How do you verify the identity $\frac{sin x + sin y}{cos x - cos y} = - cot (\frac{x - y}{2})$ $(sinx+siny)/(cosx-cosy)=-cot((x-y)/2)$ ?

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How do you verify the identity $\frac{sin x + sin y}{cos x - cos y} = - cot (\frac{x - y}{2})$ $(sinx+siny)/(cosx-cosy)=-cot((x-y)/2)$ ?

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Answer 1 · 2017-01-06T16:56:55

Please see below.

Explanation:

Using identities $sin A + sin B = 2 sin (\frac{A + B}{2}) cos (\frac{A - B}{2})$ $sinA+sinB=2sin((A+B)/2)cos((A-B)/2)$ and $cos A - cos B = 2 sin (\frac{A + B}{2}) sin (\frac{B - A}{2})$ $cosA-cosB=2sin((A+B)/2)sin((B-A)/2)$ , hence

$\frac{sin x + sin y}{cos x - cos y}$ $(sinx+siny)/(cosx-cosy)$

= $\frac{2 sin (\frac{x + y}{2}) cos (\frac{x - y}{2})}{2 sin (\frac{x + y}{2}) sin (\frac{y - x}{2})}$ $(2sin((x+y)/2)cos((x-y)/2))/(2sin((x+y)/2)sin((y-x)/2))$

= $\frac{cos (\frac{x - y}{2})}{sin (\frac{y - x}{2})}$ $(cos((x-y)/2))/(sin((y-x)/2))$

and as $sin (- A) = - sin A$ $sin(-A)=-sinA$ , we have $sin (\frac{y - x}{2}) = - sin (\frac{x - y}{2})$ $sin((y-x)/2)=-sin((x-y)/2)$

Hence above becomes

$\frac{cos (\frac{x - y}{2})}{- sin (\frac{x - y}{2})} = - cot (\frac{x - y}{2})$ $(cos((x-y)/2))/(-sin((x-y)/2))=-cot((x-y)/2)$

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How do you verify the identity $\frac{sin x + sin y}{cos x - cos y} = - cot (\frac{x - y}{2})$ $(sinx+siny)/(cosx-cosy)=-cot((x-y)/2)$ ?

1 Answer

Explanation:

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How do you verify the identity (sinx+siny)/(cosx-cosy)=-cot((x-y)/2)sinx+sinycosx−cosy=−cot(x−y2)(sinx+siny)/(cosx-cosy)=-cot((x-y)/2)?

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Explanation:

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How do you verify the identity $\frac{sin x + sin y}{cos x - cos y} = - cot (\frac{x - y}{2})$ $(sinx+siny)/(cosx-cosy)=-cot((x-y)/2)$ ?