Is sin(x)sin(y)=2sin((x+y)/2)cos((x-y)/2) ?

The answer to whether sin(x)sin(y)=2sin((x+y)/2)cos((x-y)/2) is False

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prove sin(x)sin(y)=2sin((x+y)/2)cos((x-y)/2)

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

For

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

plug in

y = 1

sin (0) sin (1) = 0

sin (0) sin (1)

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 \cdot sin (1)

= 0 \cdot sin (1)

Simplify

= 0

2 sin (\frac{0 + 1}{2}) cos (\frac{0 - 1}{2}) = 0.84147 \dots

2 sin (\frac{0 + 1}{2}) cos (\frac{0 - 1}{2})

Refine to a decimal form

= 0.84147 \dots

The two sides are not equal

\Rightarrow False

Is sin(x)sin(y)=2sin((x+y)/2)cos((x-y)/2) ?
The answer to whether sin(x)sin(y)=2sin((x+y)/2)cos((x-y)/2) is False