Show that sin 2-sin 2-cos 2-cos 2- -

sin 2α #160; +sin 2β #160; #160; cos 2α #160; cos 2β #160; #160; = 2sin( 2α +2β #160; 2 #160; ) #160; cos( 2α 2β #160; 2 ...

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Polar Representation of a Complex Number

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Question

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

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

c o s^{2} α - s i n^{2} α = t a n^{2} β .

t a n^{2} α = c o s^{2} β - s i n^{2} β .

{cos}^{2} α - {sin}^{2} α = {tan}^{2} β,

{tan}^{2} α = {cos}^{2} β - {sin}^{2} β

cos α + cos β + cos α = 0 = sin α + sin β + sin α

{cos}^{2} α + {cos}^{2} β + {cos}^{2} α = \frac{3}{α} = {sin}^{2} α + {sin}^{2} β + {sin}^{2} α

α + β - γ = π

{sin}^{2} α + {sin}^{2} β - {sin}^{2} γ = 2 sin α sin β cos γ

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