1.

Prove that:sin \(\cfrac{13\pi}3\) sin \(\cfrac{8\pi}3\) + cos \(\cfrac{2\pi}3\) sin \(\cfrac{5\pi}6\) = \(\cfrac12\)

Answer»

LHS = sin \(\cfrac{13\pi}3\) sin \(\cfrac{8\pi}3\) + cos \(\cfrac{2\pi}3\) sin \(\cfrac{5\pi}6\) 

= sin 780° sin 480° + cos 120° sin 150°

= sin (90° × 8 + 60°) sin (90° × 5 + 30°) + cos (90° × 1 + 30°) sin (90° × 1 + 60°)

We know that when n is odd, cos → sin and sin → cos.

= sin 60° cos 30° + [-sin 30°] cos 60°

= 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.



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