Cong thuc luong giac

pdf:

Code:

http://www.faidherbe.org/~pcsimath/pcsi2/doc/formultrigo.pdf

http://www.ilemaths.net/forum-sujet-197155.html

cos(3t) = 4.cos³(t)-3.cos(t)

cos(3t) = cos(2t+t)
= cos(2t).cos(t) - sin(2t).sin(t)
= [2.cos²(t)-1].cos(t) - 2.sin²(t).cos(t)
= [2.cos²(t)-1].cos(t) - 2.[1-cos²(t)].cos(t)
= 2.cos³(t)-cos(t)-2.cos(t)+2.cos³(t)
= 4.cos³(t)-3.cos(t)

cos(2t) = 2.cos²(t)-1
sin(2t) = 2.sin(t).cos(t)

cos(2t) = 1-2.sin²(t)
sin(2t) = 2.sin(t).cos(t)
----------------------------------------------
sin(3x) = - 4 sin^3 x + 3 sin x

sin(3t) = sin(2t+t)
= sin(2t).cos(t) + cos(2t).sin(t)
= 2.sin(t).cos²(t) + [1-2.sin²(t)].sin(t)
= 2.sin(t).[1-sin²(t)] + [1-2.sin²(t)].sin(t)
= sin(t).[2-2.sin²(t)] + [1-2.sin²(t)].sin(t)
= sin(t).[2-2.sin²(t)+1-2.sin²(t)]
= sin(t).[3-4.sin²(t)]
= 3.sin(t)-4.sin³(t)
= -4.sin³(t)+3.sin(t)

sin(a)+sin(b)

\Large{\sin(a)=\cos(\frac{\pi}{2}-a)}

tu poses :
a=c + pi/2
b=d+pi/2

et :
cos(a)+cos(b) = - sin(c) - sin(d)

2*cos((a-b)/2)*cos((a+b)/2) =-2*cos((c-d)/2)*sin((c+d)/2)

tu conclues

sin(c) + sin(d) = 2*cos((c-d)/2)*sin((c+d)/2)