trigonometria 28

por felipe12 Seg 30 Jul 2012, 20:03

Prove que se A+B+C=0 entao senA+senB-senC= -4scosA/2.cosB/2.senC/2

por Luck Ter 31 Jul 2012, 16:37

senA+senB-senC = x
A+B+ C = 0
A+ B = -C
sen(A+B) = sen(-C)
senA.cosB + senB.cosA = -senC
somando senA + senB :

senA.cosB + senB.cosA + senA + senB = -senC + senA + senB
x = senA.cosB + senB.cosA + senA + senB
x = senA( 1+ cosB) + senB ( 1+ cosA)
cosB = 2cos²(B/2) - 1
seny = 2sen(y/2)cos(y/2)

x = 2sen(A/2)cos(A/2) (2cos²B/2) + 2sen(B/2)cos(B/2).2cos²(A/2)
x = 4cos(B/2)cos(A/2) ( sen(A/2)cos(B/2) + sen?(B/2).cos(A/2) )
x = 4cos(B/2)cos(A/2)sen(A+B/2)
x = 4cos(B/2)cos(A/2)sen(-C/2)
x = -4cos(B/2)cos(A/2).sen(C/2)

trigonometria 28

trigonometria 28

Re: trigonometria 28

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