Given:
[tex]\cos 15^{\circ}[/tex]
To find:
The exact value of cos 15°.
Solution:
[tex]$\cos 15^{\circ}=\cos\frac{ 30^{\circ}}{2}[/tex]
Using half-angle identity:
[tex]$\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos (x)}{2}}[/tex]
[tex]$\cos \frac{30^{\circ}}{2}=\sqrt{\frac{1+\cos \left(30^{\circ}\right)}{2}}[/tex]
Using the trigonometric identity: [tex]\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}[/tex]
[tex]$=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}[/tex]
Let us first solve the fraction in the numerator.
[tex]$=\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{2}}[/tex]
Using fraction rule: [tex]\frac{\frac{a}{b} }{c}=\frac{a}{b \cdot c}[/tex]
[tex]$=\sqrt{\frac {2+\sqrt{3}}{4}}[/tex]
Apply radical rule: [tex]\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}[/tex]
[tex]$=\frac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}[/tex]
Using [tex]\sqrt{4} =2[/tex]:
[tex]$=\frac{\sqrt{2+\sqrt{3}}}{2}[/tex]
[tex]$\cos 15^\circ=\frac{\sqrt{2+\sqrt{3}}}{2}[/tex]
