- #1

- 65

- 2

*Prove for any triangle ABC that:*

[tex]cos(A)+cos(B)+cos(C)=1+4sin(\frac{1}{2}A)sin(\frac{1}{2}B)sin(\frac{1}{2}C)[/tex]

I tried using the half-angle formula on the right side to get:

[tex]1+4 \left(\frac{1-cos(A)}{2}\frac{1-cos(b)}{2}sin(\frac{1}{2}C)\right) {}[/tex]

which simplifies to

[tex]1+\bigg(1-cos(A)-cos(B)+cos(A)cos(B) \bigg)*sin(\frac{1}{2}C)[/tex]

by the product to sum rule, this simplifies to

[tex]1+\bigg(1-cos(A)-cos(B)+\frac{1}{2}\Big[cos(A+B)+cos(A-B)\Big]\bigg)*sin(\frac{1}{2}C)[/tex]

I tried expanding the sin 1/2 C, but that just made things even more complicated. How should I approach this problem?

Last edited: