Prove that: \[ \sin \frac{5\pi}{18} – \cos \frac{4\pi}{9} = \sqrt{3}\sin \frac{\pi}{9} \]
Solution
Convert the cosine term into sine form using:
\[
\cos \theta = \sin\left(\frac{\pi}{2}-\theta\right)
\]
\[
\cos \frac{4\pi}{9}
=
\sin\left(\frac{\pi}{2}-\frac{4\pi}{9}\right)
=
\sin\frac{\pi}{18}
\]
Therefore,
\[
\sin \frac{5\pi}{18} – \cos \frac{4\pi}{9}
=
\sin \frac{5\pi}{18} – \sin \frac{\pi}{18}
\]
Using the identity:
\[
\sin A – \sin B
=
2\cos\frac{A+B}{2}\sin\frac{A-B}{2}
\]
Taking
\[
A=\frac{5\pi}{18},\qquad B=\frac{\pi}{18}
\]
Then,
\[
=
2\cos\frac{\frac{5\pi}{18}+\frac{\pi}{18}}{2}
\sin\frac{\frac{5\pi}{18}-\frac{\pi}{18}}{2}
\]
\[
=
2\cos\frac{6\pi}{36}\sin\frac{4\pi}{36}
\]
\[
=
2\cos\frac{\pi}{6}\sin\frac{\pi}{9}
\]
\[
=
2\times\frac{\sqrt{3}}{2}\sin\frac{\pi}{9}
\]
\[
=
\sqrt{3}\sin\frac{\pi}{9}
\]
Hence,
\[
\boxed{
\sin \frac{5\pi}{18} – \cos \frac{4\pi}{9}
=
\sqrt{3}\sin \frac{\pi}{9}
}
\]