Tożsamości trygonometryczne

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Question język polski Answer język polski
sin" α + cos" α=
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sin" α + cos" α= in Polish
1
sin α = #cos α#
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+/- \|1 - cos" α\
cos α = #sin α#
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+/- \|1 - sin" α\
tg α= #sin & cos#
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sin α/ cos α
ctg α = #sin & cos#
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cos α / sin α
tg α * ctg α =
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1
sin" α = (cos)
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sin" α = (cos) in Polish
1 - cos" α
cos" α= (sin)
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cos" α= (sin) in Polish
1 - sin" α
sin(α+β)=
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sin α*cos β + sin β*cos α
sin(α-β)=
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sin α*cos β - sin β*cos α
cos(α+β)=
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cos α*cos β - sin α*sin β
cos(α-β)=
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cos α*cos β + sin α*sin β
cos 2α =
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cos 2α = in Polish
cos²α - sin²α = 2cos²α -1 = 1 -2sin²α = (1 -tg² α)/(1 +tg² α)
sin 2α=
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2sin α*cos α = (2tg α)/(1 +tg² α)
sin α + sin β=
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2* sin([α+β]/2) * cos([α-β]/2)
sin α - sin β=
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2* sin([α-β]/2) * cos([α+β]/2)
cos α + cos β=
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2* cos([α+β]/2) * cos([α-β]/2)
cos α - cos β=
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-2* sin([α+β]/2) * sin([α-β]/2)
tg (α+β)=
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(tg α +tg β)/(1 -tg α*tg β)
tg(α -β)=
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(tg α -tg β)/(1 +tg α*tg β)
tg 2α=
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(2tg α)/(1 -tg² α)
sin α= #x, y, r#
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y / r
cos α= #x, y, r#
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x / r
tg α= #x, y, r#
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y / x
ctg α= #x, y, r#
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x / y
sin α sin β=
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-0.5(cos(α+β)-cos(α-β))
cos α cos β=
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0.5(cos(α+β)+cos(α-β))
sin α cos β=
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0.5(sin(α+β)+sin(α-β))

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