Question

Prove the given trigonometric expression, \[\cos 4x=1-8{{\sin }^{2}}x{{\cos }^{2}}x\].

Answer 1

Hint: Let us proceed with the LHS of the given expression. We know the formula, \[\cos 2x=1-2{{\sin }^{2}}x\] . Replace x by 2x and then expand the expression. We also know that, \[\sin 2x=2\sin x.\cos x\] . Now, using this formula, substitute \[\sin 2x\] by \[2\sin x.\cos x\] and solve it further.

Complete step-by-step solution -
According to the question, we have to prove \[\cos 4x=1-8{{\sin }^{2}}x{{\cos }^{2}}x\] .
Let us proceed with the LHS of the expression. In LHS, we have \[\cos 4x\] .
We know the formula, \[\cos 2x=1-2{{\sin }^{2}}x\] ……………….(1)
Replacing x by 2x in equation (1), we get
\[\cos 2.2x=1-2{{\sin }^{2}}2x\]
\[\Rightarrow \cos 4x=1-2{{\sin }^{2}}2x\] …………………(2)
We also know that, \[\sin 2x=2\sin x.\cos x\] ………………………(3)
Now, putting the value of \[\sin 2x\] from equation (3) in equation (2), we get
\[\begin{align}
  & \cos 4x=1-2{{\sin }^{2}}2x \\
 & \Rightarrow \cos 4x=1-2{{(2\sin x\cos x)}^{2}} \\
\end{align}\]
Now, further solving the above expression we get
\[\begin{align}
  & \cos 4x=1-2(4{{\sin }^{2}}x{{\cos }^{2}}x) \\
 & \Rightarrow \cos 4x=1-8{{\sin }^{2}}x{{\cos }^{2}}x \\
\end{align}\]
So, LHS = RHS.
Hence, proved.

Note: This question can also be solved in another way.
We know the formula, \[\cos 2x=2{{\cos }^{2}}x-1\] ……………(1)
Replacing x by 2x in equation (1), we get
\[\cos 2.2x=2{{\cos }^{2}}2x-1\] …………..(2)
We know the identity, \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\] ……………..(3)
Replacing x by 2x in equation (3), we get
\[{{\sin }^{2}}2x+{{\cos }^{2}}2x=1\]
On further solving, we get
\[{{\cos }^{2}}2x=1-{{\sin }^{2}}2x\] ……………….(4)
From equation (2) and equation (4), we get
\[\begin{align}
  & \cos 2.2x=2(1-{{\sin }^{2}}2x)-1 \\
 & \Rightarrow \cos 4x=2-1-2{{\sin }^{2}}2x \\
\end{align}\]
\[\Rightarrow \cos 4x=1-2{{\sin }^{2}}2x\] ……………………..(5)
We also know that, \[\sin 2x=2\sin x.\cos x\] ………………………(6)
Now, putting the value of \[\sin 2x\] from equation (6) in equation (5), we get
\[\begin{align}
  & \cos 4x=1-2{{\sin }^{2}}2x \\
 & \Rightarrow \cos 4x=1-2{{(2\sin x\cos x)}^{2}} \\
\end{align}\]
Now, further solving the above expression we get
\[\begin{align}
  & \cos 4x=1-2(4{{\sin }^{2}}x{{\cos }^{2}}x) \\
 & \Rightarrow \cos 4x=1-8{{\sin }^{2}}x{{\cos }^{2}}x \\
\end{align}\]
So, LHS = RHS.
Hence, proved.