Question

What is the exact value of sin(105°)?ANSWER - D) sqrt(2 + sqrt3)/2

Answer 1

Answer:

$\sin 105^{\circ} = \frac{\sqrt{6}+\sqrt {2}}{4}$

Step-by-step explanation:

We can use the following trigonometrical identity:

$\sin (\alpha - \beta) = \sin \alpha \cdot \cos \beta - \cos \alpha \cdot \sin \beta$ (1)

Where $\alpha$, $\beta$ are angles, measured in sexagesimal degrees.

If we know that $\alpha = 135^{\circ}$ and $\beta = 30^{\circ}$, then the value of the given function is:

$\sin 105^{\circ} = \sin 135^{\circ}\cdot \cos 30^{\circ} - \cos 135^{\circ}\cdot \sin 30^{\circ}$

$\sin 105^{\circ} = \left(\frac{\sqrt{2}}{2} \right)\cdot \left(\frac{\sqrt{3}}{2}\right)-\left(-\frac{\sqrt{2}}{2} \right)\cdot \left(\frac{1}{2}\right)$

$\sin 105^{\circ} = \frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}$

$\sin 105^{\circ} = \sqrt{\frac{6}{16} }+\sqrt{\frac{2}{16} }$

$\sin 105^{\circ} = \sqrt{\frac{3}{8} }+\sqrt{\frac{1}{8} }$

$\sin 105^{\circ} = \frac{\sqrt{3}+1}{2\cdot \sqrt{2}}$

$\sin 105^{\circ} = \frac{\sqrt{6}+\sqrt {2}}{4}$