Question

using an addition or subtraction formula sin (43π/12)

Answer 1

$$
\sin(A+B)=\sin A\cos B+\cos A\sin B
$$
\(\cfrac{43\pi}{12}\) is \(-75^\circ\) in degrees
So
$$
\sin(-75)=\sin(15-90)=\sin (15)\cos (-90)+\cos (15)\sin (-90)\\
=-\cos (15)
$$
\(15^\circ\) is \(\cfrac{\pi}{12}\).
$$
\cfrac{\pi}{12}=\cfrac{\pi}{3}-\cfrac{\pi}{4}\\
$$
Using a similar trigonometric identity for the cos sum of angle formula, we finally arrive at our answer:
$$
-\cos(A+B)=-\cos A \cos B + \sin A \sin B\\
-\cos \cfrac{1}{12}=-\cos \cfrac{\pi}{3} \cos \cfrac{+\pi}{4} - \sin \cfrac{\pi}{3} \sin \cfrac{-\pi}{4}\\
=-\left (\cfrac{1}{2}\cfrac{\sqrt 2}{2}+\cfrac{\sqrt 3}{2}\cfrac{\sqrt 2}{2}\right )\\
=-\cfrac{1+\sqrt 3}{2\sqrt 2}
$$
Make sense?

Answer 2

yes thank you for being for helping me. Appreciate it