Sin 120-cos 150-cos 240-sin 330- is equal to

The correct option is B 1 sin #160;120^∘cos #160;150^∘ cos #160;240^∘sin #160;330^∘ =sin(90+30).cos(180 30) cos(180+60).sin(360 30) ...

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Trigonometric Ratios of Common Angles

sin 120∘cos 1...

Question

$sin 120^{\circ} cos 150^{\circ} - cos 240^{\circ} sin 330^{\circ}$ is equal to

1

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- 1

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\frac{2}{3}

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

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Prove that:

$(i) t a n 720^{\circ} - c o s 270^{\circ} - s i n 150^{\circ} c o s 120^{\circ} = \frac{1}{4}$

$(i i) s i n 780^{\circ} s i n 480^{\circ} + c o s 120^{\circ} s i n 150^{\circ} = \frac{1}{2}$

$(i i i) s i n 780^{\circ} s i n 120^{\circ} + c o s 240^{\circ} s i n 390^{\circ} = \frac{1}{2}$

$(i v) s i n 600^{\circ} c o s 390^{\circ} + c o s 480^{\circ} s i n 150^{\circ} = - 1$

$(v) t a n 250^{\circ} c o t 405^{\circ} + t a n 765^{\circ} c o t 675^{\circ} = 0$

\frac{1}{1 \cdot 2} + \frac{1 \cdot 3}{1 \cdot 2 \cdot 3 \cdot 4} + \frac{1 \cdot 3 \cdot 5}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6} + . . . . . . to \infty

\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \frac{1}{4 \cdot 5 \cdot 6}

{[\frac{- 1 + i \sqrt{3}}{2}]}^{4} + {[\frac{- 1 - i \sqrt{3}}{2}]}^{4} + {[\frac{- 1 + i \sqrt{3}}{2}]}^{5} + {[\frac{- 1 - i \sqrt{3}}{2}]}^{5}

\frac{1}{0!} + \frac{2}{1!} + \frac{3}{2!} + \frac{4}{3!} + . . .

\infty

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