If A, B, C are the angles of a triangle, prove that cos2A + cos2B + cos2 C = 1-2cosAcosBcosC
I've tried to use A+B+C=180, but I can't get the cos2 separately..
this will help u
it is done for A+B=C
if A+B=C then prove that cos2A+cos2B + cos2C=1+2cosAcosB cos
it is ur task to do the same for a+b+c=180(method used will be identical)
½[sin ( Q+ β) + sin (Q − β)] = sin Q cos β,
sin ( Q+ β) + sin ( Q− β) = 2 sin Q cos β. . . . . . (1)
Q + β = A
Q− β = B. . . . . . . . . . . . . . . .(2)
The left-hand side of line (1) then becomes
sin A + sin B.
This is now the left-hand side of (e), which is what we are trying to prove.
To complete the right−hand side of line (1), solve those simultaneous equations (2) for Q and β.
On adding them, 2Q = A + B,
Q = ½(A + B).
On subtracting those two equations, 2β = A − B,
β = ½(A − B).
On the right−hand side of line (1), substitute those expressions for and β. Line (1) then becomes
sin A + sin B = 2 sin ½(A + B) cos ½(A − B).
HERE A=2C AND B=2D
SUCH THAT sin 2C + sin 2D = 2 sin(C + D) cos (C − D).
IT IS YOUR TASK TO PROVE TH SAME FOR ADDITION OF 2 COSINES....