$\text{Dla}\alpha \in ⟨0^{\circ },{180}^{\circ }⟩:$   $\sin \left({180}^{\circ }-\alpha \right)=\sin \alpha$   $\cos \left({180}^{\circ }-\alpha \right)=-\cos \alpha$   $\text{tg}\left({180}^{\circ }-\alpha \right)=-\text{tg}\alpha$

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W rozwiązaniu będziemy korzystać z tabelki wartości funkcji trygonometrycznych dla kątów 30°, 45°, 60°:

$\alpha$	${30}^{\circ }$	${45}^{\circ }$	${60}^{\circ }$
$\sin \alpha$	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$
$\cos \alpha$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$
$\text{tg}\alpha$	$\frac{\sqrt{3}}{3}$	$1$	$\sqrt{3}$

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$\text{a)}\left(\sin {30}^{\circ }+\cos {45}^{\circ }\right)\left(\sin {150}^{\circ }+\cos {135}^{\circ }\right)=$  

$=\left(\sin {30}^{\circ }+\cos {45}^{\circ }\right)\left[\sin \left({180}^{\circ }-{30}^{\circ }\right)+\cos \left({180}^{\circ }-{45}^{\circ }\right)\right]=$  

Korzystamy ze wzorów redukcyjnych.

$=\left(\sin {30}^{\circ }+\cos {45}^{\circ }\right)\left(\sin {30}^{\circ }-\cos {45}^{\circ }\right)=$  

Podstawiamy wartości z tabeli.

$=\left(\frac{1}{2}+\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}-\frac{\sqrt{2}}{2}\right)={\left(\frac{1}{2}\right)}^2-{\left(\frac{\sqrt{2}}{2}\right)}^2=\frac{1}{4}-\frac{2}{4}=-\frac{1}{4}$  

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$\text{b)}\left(\cos {45}^{\circ }+\text{tg}{60}^{\circ }\right)\left(\text{tg}{120}^{\circ }-\cos {135}^{\circ }\right)$  

$=\left(\cos {45}^{\circ }+\text{tg}{60}^{\circ }\right)\left[\text{tg}\left({180}^{\circ }-{60}^{\circ }\right)-\cos \left({180}^{\circ }-{45}^{\circ }\right)\right]=$  

Korzystamy ze wzorów redukcyjnych.

$=\left(\cos {45}^{\circ }+\text{tg}{60}^{\circ }\right)\left(-\text{tg}{60}^{\circ }+\cos {45}^{\circ }\right)=$  

$=\left(\cos {45}^{\circ }+\text{tg}{60}^{\circ }\right)\left(\cos {45}^{\circ }-\text{tg}{60}^{\circ }\right)=$  

Podstawiamy wartości z tabeli.

$=\left(\frac{\sqrt{2}}{2}+\sqrt{3}\right)\left(\frac{\sqrt{2}}{2}-\sqrt{3}\right)={\left(\frac{\sqrt{2}}{2}\right)}^2-{\left(\sqrt{3}\right)}^2=\frac{2}{4}-3=\frac{1}{2}-3=-2\frac{1}{2}$  

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$\text{c)}\left(\cos {45}^{\circ }\right):\left(\sin {120}^{\circ }\cdot \cos {135}^{\circ }\cdot \text{tg}{150}^{\circ }\right)=$  

$=\left(\cos {45}^{\circ }\right):\left[\sin \left({180}^{\circ }-{60}^{\circ }\right)\cdot \cos \left({180}^{\circ }-{45}^{\circ }\right)\cdot \text{tg}\left({180}^{\circ }-{30}^{\circ }\right)\right]=$  

Korzystamy ze wzorów redukcyjnych.

$=\left(\cos {45}^{\circ }\right):\left[\sin {60}^{\circ }\cdot \left(-\cos {45}^{\circ }\right)\cdot \left(-\text{tg}{30}^{\circ }\right)\right]=$  

$=\left(\cos {45}^{\circ }\right):\left(\sin {60}^{\circ }\cdot \cos {45}^{\circ }\cdot \text{tg}{30}^{\circ }\right)=$  

$=\frac{\cos {45}^{\circ }}{\sin {60}^{\circ }\cdot \cos {45}^{\circ }\cdot \text{tg}{30}^{\circ }}=\frac{1}{\sin {60}^{\circ }\cdot \text{tg}{30}^{\circ }}=$  

Podstawiamy wartości z tabeli.

$=\frac{1}{\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{3}}=\frac{1}{\frac{3}{6}}=\frac{1}{\frac{1}{2}}=2$  

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$\text{d)}\left(\cos {120}^{\circ }\cdot \cos {135}^{\circ }\cdot \cos {150}^{\circ }\right):\sin {60}^{\circ }=$  

$=\left[\cos \left({180}^{\circ }-{60}^{\circ }\right)\cdot \cos \left({180}^{\circ }-{45}^{\circ }\right)\cdot \cos \left({180}^{\circ }-{30}^{\circ }\right)\right]:\sin {60}^{\circ }=$  

Korzystamy ze wzorów redukcyjnych.

$=\left[-\cos {60}^{\circ }\cdot \left(-\cos {45}^{\circ }\right)\cdot \left(-\cos {30}^{\circ }\right)\right]:\sin {60}^{\circ }=$  

$=\left(-\cos {60}^{\circ }\cdot \cos {45}^{\circ }\cdot \cos {30}^{\circ }\right):\sin {60}^{\circ }=$  

$=\frac{-\cos {60}^{\circ }\cdot \cos {45}^{\circ }\cdot \cos {30}^{\circ }}{\sin {60}^{\circ }}=$  

Podstawiamy wartości z tabeli.

$=\frac{-\frac{1}{2}\cdot \frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}}=-\frac{1}{2}\cdot \frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{4}$