jak do tego dojść? Jac:

	1
cos(arctg2)=
	√5

Mila: cos(arctg(2))=?

	π		π
arctg(2)=α ,α∊(−		,		)⇔
	2		2

tg(arctg(2))=tg(α)⇔

	sinα
2=tgα⇔		=2⇔
	cosα

sinα=2 cosα, z jedynki tryg. (2cosα)²+cos²α=1 5cos²α=1

	1
cos2α=
	5

	1		π		π
cosα=		, cosα>0 dla α∊(−		,		)
	√5		2		2

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