5.zadatak
[dispmath]\sin^2(45^\circ+\alpha)-\sin^2(30^\circ-\alpha)-\sin15^\circ\cos(15^\circ+2\alpha)[/dispmath] Dva sinusa na kvadrat posmatracu kao razliku kvadrata i rastavicu ih, a ostatak jednacine [inlmath]\sin\cdot\cos[/inlmath] transformisacu iz proizvoda u zbir...
[dispmath]\bigl(\sin(45^\circ+\alpha)-\sin(30^\circ-\alpha)\bigr)\cdot\bigl(\sin(45^\circ+\alpha)+\sin(30^\circ-\alpha)\bigr)-\frac{1}{2}\cdot[\sin(-2\alpha)+\sin(30^\circ+2\alpha)]\\
\enclose{circle}{2\cos\frac{75^\circ}{2}}\enclose{box}{\sin\frac{15^\circ+2\alpha}{2}}\cdot\enclose{circle}{2\sin\frac{75^\circ}{2}}\enclose{box}{\cos\frac{15^\circ+2\alpha}{2}}+\frac{1}{2}\sin{2\alpha}-\frac{1}{2}\sin(30^\circ+2\alpha)\\
\sin75^\circ\cdot\sin(15^\circ+2\alpha)+\frac{1}{2}\sin2\alpha-\frac{1}{2}\sin(30^\circ+2\alpha)\\
\frac{1}{2}\cdot[\cos(60^\circ-2\alpha)-\cos(90^\circ+2\alpha)]+\frac{1}{2}\sin2\alpha-\frac{1}{2}\sin(30^\circ+2\alpha)\\
\cancel{\frac{1}{2}\cdot\sin(90^\circ-60^\circ+2\alpha)}+\frac{1}{2}\cdot\sin2\alpha+\frac{1}{2}\sin2\alpha\cancel{-\frac{1}{2}\sin(30^\circ+2\alpha)}\\
\enclose{box}{\sin2\alpha}[/dispmath]