Question

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

Answer 1

Hint: To prove the given expression, we can write $\cos 6x$as$\cos 2\left( 3x \right)$. Now, apply the identity of$\cos 2\theta $ on $\cos 2\left( 3x \right)$ which will be equal to $2{{\cos }^{2}}3x-1$. Then apply the identity of $\cos 3x$ which is equal to $4{{\cos }^{3}}x-3\cos x$. And simplify.


Complete step by step solution:
The given equation that we have to prove is:
$\cos 6x=32{{\cos }^{6}}x-48{{\cos }^{4}}x+18{{\cos }^{2}}x-1$
We are going to expand $\cos 6x$ in the above equation as follows.
We can write $\cos 6x$ as $\cos 2\left( 3x \right)$ and then we will apply the identity of $\cos 2\theta $ on $\cos 2\left( 3x \right)$.
$\cos 2\left( 3x \right)=2{{\cos }^{2}}3x-1$
From the trigonometric double angle identities we know that, $\cos 3x=4{{\cos }^{3}}x-3\cos x$. Substituting this value of$\cos 3x$in the above equation we get,
$\begin{align}
  & 2{{\left( 4{{\cos }^{3}}x-3\cos x \right)}^{2}}-1 \\
 & =2\left( 16{{\cos }^{6}}x+9{{\cos }^{2}}x-24{{\cos }^{4}}x \right)-1 \\
 & =32{{\cos }^{6}}x+18{{\cos }^{4}}x-48{{\cos }^{4}}x-1 \\
\end{align}$
From the above simplification, we get the $L.H.S=32{{\cos }^{6}}x+18{{\cos }^{4}}x-48{{\cos }^{4}}x-1$
And R.H.S of the given expression is equal to $32{{\cos }^{6}}x-48{{\cos }^{4}}x+18{{\cos }^{2}}x-1$
As L.H.S = R.H.S Hence, we have proved the given expression.

Note: The other way of showing the equality of the given expression:
$\cos 6x=32{{\cos }^{6}}x-48{{\cos }^{4}}x+18{{\cos }^{2}}x-1$
We can also write $\cos 6x$ as $\cos 3(2x)$ and then apply the identity of $\cos 3\theta $ on $\cos 3(2x)$ where$\theta =2x$.
$\cos 6x=4{{\cos }^{3}}2x-3\cos 2x$
Now, substituting $\cos 2x=2{{\cos }^{2}}x-1$ in the above equation we get,
$\cos 6x=4{{\left( 2{{\cos }^{2}}x-1 \right)}^{3}}-3\left( 2{{\cos }^{2}}x-1 \right)$
Applying ${{\left( a-b \right)}^{3}}$ identity in the above equation we get,
$\begin{align}
  & \cos 6x=4\left( 8{{\cos }^{6}}x-1-3.4{{\cos }^{4}}x+3.2{{\cos }^{2}}x \right)-6{{\cos }^{2}}x+3 \\
 & \Rightarrow \cos 6x=32{{\cos }^{6}}x-4-48{{\cos }^{4}}x+24{{\cos }^{2}}x-6{{\cos }^{2}}x+3 \\
 & \Rightarrow \cos 6x=32{{\cos }^{6}}x-48{{\cos }^{4}}x+18{{\cos }^{2}}x-1 \\
\end{align}$
Now, the simplification of $\cos 6x$ yields the same result as given in the R.H.S of the given expression.
Hence, we have proved L.H.S = R.H.S by making a little change in writing the trigonometric function$\cos 6x$.
You must know the double and triple angle identities of cosine and sine as it has many applications in proving the trigonometric expressions just as we have shown above.