If cos(α+β) = 0, then sin(α-β) can be reduced to

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

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