Question

Prove that $\cos \left(3x\right)=4{\cos }^3\left(x\right)-3\cos \left(x\right)$

Answer 1

Solve for the required proof

Given that $\cos \left(3x\right)=4{\cos }^3\left(x\right)-3\cos \left(x\right)$

Consider $\cos \left(3x\right)=\cos \left(2x+x\right)$

We know that cosine of summation of angles is given as

$\cos \left(A+B\right)=\cos \left(A\right)\cos \left(B\right)-\sin \left(A\right)\sin \left(B\right)$

Substituting $A=2x.B=x$ we get

$\cos \left(2x+x\right)=\cos \left(2x\right)\cos \left(x\right)-\sin \left(2x\right)\sin \left(x\right)$

Using double angle identity we know that

$\cos \left(2x\right)=2{\cos }^2\left(x\right)-1$ and $\sin \left(2x\right)=2\sin \left(x\right)\cos \left(x\right)$

Substituting these values we get

$\cos \left(3x\right)=\left(2{\cos }^2\left(x\right)-1\right)\cos \left(x\right)-2\sin \left(x\right)\cos \left(x\right)\sin \left(x\right)$

$=\cos \left(x\right)\left[\left(2{\cos }^2\left(x\right)-1\right)-2{\sin }^2\left(x\right)\right]$

$=\cos \left(x\right)\left[2\left({\cos }^2\left(x\right)-{\sin }^2\left(x\right)\right)-1\right]$

$=\cos \left(x\right)\left[2\left({\cos }^2\left(x\right)-\left(1-{\cos }^2\left(x\right)\right)\right)-1\right]$

$=\cos \left(x\right)\left[2\left(2{\cos }^2\left(x\right)-1\right)-1\right]$

$=\cos \left(x\right)\left(4{\cos }^2\left(x\right)-3\right)$

$\Rightarrow$$\cos \left(3x\right)=4{\cos }^3\left(x\right)-3\cos \left(x\right)$

Hence, it is proved that $\cos \left(3x\right)=4{\cos }^3\left(x\right)-3\cos \left(x\right)$.