Question

2sin2pi/3cospi/2 Express the following as a sum or difference of two trigonometric functions​

Answer 1

Answer:

$\displaystyle{2\sin\dfrac{2\pi}{3} \ \cos\dfrac{3\pi}{2}}\\\\\\\displaystyle{\implies\left[\sin\left(\dfrac{13\pi}{6}\right)-\sin\left(\dfrac{5\pi}{6}\right)\right]}$

Step-by-step explanation:

$\displaystyle{2\sin\dfrac{2\pi}{3} \ \cos\dfrac{3\pi}{2}}\\\\\\\displaystyle{Using \ formula}\\\\\\\displaystyle{2\sin A\cos B=\sin(A+B)+\sin(A-B)}\\\\\\\displaystyle{2\sin\dfrac{2\pi}{3}\cos\dfrac{3\pi}{2}}\\\\\\\displaystyle{\implies\sin\left(\dfrac{2\pi}{3}+\dfrac{3\pi}{2}\right)+\sin\left(\dfrac{2\pi}{3}-\dfrac{3\pi}{2}\right)}\\\\\\\displaystyle{\implies\sin\left(\dfrac{13\pi}{6}\right)+\sin\left(\dfrac{-5\pi}{6}\right)}$

$\displaystyle{\implies\left[\sin\left(\dfrac{13\pi}{6}\right)-\sin\left(\dfrac{5\pi}{6}\right)\right]}$

Thus we get answer .