W zadaniu skorzystamy z wartości funkcji trygonometrycznych podanych w tabeli na stronie 32.

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$\text{a)}5\cdot \sin {30}^{\circ }+4\cdot \cos {60}^{\circ }+\text{tg}{45}^{\circ }=5\cdot \frac{1}{2}+4\cdot \frac{1}{2}+1=\frac{5}{2}+\frac{4}{2}+1=\frac{9}{2}+1=4,5+1=5,5$  

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$\text{b)}3\sin {60}^{\circ }-5\text{tg}{45}^{\circ }+3\text{tg}{30}^{\circ }=3\cdot \frac{\sqrt{3}}{2}-5\cdot 1+3\cdot \frac{\sqrt{3}}{3}=\frac{3\sqrt{3}}{2}-5+\sqrt{3}=\frac{3\sqrt{3}}{2}+\frac{2\sqrt{3}}{2}-5=\frac{5\sqrt{3}}{2}-5$  

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$\text{c)}{\sin }^2{45}^{\circ }+{\cos }^2{45}^{\circ }+{\text{tg}}^2{45}^{\circ }={\left(\frac{\sqrt{2}}{2}\right)}^2+{\left(\frac{\sqrt{2}}{2}\right)}^2+1^2=\frac{2}{4}+\frac{2}{4}+1=\frac{4}{4}+1=1+1=2$  

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$\text{d)}\frac{2-{\text{tg}}^2{60}^{\circ }}{\sin {30}^{\circ }\cdot \cos {60}^{\circ }}=\frac{2-{\left(\sqrt{3}\right)}^2}{\frac{1}{2}\cdot \frac{1}{2}}=\frac{2-3}{\frac{1}{4}}=-1\cdot 4=-4$