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


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

$\cos 2 x = {2 \cos}^{2} - 1$ $\cos 3 x = {4 \cos}^{3} x - 3 \cos x$ $\cos 4 x = {8 \cos}^{4} x - {8 \cos}^{2} x + 1$ $\cos 5 x = {16 \cos}^{5} x - {20 \cos}^{3} + 5 \cos x$ $\cos 6 x = {32 \cos}^{6} x - {48 \cos}^{4} x + {18 \cos}^{2} x - 1$ $\cos 7 x = {64 \cos}^{7} x - {112 \cos}^{5} x + {56 \cos}^{3} x - 4 \cos x$ $\cos 8 x = {128 \cos}^{8} x - {256 \cos}^{6} x + {160 \cos}^{4} x - {32 \cos}^{2} x + 1$
Commented by malwaan last updated on 09/Sep/18

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Commented by Rauny last updated on 09/Sep/18

$∵ \cos (a + b) = \cos a \cos b - \sin a \sin b$ $\sin (a + b) = \sin a \cos b + \cos a \sin b$
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