If $\sin α = \sin β a n d \cos α = \cos β$ then

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$\cos (\frac{α - β}{2}) = 0$

$\cos (\frac{α + β}{2}) = 0$

$\sin (\frac{α + β}{2}) = 0$

$\sin (\frac{α - β}{2}) = 0$

answer is C.

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Detailed Solution

$\sin α = \sin β \cos α = \cos β$

$\sin α - \sin β = 2 \cos (\frac{α + β}{2}) \sin (\frac{α - β}{2}) = 0 - 1 \cos α - \cos β = 2 \sin (\frac{α + β}{2}) \sin (\frac{α - β}{2}) = 0 - 2$

Compare the equation 1 from 2 , we get

$\sin α - \sin β = \cos α - \cos β 2 \cos (\frac{α + β}{2}) \sin (\frac{α - β}{2}) = 2 \sin (\frac{α + β}{2}) \sin (\frac{α - β}{2})$

$\sin (\frac{α - β}{2}) = 0$

So, option 3 is correct.

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