Answer:
False
sin x cos y [tex]\neq\frac{\bf 1}{\bf 2} [/tex] (sin (x + y) + cos (x - y))
Step-by-step explanation:
Given [tex]\sin x \cos y = 1/2(\sin(x + y) + \cos(x - y))\hfill (1)[/tex]
To verify that the equality is true or false
[tex]\sin (x+y)=\sin x \cos y+ \cos x \sin y[/tex] and
[tex]\cos (x-y)=\cos x \cos y+ \sin x \sin y[/tex]
Now adding the above equations we get
[tex]\sin (x+y)+ \cos (x-y)= \sin x \cos y+ \cos x \sin y+ \cos x \cos y+ \sin x \sin y\hfill (2)[/tex]
Comparing the equations (1) and (2) we get
[tex]\sin x \cos y \neq \frac {1}{2} \(sin (x + y) + \cos (x - y))[/tex]
Therefore the given equality is not true (ie, false)
