Prove that :
\(sin\cfrac{8\pi}3cos\cfrac{23\pi}6+cos\cfrac{13\pi}3sin\cfrac{35\pi}6=\cfrac12\)
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Prove that :
\(sin\cfrac{8\pi}3cos\cfrac{23\pi}6+cos\cfrac{13\pi}3sin\cfrac{35\pi}6=\cfrac12\)
Prove that :
\(sin\cfrac{8\pi}3cos\cfrac{23\pi}6+cos\cfrac{13\pi}3sin\cfrac{35\pi}6=\cfrac12\)
LHS = \(sin\cfrac{8\pi}3cos\cfrac{23\pi}6+cos\cfrac{13\pi}3sin\cfrac{35\pi}6\)
= sin 480° cos 690° + cos 780° sin 1050°
= sin (90° × 5 + 30°) cos (90° × 7 + 60°) + cos (90° × 8 + 60°) sin (90° × 11 + 60°)
We know that when n is odd, sin → cos and cos → sin.
= cos 30° sin 60° + cos 60° [-cos 60°]
= \(\cfrac{\sqrt3}2\times\cfrac{\sqrt3}2-\cfrac12\times\cfrac12\)
= 3/4 – 1/4
= 2/4
= 1/2
= RHS
Hence proved.