I Need Some Clever Manipulation

Note first that, by using the relevant sum to product identity , we have that the left-hand side of the given equation can be written as

$A = (\cos(12) + \cos(84)) + \cos(60) = 2\cos(36)\cos(48) + \dfrac{1}{2}$ .

Then using the identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$ we see that

$2\sin(36)*A = 2\sin(72)\cos(48) + \sin(36) = \sin(120) + \sin(24) + \sin(36)$ ,

this last step coming as a result of one of the product to sum identities listed in the aforementioned link. Two more applications of sum to product formulas, along with the facts that $\sin(x) = \sin(180 - x)$ and $\sin(x) = \cos(90 - x)$ , yields that

$2\sin(36)*A = \sin(60) + (\sin(24) + \sin(36)) = \cos(30) + 2\sin(30)\cos(6) =$

$\cos(30) + \cos(6) = 2\cos(18)\cos(12)$ .

Resorting to yet more sum to product/product to sum/double-angle 'trickery', we find that

$2\sin(36)\cos(36)*A = 2\cos(36)\cos(18)\cos(12) \Longrightarrow \sin(72)*A = 2\cos(36)\cos(18)\cos(12)$

$\Longrightarrow \cos(18)*A = \cos(18)*(2\cos(36)\cos(12)) \Longrightarrow A = \cos(24) + \cos(48)$ ,

thus proving that the given equation is indeed $\boxed{true}$ .

cos 60° = 0.5

cos 12° = 0.97815

cos 84° = 0.10453

L.H.S= cos 60°+ cos 12° +cos 84°

     = 0.5 + 0.97815 + 0.10453    

     = 1.58268

cos 24° = 0.91355

cos 48° = 0.66913

R.H.S= cos 24° + cos 48°

      = 0.91355 + 0.66913

      = 1.58268

Therefore, L.H.S=R.H.S