If in a triangle ABC, ab = c2 prove that
cos(A - B) + cos C + cos 2C = 1
NB - could only solve if a and b are equal to c [equilateal]...but not sure if correct reasoning.
Hope you are having a good day...regards
From the sine rule asin(A)=bsin(B)=csin(C),
we can say that
absin(A)sin(B)=c2sin2(C),
and since we are told that ab=c2,
it follows that
sin(A)sin(B)=sin2(C)…(1)
Expanding the LHS of the (supposed) identity using standard trig identities, we have
cos(A)cos(B)+sin(A)sin(B)+cos(C)+1−2sin2(C)
which equals, (making use of (1)),
cos(A)cos(B)−sin(A)sin(B)+cos(C)+1
=cos(A+B)+cos(C)+1
=cos(180−C)+cos(C)+1
=−cos(C)+cos(C)+1
=1.