Question

cos +cos B +cos C =1+4*sin(A/2)sin(B/2)sin(C/2)

Answer 1

prove

Answer 2

\[cosA+cosB+cosC=1+4(sin(\frac{A}2)sin(\frac{B}2{})sin(\frac{C}2{}))\]

Answer 3

ya

Answer 4

james has proven capable of the impossible in the past :)

Answer 5

note A+B+C=180

Answer 6

This identity is false. Let A = B = C = pi/2. Then the LHS = 0 + 0 + 0 = 0. But the RHS = 1 + 4/2.sqrt(2) which is not zero.

Answer 7

Ah ... that makes a difference.

Answer 8

:)

Answer 9

lol .. its a question in progress :)

Answer 10

1+0+0= 1 if we go 0, 90, 90

Answer 11

prove as in its an identity? or prove that there exists some ....

Answer 12

1+4*sin(0/2)sin(45)sin(45)

1+4*0sin(45)sin(45)=1

Answer 13

prove by steps ya

Answer 14

First: for all x, y and z
cos x + cos y + cos z + cos (x+y+z)= 4cos(x+y/2) cos(y+z/2) cos(z+x/2)

Proof:
cos a + cos b = 2 cos ((a+b)/2) cos ((a-b)/2)

So
cos x + cos y + cos z + cos (x+y+z)=
= 2 cos ((x+y)/2) cos ((x-y)/2) + 2 cos ((x+y+2z)/2) cos ((x+y)/2)=
= 2 cos ((x+y)/2) * ( cos ((x-y)/2)+ cos ((x+y+2z)/2) )=
= 2 cos ((x+y)/2) * 2 cos ((x+z)/2) cos ((x+z)/2)=
= 4cos((x+y)/2) cos((y+z)/2) cos((z+x)/2)

Now x + y + z = pi, so cos of that is -1 and cos((x+y)/2) = cos(pi/2-z) = sin(z)

So now it all falls out.

Answer 15

Sorry ... last terms are cos((x+y)/2) = cos(pi/2-z/2) = sin(z/2)

Hence

cos x + cos y + cos z + cos (x+y+z)= 4cos(x+y/2) cos(y+z/2) cos(z+x/2)

=> cos x + cos y + cos z - 1 = 4.sin(z/2).sin(x/2).sin(y/2)

=> cos x + cos y + cos z = 1 + 4.sin(x/2).sin(y/2).sin(z/2)