Udowodnij tożsamość: (sin a+cos a)^2 + (sin a-cos a)^2 = 2
Związki między funkcjami trygonometrycznymi tego samego kąta:
sin²α+cos²α=1 (jedynka trygonometryczna)
tgα=sinα/cosα
ctgα=cosα/sinα
tgα*ctgα=1
===============================================
(sinα+cosα)²+(sinα-cosα)²=2
L=(sinα+cosα)²+(sinα-cosα)²=
=sin²α+2sinαcosα+cos²α+sin²α-2sinαcosα+cos²α=
=sin²α+cos²α+sin²α+cos²α=
=2sin²α+2cos²α=
=2(sin²α+cos²α)=
=2*1=
=2=P c.n.d
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Związki między funkcjami trygonometrycznymi tego samego kąta:
sin²α+cos²α=1 (jedynka trygonometryczna)
tgα=sinα/cosα
ctgα=cosα/sinα
tgα*ctgα=1
===============================================
(sinα+cosα)²+(sinα-cosα)²=2
L=(sinα+cosα)²+(sinα-cosα)²=
=sin²α+2sinαcosα+cos²α+sin²α-2sinαcosα+cos²α=
=sin²α+cos²α+sin²α+cos²α=
=2sin²α+2cos²α=
=2(sin²α+cos²α)=
=2*1=
=2=P c.n.d