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Question Number 43267 by Rauny last updated on 09/Sep/18

cos 2x=2cos^2 −1 cos 3x=4cos^3 x−3cos x cos 4x=8cos^4 x−8cos^2 x+1 cos 5x=16cos^5 x−20cos^3 +5cos x cos 6x=32cos^6 x−48cos^4 x+18cos^2 x−1 cos 7x=64cos^7 x−112cos^5 x+56cos^3 x−4cos x cos 8x=128cos^8 x−256cos^6 x+160cos^4 x−32cos^2 x+1

$$\mathrm{cos}\:\mathrm{2}{x}=\mathrm{2cos}^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{cos}\:\mathrm{3}{x}=\mathrm{4cos}^{\mathrm{3}} \:{x}−\mathrm{3cos}\:{x} \\ $$$$\mathrm{cos}\:\mathrm{4}{x}=\mathrm{8cos}^{\mathrm{4}} \:{x}−\mathrm{8cos}^{\mathrm{2}} \:{x}+\mathrm{1} \\ $$$$\mathrm{cos}\:\mathrm{5}{x}=\mathrm{16cos}^{\mathrm{5}} \:{x}−\mathrm{20cos}^{\mathrm{3}} +\mathrm{5cos}\:{x}\: \\ $$$$\mathrm{cos}\:\mathrm{6}{x}=\mathrm{32cos}^{\mathrm{6}} \:{x}−\mathrm{48cos}^{\mathrm{4}} \:{x}+\mathrm{18cos}^{\mathrm{2}} \:{x}−\mathrm{1} \\ $$$$\mathrm{cos}\:\mathrm{7}{x}=\mathrm{64cos}^{\mathrm{7}} \:{x}−\mathrm{112cos}^{\mathrm{5}} \:{x}+\mathrm{56cos}^{\mathrm{3}} \:{x}−\mathrm{4cos}\:{x} \\ $$$$\mathrm{cos}\:\mathrm{8}{x}=\mathrm{128cos}^{\mathrm{8}} \:{x}−\mathrm{256cos}^{\mathrm{6}} \:{x}+\mathrm{160cos}^{\mathrm{4}} \:{x}−\mathrm{32cos}^{\mathrm{2}} \:{x}+\mathrm{1} \\ $$

Commented by malwaan last updated on 09/Sep/18

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Commented by Rauny last updated on 09/Sep/18

∵cos (a+b)=cos acos b−sin asin b sin (a+b)=sin acos b+cos asin b

$$\because\mathrm{cos}\:\left({a}+{b}\right)=\mathrm{cos}\:{a}\mathrm{cos}\:{b}−\mathrm{sin}\:{a}\mathrm{sin}\:{b} \\ $$$$\mathrm{sin}\:\left({a}+{b}\right)=\mathrm{sin}\:{a}\mathrm{cos}\:{b}+\mathrm{cos}\:{a}\mathrm{sin}\:{b} \\ $$

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