Odpowiedź :
Odpowiedź:
[tex]sin(15^o) = sin(45^o-30^o) = sin(45^o)cos(30^o) - cos(45^o)sin(30^o) = \frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}[/tex]
[tex]cos(15^o) = cos(45^o-30^o) = cos(45^o)cos(30^o) +sin(45^o)sin(30^o) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{\sqrt{3}+1}{2 \sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\left(\sqrt{3}+1\right) \sqrt{2}}{4} = \frac{\sqrt{6}+\sqrt{2}}{4}[/tex]
[tex]tg(15^o) = \frac{sin(15^o)}{cos(15^o)} = \frac{\sqrt{6} - \sqrt{2}}{4} \cdot \frac{2\sqrt{2}}{\sqrt{3} + 1} =\frac{2\sqrt{2}(\sqrt{6}-\sqrt{2})}{4(\sqrt{3}+1)} = \frac{2\left(\sqrt{3}-1\right)}{2 \left(\sqrt{3}+1\right)} = \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{\left(\sqrt{3}-1\right) \left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right) \left(\sqrt{3}-1\right)} = \frac{\left(\sqrt{3}-1\right)^{2}}{2} = \frac{4-2 \sqrt{3}}{2} = 2 - \sqrt{3}[/tex]
[tex]cos(15^o) = \frac{1}{tg(15^o)} = \frac{1}{2-\sqrt{3}} = \frac{\sqrt{3}+2}{\left(2-\sqrt{3}\right) \left(\sqrt{3}+2\right)} = 2 + \sqrt{3}[/tex]