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Show that ( 3A - A)/( 3A + A) = 2A

Mathematics

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3. (i)
Self-correction: The problem asks to show that sin⁡3A−sin⁡Acos⁡3A+cos⁡A=tan⁡2A\frac{\sin 3A - \sin A}{\cos 3A + \cos A} = \tan 2A.
Using sum-to-product identities:
sin⁡X−sin⁡Y=2cos⁡(X+Y2)sin⁡(X−Y2)\sin X - \sin Y = 2 \cos \left(\frac{X+Y}{2}\right) \sin \left(\frac{X-Y}{2}\right)
cos⁡X+cos⁡Y=2cos⁡(X+Y2)cos⁡(X−Y2)\cos X + \cos Y = 2 \cos \left(\frac{X+Y}{2}\right) \cos \left(\frac{X-Y}{2}\right)

Let X=3AX = 3A and Y=AY = A.

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