Prove that:
sin \(\cfrac{13\pi}3\) sin \(\cfrac{2\pi}3\) + cos \(\cfrac{4\pi}3\) sin \(\cfrac{13\pi}6\) = \(\cfrac12\)
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Prove that:
sin \(\cfrac{13\pi}3\) sin \(\cfrac{2\pi}3\) + cos \(\cfrac{4\pi}3\) sin \(\cfrac{13\pi}6\) = \(\cfrac12\)
Prove that:
sin \(\cfrac{13\pi}3\) sin \(\cfrac{2\pi}3\) + cos \(\cfrac{4\pi}3\) sin \(\cfrac{13\pi}6\) = \(\cfrac12\)
LHS = sin \(\cfrac{13\pi}3\) sin \(\cfrac{2\pi}3\) + cos \(\cfrac{4\pi}3\) sin \(\cfrac{13\pi}6\)
= sin 780° sin 120° + cos 240° sin 390°
= sin (90° × 8 + 60°) sin (90° × 1 + 30°) + cos (90° × 2 + 60°) sin (90° × 4 + 30°)
We know that when n is odd, sin → cos.
= sin 60° cos 30° + [-cos 60°] sin 30°
= sin 60° cos 30° – sin 30° cos 60°
We know that sin A cos B – cos A sin B = sin (A – B)
= sin (60° – 30°)
= sin 30°
= 1/2
= RHS
Hence proved.