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.