40 POINTS!Find the exact value by using a half-angle identity.⬇️⬇️⬇️

[tex]\cos\dfrac{5\pi}{12}=\cos\left(\dfrac{6\pi}{12}-\dfrac{\pi}{12}\right)=\cos\left(\dfrac{\pi}{2}-\dfrac{\pi}{12}\right)=(*)[/tex]

[tex]Use: \cos(\alpha-\beta)=\cos\alpha \cos\beta+\sin\alpha \cos\beta[/tex]

[tex](*)=\cos\dfrac{\pi}{2} \cos\dfrac{\pi}{12}+\sin\dfrac{\pi}{2} \sin\dfrac{\pi}{12}\\\\=0\cdot\cos\dfrac{\pi}{12}+1\cdot\sin\dfrac{\pi}{12}=\sin\dfrac{\pi}{12}[/tex]

[tex]\dfrac{\pi}{12}=\dfrac{\dfrac{\pi}{6}}{2}[/tex]

[tex]The\ half-angle\ identity\ \sin^2\dfrac{\alpha}{2}=\dfrac{1}{2}\left(1-\cos\alpha\right)[/tex]

[tex]Using\ the\ above\ formulas,\ we\ get:\\\\\sin^2\dfrac{\pi}{12}=\dfrac{1}{2}\left(1-\cos\dfrac{\pi}{6}\right)\\\\\sin^2\dfrac{\pi}{12}=\dfrac{1}{2}\left(1-\dfrac{\sqrt3}{2}\right)\\\\\sin^2\dfrac{\pi}{12}=\dfrac{1}{2}-\dfrac{\sqrt3}{4}[/tex]

[tex]Since\ 0 \ \textless \ \dfrac{\pi}{12} \ \textless \ \pi,\ then\ \sin\dfrac{\pi}{12}\ is\ a\ positive\ number.\\\\ Therefore,\ we\ have:\\\\\sin\dfrac{\pi}{12}=\sqrt{\dfrac{1}{2}-\dfrac{\sqrt3}{4}}=\sqrt{\dfrac{2-\sqrt3}{4}}=\dfrac{\sqrt{2-\sqrt3}}{2}[/tex]

[tex]Answer:\ \boxed{\cos\dfrac{5\pi}{12}=\dfrac{\sqrt{2-\sqrt3}}{2}}[/tex]