Prove that: sin(4π/9 + θ) cos(π/9 + θ) − cos(4π/9 + θ) sin(π/9 + θ) = √3/2

Question

Prove that:

\[ \sin\left(\frac{4\pi}{9}+\theta\right)\cos\left(\frac{\pi}{9}+\theta\right) – \cos\left(\frac{4\pi}{9}+\theta\right)\sin\left(\frac{\pi}{9}+\theta\right) = \frac{\sqrt{3}}{2} \]

Proof

Consider the left-hand side:

\[ \sin\left(\frac{4\pi}{9}+\theta\right)\cos\left(\frac{\pi}{9}+\theta\right) – \cos\left(\frac{4\pi}{9}+\theta\right)\sin\left(\frac{\pi}{9}+\theta\right) \]

Using the identity:

\[ \sin A\cos B-\cos A\sin B=\sin(A-B) \]

Let

\[ A=\frac{4\pi}{9}+\theta, \qquad B=\frac{\pi}{9}+\theta \]

Then,

\[ = \sin\left[\left(\frac{4\pi}{9}+\theta\right)-\left(\frac{\pi}{9}+\theta\right)\right] \]

\[ = \sin\left(\frac{4\pi}{9}-\frac{\pi}{9}\right) \]

\[ = \sin\left(\frac{3\pi}{9}\right) \]

\[ = \sin\left(\frac{\pi}{3}\right) \]

We know that:

\[ \sin\frac{\pi}{3}=\frac{\sqrt{3}}{2} \]

Therefore,

\[ \sin\left(\frac{4\pi}{9}+\theta\right)\cos\left(\frac{\pi}{9}+\theta\right) – \cos\left(\frac{4\pi}{9}+\theta\right)\sin\left(\frac{\pi}{9}+\theta\right) = \frac{\sqrt{3}}{2} \]

Hence proved.