If in a triangle ABC, ab = c^{2} 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

OldTimer May 26, 2019

#1**+3 **

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)),

\(\displaystyle \cos(A)\cos(B)-\sin(A)\sin(B)+\cos(C)+1\)

\(\displaystyle =\cos(A+B)+\cos(C)+1\)

\(\displaystyle =\cos(180-C)+\cos(C)+1\)

\(\displaystyle =-\cos(C)+\cos(C)+1 \)

\(\displaystyle =1.\)

.Tiggsy May 27, 2019