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11TH · MATH

Question 4

menu_bookSolutionmenu_bookTheorymenu_book•(i)cos⁡15∘+cos⁡105∘+cos⁡195∘+cos⁡160∘+cos⁡285∘=0\cos 15^\circ + \cos 105^\circ + \cos 195^\circ + \cos 160^\circ + \cos 285^\circ = 0cos15∘+cos105∘+cos195∘+cos160∘+cos285∘=0menu_book•(ii)sin⁡2θ+sin⁡4θ+sin⁡6θ+sin⁡8θcos⁡2θ+cos⁡4θ+cos⁡6θ+cos⁡8θ=tan⁡5θ\dfrac{\sin 2\theta + \sin 4\theta + \sin 6\theta + \sin 8\theta}{\cos 2\theta + \cos 4\theta + \cos 6\theta + \cos 8\theta} = \tan 5\thetacos2θ+cos4θ+cos6θ+cos8θsin2θ+sin4θ+sin6θ+sin8θ​=tan5θ
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11TH · MATH

Question 4

menu_bookSolutionmenu_bookTheorymenu_book•(i)cos⁡15∘+cos⁡105∘+cos⁡195∘+cos⁡160∘+cos⁡285∘=0\cos 15^\circ + \cos 105^\circ + \cos 195^\circ + \cos 160^\circ + \cos 285^\circ = 0cos15∘+cos105∘+cos195∘+cos160∘+cos285∘=0menu_book•(ii)sin⁡2θ+sin⁡4θ+sin⁡6θ+sin⁡8θcos⁡2θ+cos⁡4θ+cos⁡6θ+cos⁡8θ=tan⁡5θ\dfrac{\sin 2\theta + \sin 4\theta + \sin 6\theta + \sin 8\theta}{\cos 2\theta + \cos 4\theta + \cos 6\theta + \cos 8\theta} = \tan 5\thetacos2θ+cos4θ+cos6θ+cos8θsin2θ+sin4θ+sin6θ+sin8θ​=tan5θ
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