Respuesta :
Answer:
The value of given expression is [tex]-\frac{\sqrt{2}-\sqrt{6}}{4}[/tex].
Step-by-step explanation:
The given expression is
[tex]\cos\left(\dfrac{19\pi}{12}\right)[/tex]
The trigonometric ratios are not defined for [tex]\dfrac{19\pi}{12}[/tex].
[tex]\dfrac{19\pi}{12}[/tex] can be split into [tex]\frac{5\pi}{4}+\frac{\pi}{3}[/tex].
[tex]\cos\left(\dfrac{19\pi}{12}\right)=\cos (\frac{5\pi}{4}+\frac{\pi}{3})[/tex]
Using the addition formula
[tex]\cos (A+B)=\cos A\cos B-\sin A\sin B[/tex]
[tex]\cos (\frac{5\pi}{4}+\frac{\pi}{3})=\cos( \frac{\pi}{3})\cdot \cos (\frac{5\pi}{4})-\sin( \frac{\pi}{3})\cdot \sin (\frac{5\pi}{4})[/tex]
We know that, [tex]\cos(\frac{\pi}{3})=\frac{1}{2}[/tex] and [tex]\sin (\frac{\pi}{3})=\frac{\sqrt{3}}{2}[/tex]
[tex]\cos\left(\dfrac{19\pi}{12}\right)=\frac{1}{2}\cdot \cos (\frac{5\pi}{4})-\frac{\sqrt{3}}{2}\cdot \sin (\frac{5\pi}{4})[/tex]
[tex]\frac{5\pi}{4}[/tex] lies in third quadrant, by using reference angle properties,
[tex]\cos(\frac{5\pi}{4})=-\cos(\frac{\pi}{4})=-\frac{\sqrt{2}}{2}[/tex]
[tex]\sin(\frac{5\pi}{4})=-\sin(\frac{\pi}{4})=-\frac{\sqrt{2}}{2}[/tex]
[tex]\cos\left(\dfrac{19\pi}{12}\right)=\frac{1}{2}\cdot (-\frac{\sqrt{2}}{2})-\frac{\sqrt{3}}{2}\cdot (-\frac{\sqrt{2}}{2})[/tex]
[tex]\cos\left(\dfrac{19\pi}{12}\right)=-\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}[/tex]
[tex]\cos\left(\dfrac{19\pi}{12}\right)=-\frac{(\sqrt{2}-\sqrt{6})}{4}[/tex]
Therefore the value of given expression is [tex]-\frac{\sqrt{2}-\sqrt{6}}{4}[/tex].
