Kinley Richmond

2023-02-19

What is the formula of $\mathrm{cos}\left(4x\right)$?

$\mathrm{cos}4x=\mathrm{cos}\left(2x+2x\right)⇒\mathrm{cos}\left(4x\right)=\mathrm{cos}2x.\mathrm{cos}2x-\mathrm{sin}2x.\mathrm{sin}2x\left[\because cos\left(a+b\right)=cosa.cosb-sina.sinb\right]⇒\mathrm{cos}\left(4x\right)=\left(2{\mathrm{cos}}^{2}x-1\right).\left(2{\mathrm{cos}}^{2}x-1\right)-\left(2\mathrm{sin}x.\mathrm{cos}x\right)\left(2\mathrm{sin}x.\mathrm{cos}x\right)\left[\because sin2x=2sinx.cosx\right]⇒\mathrm{cos}\left(4x\right)={\left(2{\mathrm{cos}}^{2}x-1\right)}^{2}-{\left(2\mathrm{sin}x.\mathrm{cos}x\right)}^{2}\left(\because cos2x=2co{s}^{2}x-1\right)⇒\mathrm{cos}\left(4x\right)=\left(4{\mathrm{cos}}^{4}x-4{\mathrm{cos}}^{2}x+1\right)-\left(4{\mathrm{sin}}^{2}x.{\mathrm{cos}}^{2}x\right)⇒\mathrm{cos}\left(4x\right)=\left(4{\mathrm{cos}}^{4}x-4{\mathrm{cos}}^{2}x+1\right)-\left(4\left(1-{\mathrm{cos}}^{2}x\right).{\mathrm{cos}}^{2}x\right)⇒\mathrm{cos}\left(4x\right)=\left(8{\mathrm{cos}}^{4}x-8{\mathrm{cos}}^{2}x+1\right)$