According to the trigonometric properties, we have the following
1.
[tex]\sin (x+y)=\sin x\cdot\cos y+\cos x\cdot\sin y[/tex]
2.
[tex]\cos (x-y)=\cos x\cdot\cos y+\sin x\cdot\sin y[/tex]
we replace the above into the right hand side of the equation
[tex]\frac{1}{2}((\sin x\cdot\cos y+\cos x\cdot\sin y)\cdot(\cos x\cdot\cos y+\sin x\cdot\sin y))[/tex]
1/2now, let's continue expanding the above
[tex]\frac{1}{2}((\sin x\cdot\cos y\cdot\cos x\cdot\cos y)+(\sin x\cdot\cos y\cdot\sin x\cdot\sin y)+(\cos x\cdot\sin y\cdot\cos x\cdot\cos y)+(\cos x\cdot\sin y\cdot\sin x\cdot\sin y))[/tex]
if we simplify this, we will obtain
[tex]=\frac{1}{2}\cos \mleft(x-y\mright)\sin \mleft(x+y\mright)[/tex]
on the other hand,
From the trigonometric property sinx * cosy, this is the same as
[tex]\sin x\cdot\cos y=\frac{1}{2}((\sin x-y)+\sin (x+y))[/tex]
Therefore, the statement must be false
