Answer:
[tex]\sin 105^{\circ} = \frac{\sqrt{6}+\sqrt {2}}{4}[/tex]
Step-by-step explanation:
We can use the following trigonometrical identity:
[tex]\sin (\alpha - \beta) = \sin \alpha \cdot \cos \beta - \cos \alpha \cdot \sin \beta[/tex] (1)
Where [tex]\alpha[/tex], [tex]\beta[/tex] are angles, measured in sexagesimal degrees.
If we know that [tex]\alpha = 135^{\circ}[/tex] and [tex]\beta = 30^{\circ}[/tex], then the value of the given function is:
[tex]\sin 105^{\circ} = \sin 135^{\circ}\cdot \cos 30^{\circ} - \cos 135^{\circ}\cdot \sin 30^{\circ}[/tex]
[tex]\sin 105^{\circ} = \left(\frac{\sqrt{2}}{2} \right)\cdot \left(\frac{\sqrt{3}}{2}\right)-\left(-\frac{\sqrt{2}}{2} \right)\cdot \left(\frac{1}{2}\right)[/tex]
[tex]\sin 105^{\circ} = \frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}[/tex]
[tex]\sin 105^{\circ} = \sqrt{\frac{6}{16} }+\sqrt{\frac{2}{16} }[/tex]
[tex]\sin 105^{\circ} = \sqrt{\frac{3}{8} }+\sqrt{\frac{1}{8} }[/tex]
[tex]\sin 105^{\circ} = \frac{\sqrt{3}+1}{2\cdot \sqrt{2}}[/tex]
[tex]\sin 105^{\circ} = \frac{\sqrt{6}+\sqrt {2}}{4}[/tex]
