Relações Trigonométricas
Por: eduardamaia17 • 15/4/2018 • 1.210 Palavras (5 Páginas) • 367 Visualizações
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cos α = 1 − 2 sen2 α/2
tan α = √[ (1/cos2 α) − 1 ]
cot α = √[ (1/sen2 α) − 1 ]
sen α = √[ (1 − cos 2α) / 2 ]
cos α = √[ (1 + cos 2α) / 2 ]
tan α = [ √(1 − cos2 α) ] / cos α
cot α = [ √(1 − sen2 α) ] / sen α
sen α = 1 / √(1 + cot2 α)
cos α = 1 / √(1 + tan2 α)
sen 2α = 2 sen α cos α
cos 2α = cos2 α − sen2 α
cos 2α = 2 cos2 α − 1
cos 2α = 1 − 2 sen2 α
tan 2α = 2 tan α / (1 − tan2 α)
tan 2α = 2 / (cot α − tan α)
cot 2α = (cot2 α − 1) / (2 cot α)
cot 2α = (1/2) cot α − (1/2) tan α
sen α/2 = √[ (1 − cos α) / 2 ]
cos α/2 = √[ (1 + cos α) / 2 ]
tan α/2 = sen α / (1 + cos α)
cot α/2 = sen α / (1 − cos α)
tan α/2 = (1 − cos α) / sen α
cot α/2 = (1 + cos α) / sen α
tan α/2 = √[ (1 − cos α) / (1 + cos α) ]
cot α/2 = √[ (1 + cos α) / (1 − cos α) ]
Relações de funções inversas
Relações básicasarccos x = π/2 − arcsen x
arccot x = π/2 − arctan x
Relações de soma / subtraçãoarcsen a [pic 15] arcsen b = arcsen [ a √(1 − b2) [pic 16] b √(1 − a2) ]
arccos a [pic 17] arccos b = arccos [ a b [pic 18] √(1 − a2) √(1 − b2) ]
arctan a [pic 19] arctan b = arctan [ (a [pic 20] b) / (1 [pic 21] a b) ]
arccot a [pic 22] arccot b = arccot [ (a b [pic 23] 1) / (b [pic 24] a) ]
Relações entre funções (para x > 0)
arcsen x = arccos √(1 − x2) = arctan [ x / √(1 − x2) ] = arccot [ √(1 − x2) / x ]
arccso x = arcsen √(1 − x2) = arctan [ √(1 − x2) / x ] = arccot [ x / √(1 − x2) ]
arctan x = arcsen [ x / √(1 + x2) ] = arccos [ 1 / √(1 + x2) ] = arccot [ 1 / x ]
arccot x = arcsen [ 1 / √(1 + x2) ] = arccos [ x / √(1 + x2) ] = arctan [ 1 / x ]
Para x , valem as relações:
arcsen (−x) = − arcsen x
arccos (−x) = π − arccos x
arctan (−x) = − arctan x
arccot (−x) = π − arccot x
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