Solution : The value of cos 120 degrees is \(-1\over 2\). Proof : We Know that sin 60 = \(\sqrt{3}\over 2\) and cos 60 = \(1\over 2\) By using cos 2A formula, cos 120 = \(cos^2 60\) โ \(sin^2 60\) = \(1\over 4\) โ \(3\over 4\) \(\implies\)ย cos 120 = \(-1\over 2\)
Solution : The value of sin 120 degrees is \(\sqrt{3}\over 2\). Proof : We Know that sin 60 = \(\sqrt{3}\over 2\) and cos 60 = \(1\over 2\) By using the formula of sin 2A, sin 120 = 2 sin 60 cos 60 = 2 \(\times\) \(\sqrt{3}\over 2\) \(\times\) \(1\over 2\) \(\implies\)ย sin 120 = \(\sqrt{3}\over โฆ
Solution : The formula of cos C โ cos D isย 2 sin(\(C + D\over 2\)) sin(\(D โ C\over 2\)). Example : Express cos \(6\theta\) โ cos \(8\theta\) as a product. Solution : We have, cos \(6\theta\) โ cos \(8\theta\) = 2 sin(\(6\theta + 8\theta\over 2\)) sin(\(8\theta โ 6 \theta\over 2\)) = 2 sin \(7\theta\) sin โฆ
Solution : The formula of cos C + cos D is 2 cos(\(C + D\over 2\)) cos(\(C โ D\over 2\)). Example : Express cos \(4\theta\) + cos \(8\theta\) as a product. Solution : We have, cos \(4\theta\) + cos \(8\theta\) = 2 cos(\(8\theta + 4\theta\over 2\)) cos(\(8\theta โ 4\theta\over 2\)) = 2 cos \(6\theta\) cos โฆ
Solution : The formula of sin C โ sin D is 2 sin(\(C โ D\over 2\)) cos(\(C + D\over 2\)). Example : Express sin \(6\theta\) โ sin \(2\theta\) as a product. Solution : We have, sin \(6\theta\) โ sin \(2\theta\) = 2 sin(\(6\theta โ 2\theta\over 2\)) cos(\(6\theta + 2\theta\over 2\)) = 2 sin \(2\theta\) cos โฆ
Solution : The formula of sin C + sin D is 2 sin(\(C + D\over 2\)) cos(\(C โ D\over 2\)). Example : Express sin \(4\theta\) + sin \(2\theta\) as a product. Solution : We have, sin \(4\theta\) + sin \(2\theta\) = 2 sin(\(4\theta + 2\theta\over 2\)) cos(\(4\theta โ 2\theta\over 2\)) = 2 sin \(3\theta\) cos โฆ
Solution : The formula of 2 sin A sin B is cos (A โ B) โ cos (A + B). Example : Solve : 2 sin 75 sin 15. Solution : We have, 2 sin 75 sin 15 = cos (75 โ 15) โ cos (75 + 15) = cos 60 โ cos 90 = โฆ
Solution : The formula of 2 cos A cos B is cos (A + B) + cos (A โ B). Example : Solve : 2 cos 75 cos 15. Solution : We have, 2 cos 75 cos 15 = cos (75 + 15) + cos (75 โ 15) = cos 90 + cos 60 = โฆ
Solution : The formula of 2 cos A sin B is sin (A + B) โ sin (A โ B). Example : Solve : 2 cos 75 sin 15. Solution : We have, 2 cos 75 sin 15 = sin (75 + 15) โ sin (75 โ 15) = sin 90 โ sin 60 = โฆ
Solution : The formula of 2 sin A cos B is sin (A + B) + sin (A โ B). Example : Solve : 2 sin 75 cos 15. Solution : We have, 2 sin 75 cos 15 = sin (75 + 15) + sin (75 โ 15) = sin 90 + sin 60 = โฆ
