Question

Prove that: $cos4x=1-8si{n}^{2}xco{s}^{2}x$

Answer 1

Determine the proving of the $cos4x=1-8si{n}^{2}xco{s}^{2}x$

Using formula:

$\cos 2x=2{\cos }^{2}x–1$

replacing $x$ by $2x$, we get

$$\begin{gathered} \cos 2\left(2x\right)=2{\cos }^2\left(2x\right)–1 \\ \Rightarrow \cos 4x=2{\cos }^{2}2x–1OR \\ \Rightarrow =2{(\cos 2x)}^2-1\mathbf{\because }\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{x}=2{\cos }^{2}x–1 \\ \Rightarrow =2{(2{\cos }^{2}x–1)}^2-1 \\ \Rightarrow =2{[{(2{\cos }^{2}x)}^2+(1)}^2-2.2{\cos }^{2}x.1]-1 \\ \Rightarrow =2\left[\right(4{\cos }^{4}x)+1-4{\cos }^{2}x]-1 \\ \Rightarrow =8{\cos }^{4}x+2-8{\cos }^{2}x-1 \\ \Rightarrow =8{\cos }^{2}x({\cos }^{2}x-1)+1 \\ \Rightarrow =8{\cos }^{2}x[-(1-{\cos }^{2}x\left)\right]+1 \\ \Rightarrow =-8{\cos }^{2}x\left[\right(1-{\cos }^{2}x\left)\right]+1 \\ \Rightarrow =-8{\cos }^{2}x{\sin }^{2}x+1[\mathbf{\because }\mathbf{}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathbf{}\mathit{x}=1-{\cos }^{2}x] \\ \Rightarrow =1-8{\cos }^{2}x{\sin }^{2}x \end{gathered}$$

Hence, the given expression is true.