1.

Prove that cos 510° cos 330° + sin 390° cos 120° = -1.

Answer»

The vales must be following:

Cos 510° = cos (90° × 5 + 60° = - sin 60°) = \(\frac{-\sqrt 3}{2}\)

Cos 330° = cos (90° × 3 + 60° = sin 60°) = sin 60° = \(\frac{-\sqrt 3}{2}\)

Sin 390° = sin (90° × 4 + 30°) = sin 30° = \(\frac{1}{2}\)

Cos 120° = cos (90° × 1 + 30°) = – sin 30° = \(\frac{-1}{2}\)

Now,

L.H.S = cos 510° cos 330° + sin 390° cos 120°

\((\frac{-\sqrt 2}{2})\)\((\frac{\sqrt 3}{2})\)+\((\frac{1}{2})\)\((\frac{-1}{2})\)

\((\frac{-3}{4})+(\frac{-1}{4})\)

\((\frac{-3}{4}+\frac{-1}{4})\)

= − 1 = R. H. S



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