MATH SOLVE

2 months ago

Q:
# Describe how you would convince a friend that sin(x + y) does not equal sin x + sin y.

Accepted Solution

A:

If the claim is that [tex]\sin(x+y)=\sin x+\sin y[/tex] for all [tex]x,y[/tex], then all you need to refute it is a single example for which the equality doesn't hold. Take [tex]x=y=\dfrac\pi2[/tex]. Then [tex]\sin\left(\dfrac\pi2+\dfrac\pi2\right)=\sin\pi=0[/tex], but [tex]\sin\dfrac\pi2+\sin\dfrac\pi2=1+1=2[/tex].

If the claim is that there is some choice of [tex]x,y[/tex] for which equality holds, then we can recall the angle sum identity for sine:

[tex]\sin(x+y)=\sin x\cos y+\cos x\sin y[/tex]

which means that, in order to have equality, we need to choose [tex]x,y[/tex] such that both [tex]\cos y=1[/tex] and [tex]\cos x=1[/tex]. We can do that; [tex]\cos x=1[/tex] only when [tex]x[/tex] is an even multiple of [tex]\pi[/tex], i.e. [tex]x\in\{0,\pm2\pi,\pm4\pi,\ldots\}[/tex]. Take any two of these and add them together, and you'd end up with yet another even multiple of [tex]\pi[/tex], and so both sides will reduce to 0. Thus equality holds only when [tex]x=2n\pi[/tex] and [tex]y=2m\pi[/tex] for some integers [tex]m,n[/tex].

If the claim is that there is some choice of [tex]x,y[/tex] for which equality holds, then we can recall the angle sum identity for sine:

[tex]\sin(x+y)=\sin x\cos y+\cos x\sin y[/tex]

which means that, in order to have equality, we need to choose [tex]x,y[/tex] such that both [tex]\cos y=1[/tex] and [tex]\cos x=1[/tex]. We can do that; [tex]\cos x=1[/tex] only when [tex]x[/tex] is an even multiple of [tex]\pi[/tex], i.e. [tex]x\in\{0,\pm2\pi,\pm4\pi,\ldots\}[/tex]. Take any two of these and add them together, and you'd end up with yet another even multiple of [tex]\pi[/tex], and so both sides will reduce to 0. Thus equality holds only when [tex]x=2n\pi[/tex] and [tex]y=2m\pi[/tex] for some integers [tex]m,n[/tex].