Value of sin 270 degrees

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Solution

sin 270° = sin (3 × 90°) = – sin 90° = – 1

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sin (270^{o})

sin (- 270^{o})

cos (270^{o} + θ) cos (90^{o} - θ) - sin (270^{o} - θ) cos θ

180 < θ < 270

{sin}^{2} 30^{\circ} + {sin}^{2} 60^{\circ} + 3 ({cos}^{2} 70^{\circ} + {cos}^{2} 20^{\circ}) = 4 K, t h e n t h e v a l u e o f K i s

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