If cos ($\alpha$ + $\beta$) = 0, then sin ($\alpha$ – $\beta$) can be reduced to ______.

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If cos ($\alpha$ + $\beta$) = 0, then sin ($\alpha$ – $\beta$) can be reduced to cos 2$\beta$. Explanation: According to the question, cos($\alpha$ + $\beta$) = 0 Since, cos $90^\circ$ = 0 We can write, cos($\alpha$ + $\beta$) = cos $90^\circ$ By comparing cosine equation on L.H.S and R.H.S, We get, ($\alpha$ + $\beta$) = $90^\circ$ $\alpha$ = $90^\circ$ – $\beta$ Now we need to reduce sin($\alpha$ – $\beta$), So, we take, sin($\alpha$ – $\beta$) = sin($90^\circ$ – $\beta$ – $\beta$) = sin($90^\circ$ – 2$\beta$) sin($90^\circ$ – $\theta$) = cos $\theta$ So, sin($90^\circ$ – 2$\beta$) = cos 2$\beta$ Therefore, sin($\alpha$ – $\beta$) = cos 2$\beta$

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