Answer:
Proof in the explanation
Step-by-step explanation:
Trigonometric Equalities
Those are expressions involving trigonometric functions which must be proven, generally working on only one side of the equality
For this particular equality, we'll use the following equation
[tex]\displaystyle cos(x-y)=cos\ x\ cos\ y+sin\ x\ sin\ y[/tex]
The equality we want to prove is
[tex]\displaystyle arccos\ \sqrt{\frac{2}{3}}-arccos\left(\frac{1+\sqrt{6}}{2\sqrt{3}}\right)=\frac{\pi}{6}[/tex]
Let's set the following variables:
[tex]\displaystyle x=arccos\ \sqrt{\frac{2}{3}},\ y=arccos(\frac{1+\sqrt{6}}{2\sqrt{3}})[/tex]
And modify the first variable:
[tex]\displaystyle x=arccos\ \frac{\sqrt{6}}{3}}=>\ cos\ x= \frac{\sqrt{6}}{3}}[/tex]
Now with the second variable
[tex]\displaystyle y=arccos\ \frac{1+\sqrt{6}}{2\sqrt{3}}=>cos\ y=\frac{1+\sqrt{6}}{2\sqrt{3}}=\frac{\sqrt{3}+3\sqrt{2}}{6}[/tex]
Knowing that
[tex]sin^2x+cos^2x=1[/tex]
We compute the other two trigonometric functions of X and Y
[tex]\displaystyle sin \ x=\sqrt{1-cos^2\ x}=\sqrt{1-(\frac{\sqrt{6}}{3})^2}=\sqrt{1-\frac{6}{9}}=\frac{\sqrt{3}}{3}[/tex]
[tex]\displaystyle sin\ y=\sqrt{1-cos^2y}=\sqrt{1-\frac{(\sqrt{3}+3\sqrt{2})^2}{36}}}[/tex]
[tex]\displaystyle sin\ y=\sqrt{\frac{36-(3+6\sqrt{6}+18)}{36}}=\sqrt{\frac{15-6\sqrt{6}}{36}}[/tex]
Computing
[tex]15-6\sqrt{6}=(3-\sqrt{6})^2[/tex]
Then
[tex]\displaystyle sin\ y=\frac{3-\sqrt{6}}{6}[/tex]
Now we replace all in the first equality:
[tex]\displaystyle cos(x-y)=\frac{\sqrt{6}}{3}.\frac{\sqrt{3}+3\sqrt{2}}{6}+\frac{\sqrt{3}}{3}.\frac{3-\sqrt{6}}{6}[/tex]
[tex]\displaystyle cos(x-y)=\frac{3\sqrt{2}+6\sqrt{3}}{18}+\frac{3\sqrt{3}-3\sqrt{2}}{18}[/tex]
[tex]\displaystyle cos(x-y)=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}=cos\ \pi/6[/tex]
Thus, proven
