Wiemy, że:

$\alpha ={15}^{\circ },\beta ={30}^{\circ }$  

zatem:

$\gamma ={180}^{\circ }-{15}^{\circ }-{30}^{\circ }={135}^{\circ }$  

 

Korzystamy ze wzorów redukcyjnych:

$\cos \gamma =\cos {135}^{\circ }=\cos \left({180}^{\circ }-{45}^{\circ }\right)=-\cos {45}^{\circ }=-\frac{\sqrt{2}}{2}$  

więc:

$A.-\sqrt{2}\cdot \sin {30}^{\circ }=-\sqrt{2}\cdot \frac{1}{2}=-\frac{\sqrt{2}}{2}$  

$B.\sin \left({15}^{\circ }+{30}^{\circ }\right)=\sin {45}^{\circ }=\frac{\sqrt{2}}{2}$  

$C.-0,5\cdot \mathrm{ctg}\left({15}^{\circ }+{30}^{\circ }\right)=-0,5\cdot \mathrm{ctg}{45}^{\circ }=-0,5\cdot 1=-0,5$  

$D.-\mathrm{tg}\left(2\cdot {30}^{\circ }\right)=-\mathrm{tg}{60}^{\circ }=-\sqrt{3}\ne \sqrt{3}\cdot \left(-\frac{\sqrt{2}}{2}\right)$  

Prawidłowa odpowiedź to A.