1. Zadatak

Zapišimo dane jednakosti u obliku

$$2\cos a+9\cos d=-6\cos b-7\cos c, \qquad 2\sin a-9\sin d=6\sin b-7\sin c$$

i kvadrirajmo ih. Dobit ćemo

$$\begin{aligned} 4\cos^2a+36\cos a\cos d+81\cos^2d &=36\cos^2b+84\cos b\cos c+49\cos^2c,\\ 4\sin^2a-36\sin a\sin d+81\sin^2d &=36\sin^2b-84\sin b\sin c+49\sin^2c. \end{aligned}$$

Zbrajanjem i korištenjem identiteta $\sin^2x+\cos^2x=1$ (za $x=a,b,c,d$) dobivamo

$$4+81+36(\cos a\cos d-\sin a\sin d) =36+49+84(\cos b\cos c-\sin b\sin c),$$ $$3(\cos a\cos d-\sin a\sin d) =7(\cos b\cos c-\sin b\sin c),$$

što primjenom adicijskih formula za kosinus prelazi u

$$3\cos(a+d)=7\cos(b+c), \qquad\text{tj.}\qquad \frac{\cos(a+d)}{\cos(b+c)}=\frac73.$$