Question

Find an exact value. sine of negative eleven pi divided by twelve.

Answer 1

sin(-11pi/12) = -sin(11pi/12) = -sin(pi - pi/12) = -sin(pi/12) = -sin( (pi/6) / 2)

= - sqrt( (1-cos(pi/6) ) / 2) = -sqrt( (1-√3/2) / 2 ) = -(√3-1) / 2√2

Answer 2

Answer:

The exact value is:

$\sin (\dfrac{-11\pi}{12})=-(\dfrac{\sqrt{3}-1}{2\sqrt{2}})$

Step-by-step explanation:

We are asked to find the value of:

$\sin (\dfrac{-11\pi}{12})$

We know that for any angle theta (θ) :

$\sin (-\theta)=-\sin \theta$

Hence, we get:

$\sin (\dfrac{-11\pi}{12})=-\sin (\dfrac{11\pi}{12})$

Hence,

$\sin (\dfrac{-11\pi}{12})=-\sin (\pi-\dfrac{\pi}{12})\\\\\\i.e.\\\\\\\sin (\dfrac{-11\pi}{12})=-\sin (\dfrac{\pi}{12})$

Since,

$\sin (\pi-\theta)=\sin \theta$

Also,

$\sin (\dfrac{\pi}{12})=\sin (\dfrac{\pi}{3}-\dfrac{\pi}{4})$

Hence,

on using the formula:

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

Hence, we get:

$\sin (\dfrac{\pi}{12})=\sin (\dfrac{\pi}{3})\cos (\dfrac{\pi}{4})-\cos (\dfrac{\pi}{3})\sin (\dfrac{\pi}{4})\\\\\\\sin (\dfrac{\pi}{12})=\dfrac{\sqrt{3}}{2}\cdot \dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\cdot \dfrac{1}{\sqrt{2}}\\\\\\\sin (\dfrac{\pi}{12})=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$

Hence, the value is:

$\sin (\dfrac{-11\pi}{12})=-(\dfrac{\sqrt{3}-1}{2\sqrt{2}})$