution

True

Solution steps

2 sin (2 x) sin (2 x - \frac{π}{2}) = - cos (4 x - \frac{π}{2})

Manipulating left side

2 sin (2 x) sin (2 x - \frac{π}{2})

Rewrite using trig identities

sin (2 x - \frac{π}{2})

Use the Angle Difference identity:

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

= sin (2 x) cos (\frac{π}{2}) - cos (2 x) sin (\frac{π}{2})

Simplify

sin (2 x) cos (\frac{π}{2}) - cos (2 x) sin (\frac{π}{2}) : - cos (2 x)

sin (2 x) cos (\frac{π}{2}) - cos (2 x) sin (\frac{π}{2})

sin (2 x) cos (\frac{π}{2}) = 0

sin (2 x) cos (\frac{π}{2})

Simplify

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 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}

= 0

= 0 \cdot sin (2 x)

Apply rule

0 \cdot a = 0

= 0

cos (2 x) sin (\frac{π}{2}) = cos (2 x)

cos (2 x) sin (\frac{π}{2})

Simplify

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

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}

= 1

= 1 \cdot cos (2 x)

Multiply:

cos (2 x) \cdot 1 = cos (2 x)

= cos (2 x)

= 0 - cos (2 x)

0 - cos (2 x) = - cos (2 x)

= - cos (2 x)

= - cos (2 x)

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

Remove parentheses:

(- a) = - a

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

Rewrite using trig identities

- 2 sin (2 x) cos (2 x)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

= - sin (2 \cdot 2 x)

Simplify

= - sin (4 x)

= - sin (4 x)

Manipulating right side

- cos (4 x - \frac{π}{2})

Rewrite using trig identities

cos (4 x - \frac{π}{2})

Use the Angle Difference identity:

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

= cos (4 x) cos (\frac{π}{2}) + sin (4 x) sin (\frac{π}{2})

Simplify

cos (4 x) cos (\frac{π}{2}) + sin (4 x) sin (\frac{π}{2}) : sin (4 x)

cos (4 x) cos (\frac{π}{2}) + sin (4 x) sin (\frac{π}{2})

cos (4 x) cos (\frac{π}{2}) = 0

cos (4 x) cos (\frac{π}{2})

Simplify

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 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}

= 0

= 0 \cdot cos (4 x)

Apply rule

0 \cdot a = 0

= 0

sin (4 x) sin (\frac{π}{2}) = sin (4 x)

sin (4 x) sin (\frac{π}{2})

Simplify

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

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}

= 1

= 1 \cdot sin (4 x)

Multiply:

sin (4 x) \cdot 1 = sin (4 x)

= sin (4 x)

= 0 + sin (4 x)

0 + sin (4 x) = sin (4 x)

= sin (4 x)

= sin (4 x)

= - sin (4 x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is 2sin(2x)sin(2x-pi/2)=-cos(4x-pi/2) ?
The answer to whether 2sin(2x)sin(2x-pi/2)=-cos(4x-pi/2) is True