Nathaly Rivers

2022-12-29

What is the formula of $\mathrm{cos}3x$ ?

mosellasv2

Expert

We know that,
$\mathrm{cos}\left(A+B\right)=\mathrm{cos}A\mathrm{cos}B-\mathrm{sin}A\mathrm{sin}B$
Using the aforementioned property, we get
$\mathrm{cos}3x=\mathrm{cos}\left(2x+x\right)$
$\mathrm{cos}3x=\mathrm{cos}2x\mathrm{cos}x-\mathrm{sin}2x\mathrm{sin}x$
$\mathrm{cos}3x=2{\mathrm{cos}}^{3}x-\mathrm{cos}x-2{\mathrm{sin}}^{2}x\mathrm{cos}x$
$\mathrm{cos}3x=2{\mathrm{cos}}^{3}x-\mathrm{cos}x-2\left(1-{\mathrm{cos}}^{2}x\right)\mathrm{cos}x$
$\mathrm{cos}3x=2{\mathrm{cos}}^{3}x-\mathrm{cos}x-2\mathrm{cos}x+2{\mathrm{cos}}^{3}x$
$\mathrm{cos}3x=4{\mathrm{cos}}^{3}x-3\mathrm{cos}x$
Thus the formula $\mathrm{cos}3x=4{\mathrm{cos}}^{3}x-3\mathrm{cos}x$

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