Question

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

Answer 1

Hint : We use the formulas for the trigonometric function of multiple angles where $\cos 3x=4{{\cos }^{3}}x-3\cos x$ and $\cos 2x=2{{\cos }^{2}}x-1$ . We also use the cubic form of ${{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}$ . We complete the multiplication to find the proof of the given theorem.

Complete step by step solution:
We have to prove that $\cos 6x=32{{\cos }^{6}}x-48{{\cos }^{4}}x+18{{\cos }^{2}}x-1$ .
We apply the multiple angle formula for the cos where $\cos 3x=4{{\cos }^{3}}x-3\cos x$ .
We take $2x$ in place of $x$ . We get $\cos 6x=4{{\cos }^{3}}\left( 2x \right)-3\cos \left( 2x \right)$ .
Now we replace the value of $\cos 2x$ with the theorem of $\cos 2x=2{{\cos }^{2}}x-1$ .
We have $\cos 6x=4{{\cos }^{3}}\left( 2x \right)-3\cos \left( 2x \right)=4{{\left( 2{{\cos }^{2}}x-1 \right)}^{3}}-3\left( 2{{\cos }^{2}}x-1 \right)$ .
We now break the higher power terms into their simplest form.
We have ${{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}$ .
Therefore, ${{\left( 2{{\cos }^{2}}x-1 \right)}^{3}}=8{{\cos }^{6}}x-12{{\cos }^{4}}x+6{{\cos }^{2}}x-1$ .

$$\cos 6x =4{{\left( 2{{\cos }^{2}}x-1 \right)}^{3}}-3\left( 2{{\cos }^{2}}x-1 \right) \\\ =4\left( 8{{\cos }^{6}}x-12{{\cos }^{4}}x+6{{\cos }^{2}}x-1 \right)-3\left( 2{{\cos }^{2}}x-1 \right) \\\ =32{{\cos }^{6}}x-48{{\cos }^{4}}x+24{{\cos }^{2}}x-4-6{{\cos }^{2}}x+3 \\\ =32{{\cos }^{6}}x-48{{\cos }^{4}}x+18{{\cos }^{2}}x-1 \\\$$

Thus proved, $\cos 6x=32{{\cos }^{6}}x-48{{\cos }^{4}}x+18{{\cos }^{2}}x-1$ .

Note : instead of breaking the multiple angle theorem as \cos 6x=\cos \left\\{ 3\times \left( 2x \right) \right\\} , we can take \cos 6x=\cos \left\\{ 2\times \left( 3x \right) \right\\} . It is the same in both cases and the final proof is also the same.