2 Answers

Kolen · Answer 1 · 2022-09-13T14:48:52+0000

Answer:

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

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:

$\sin 105^(\circ) = \sin 135^(\circ)\cdot \cos 30^(\circ) - \cos 135^(\circ)\cdot \sin 30^(\circ)$

$\sin 105^(\circ) = \left((√(2))/(2) \right)\cdot \left((√(3))/(2)\right)-\left(-(√(2))/(2) \right)\cdot \left((1)/(2)\right)$

$\sin 105^(\circ) = (√(6))/(4)+(√(2))/(4)$

$\sin 105^(\circ) = \sqrt{(6)/(16) }+\sqrt{(2)/(16) }$

$\sin 105^(\circ) = \sqrt{(3)/(8) }+\sqrt{(1)/(8) }$

$\sin 105^(\circ) = (√(3)+1)/(2\cdot √(2))$

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

Please log in or register to add a comment.