Question

Demonstre que: (cos α – cos β) · (cos α + cos β) + (sen α – sen β) · (sen α + sen β) = 0

Answer 1

(cos α – cos β) · (cos α + cos β) + (sen α – sen β) · (sen α + sen β) = 0

cos² α – cos² β) + sen² α + sen² β = 0

{sen² α + cos² α } + sen² β – cos² β  = 0

  1 +  sen² β – cos² β  = 0

cos² β  = 1 +  sen² β

sen² β + cos² β = 1 ==> cos² β =  1 - sen² β
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Substituindo :


1 +  sen² β = 1 -  sen² β

sen² β + sen² β = 1 - 1

2sen² β = 0

senβ = 0
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cos² β  = 1 +  sen² β

cos² β  = 1 + 0 ==>  cos² β  = 1 ==> cos β  = 0

Answer 2

$(cos \alpha -cos \beta).(cos \alpha + cos \beta)+(sen \alpha-sen \beta).(sen \alpha+sen \beta)=0 \\ cos^{2} \alpha+cos \alpha.cos \beta-cos \alpha .cos \beta -cos^{2} \beta + \\ + sen^{2} \alpha +sen \alpha.sen \beta-sen \alpha.sen \beta-sen^{2} \beta =0$

Como:
$sen^{2}x+cos^{2}x=1 \\ cos^{2}x=1-sen^{2}x \\ -sen^{2}x=-1+cos^{2}x$

Logo:
$1-sen^{2}\alpha-cos^{2}\beta+sen^{2}\alpha-1+cos^{2}\beta=0 \\ 1-1-sen^{2}\alpha+sen^{2}\alpha-cos^{2}\beta+cos^{2}\beta=0 \\ 0=0$

Espero ajudar... (: