Question

Find \cos\left(\dfrac{19\pi}{12}\right)cos( 12 19π ​ )cosine, left parenthesis, start fraction, 19, pi, divided by, 12, end fraction, right parenthesis exactly using an angle addition or subtraction formula.

Answer 1

Answer:

The value of given expression is $-\frac{\sqrt{2}-\sqrt{6}}{4}$.

Step-by-step explanation:

The given expression is

$\cos\left(\dfrac{19\pi}{12}\right)$

The trigonometric ratios are not defined for $\dfrac{19\pi}{12}$.

$\dfrac{19\pi}{12}$ can be split into $\frac{5\pi}{4}+\frac{\pi}{3}$.

$\cos\left(\dfrac{19\pi}{12}\right)=\cos (\frac{5\pi}{4}+\frac{\pi}{3})$

Using the addition formula

$\cos (A+B)=\cos A\cos B-\sin A\sin B$

$\cos (\frac{5\pi}{4}+\frac{\pi}{3})=\cos( \frac{\pi}{3})\cdot \cos (\frac{5\pi}{4})-\sin( \frac{\pi}{3})\cdot \sin (\frac{5\pi}{4})$

We know that, $\cos(\frac{\pi}{3})=\frac{1}{2}$ and $\sin (\frac{\pi}{3})=\frac{\sqrt{3}}{2}$

$\cos\left(\dfrac{19\pi}{12}\right)=\frac{1}{2}\cdot \cos (\frac{5\pi}{4})-\frac{\sqrt{3}}{2}\cdot \sin (\frac{5\pi}{4})$

$\frac{5\pi}{4}$ lies in third quadrant, by using reference angle properties,

$\cos(\frac{5\pi}{4})=-\cos(\frac{\pi}{4})=-\frac{\sqrt{2}}{2}$

$\sin(\frac{5\pi}{4})=-\sin(\frac{\pi}{4})=-\frac{\sqrt{2}}{2}$

$\cos\left(\dfrac{19\pi}{12}\right)=\frac{1}{2}\cdot (-\frac{\sqrt{2}}{2})-\frac{\sqrt{3}}{2}\cdot (-\frac{\sqrt{2}}{2})$

$\cos\left(\dfrac{19\pi}{12}\right)=-\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}$

$\cos\left(\dfrac{19\pi}{12}\right)=-\frac{(\sqrt{2}-\sqrt{6})}{4}$

Therefore the value of given expression is $-\frac{\sqrt{2}-\sqrt{6}}{4}$.