Question

Show that $\cos \left(6\mathrm{x}\right)=32{\cos }^6\left(\mathrm{x}\right)-48{\cos }^4\left(\mathrm{x}\right)+18{\cos }^2\left(\mathrm{x}\right)-1$

Answer 1

To prove :$\cos \left(6\mathrm{x}\right)=32{\cos }^6\left(\mathrm{x}\right)-48{\cos }^4\left(\mathrm{x}\right)+18{\cos }^2\left(\mathrm{x}\right)-1$

Proof:

Consider $\mathrm{L}.\mathrm{H}.\mathrm{S}=\cos 6\mathrm{x}$

$\cos 6\mathrm{x}$can be expressed as $\cos 3\left(2\mathrm{x}\right)$

We know that

$\cos 3\mathrm{A}=4{\cos }^3\mathrm{A}–3\cos \mathrm{A}$

$\cos 6\mathrm{x}$ $=4{\cos }^{3}2\mathrm{x}–3\cos 2\mathrm{x}$ (Here $\mathrm{A}=2\mathrm{x}$)

$=4{(2{\cos }^2\mathrm{x}-1)}^3$$-3(2{\cos }^2\mathrm{x}-1)$ [Identity used: $\cos 2\mathrm{x}=2{\cos }^2\mathrm{x}–1$& ${\cos }^{3}2\mathrm{x}={(2{\cos }^2\mathrm{x}–1)}^3$]

$=4[{\left(2{\cos }^2\mathrm{x}\right)}^3-(1{)}^3-3\times {\left(2{\cos }^2\mathrm{x}\right)}^2\times 1+3\times \left(2{\cos }^2\mathrm{x}\right)\times {\left(1\right)}^2]-6{\cos }^2\mathrm{x}+3$ [Identiy used: ${(\mathrm{a}+\mathrm{b})}^3={\mathrm{a}}^3+{\mathrm{b}}^3+3{\mathrm{a}}^2\mathrm{b}+3{\mathrm{ab}}^2$]

$=4[8{\cos }^6\mathrm{x}-1-12{\cos }^4\mathrm{x}+6{\cos }^2\mathrm{x}]$$-$ $6{\cos }^2\mathrm{x}+3$

$=32{\cos }^6\mathrm{x}-48{\cos }^4\mathrm{x}+24{\cos }^2\mathrm{x}-4$ $-$$6{\cos }^2\mathrm{x}+3$

$=32{\cos }^6\mathrm{x}-48{\cos }^4\mathrm{x}+18{\cos }^2\mathrm{x}-1$

⇐$\mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}$

Thus,$\cos \left(6\mathrm{x}\right)=32{\cos }^6\left(\mathrm{x}\right)-48{\cos }^4\left(\mathrm{x}\right)+18{\cos }^2\left(\mathrm{x}\right)-1$. Hence proved