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 \(\displaystyle \frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)},\)
we can say that
\(\displaystyle \frac{ab}{\sin(A)\sin(B)}=\frac{c^{2}}{\sin^{2}(C)},\)
and since we are told that \(\displaystyle ab=c^{2},\)
it follows that
\(\displaystyle \sin(A)\sin(B)=\sin^{2}(C) \dots(1)\)
Expanding the LHS of the (supposed) identity using standard trig identities, we have
\(\displaystyle \cos(A)\cos(B)+\sin(A)\sin(B)+\cos(C)+1-2\sin^{2}(C)\)
which equals, (making use of (1)),
\(\displaystyle \cos(A)\cos(B)-\sin(A)\sin(B)+\cos(C)+1\)
\(\displaystyle =\cos(A+B)+\cos(C)+1\)
\(\displaystyle =\cos(180-C)+\cos(C)+1\)
\(\displaystyle =-\cos(C)+\cos(C)+1 \)
\(\displaystyle =1.\)