Відповідь:

cos²67°30'-sin²67°30' =cos135° = -1/√2

[tex]\displaystyle\bf\\Cos^{2} 67^\circ 30'-Sin^{2} 67^\circ 30'=Cos(2\cdot 67^\circ 30')=Cos135^\circ=\\\\\\=Cos(90^\circ+45^\circ)=-Sin45^\circ=-\frac{\sqrt{2} }{2}[/tex]