Given: cos(α+β) = 0

We can write, cos(α+β)= cos 90° (∵ , cos 90° = 0)

By comparing cosine equation on either sides,

We get(α+β)= 90°

⇒ α = 90°-β

Now we need to reduce sin (α -β )

So, sin(α-β) = sin(90°-β-β) (∵, we have got the value of α, which is α = 90°-β)

= sin(90°-2β)

= cos 2β (∵, sin(90°-θ) = cos θ)

Therefore, sin(α-β) = cos 2β