Answer:
[tex]\sin(105) = \frac{\sqrt 2 + \sqrt 6}{4}[/tex]
Step-by-step explanation:
Given
[tex]\sin(105^o)[/tex]
Required
Solve
Using sine rule, we have:
[tex]\sin(A + B) = \sin(A)\cos(B) + \sin(B)\cos(A)[/tex]
This gives:
[tex]\sin(105^o) = \sin(60 + 45)[/tex]
So, we have:
[tex]\sin(60 + 45) = \sin(60)\cos(45) + \sin(45)\cos(60)[/tex]
In radical forms, we have:
[tex]\sin(60 + 45) = \frac{\sqrt 3}{2} * \frac{\sqrt 2}{2} + \frac{\sqrt 2}{2} * \frac{1}{2}[/tex]
[tex]\sin(60 + 45) = \frac{\sqrt 6}{4} + \frac{\sqrt 2}{4}[/tex]
Take LCM
[tex]\sin(60 + 45) = \frac{\sqrt 6 + \sqrt 2}{4}[/tex]
Rewrite as:
[tex]\sin(60 + 45) = \frac{\sqrt 2 + \sqrt 6}{4}[/tex]
Hence:
[tex]\sin(105) = \frac{\sqrt 2 + \sqrt 6}{4}[/tex]
