Question

Prove the given trigonometric equation such that:
$\cos 4x=1-8{{\sin }^{2}}x {{\cos }^{2}}x$

Answer 1

Hint: For the given questions we will use the trigonometric identity as follows:
$\cos 2A={{\cos }^{2}}A-{{\sin }^{2}}A=2{{\cos }^{2}}A-1$. So we can use the above formula with cos4x to get the required expression.

Complete step-by-step solution -
We have been asked to prove $\cos 4x=1-8{{\sin }^{2}}x {{\cos }^{2}}x$.
We know that $\cos 2A=2{{\cos }^{2}}A-1$.
Now taking left hand side $=\cos 4x$
$\Rightarrow \cos 4x=\cos 2(2x)$
Since it is in the form of cos2A, here $A=2x$
$\Rightarrow \cos 4x=2{{\cos }^{2}}(2x)-1$
We know that $\cos 2x=2{{\cos }^{2}}x-1$. So by substituting the values of cos2x in the above expression, we get as follows:
$\Rightarrow \cos 4x=2{{\left( 2{{\cos }^{2}}x-1 \right)}^{2}}-1$
By using ${{\left( a-b \right)}^{2}}={{a}^{2}}+2ab-{{b}^{2}}$ we get as follows:

& \Rightarrow \cos 4x=2\left[ {{\left( 2{{\cos }^{2}}x \right)}^{2}}-2\left( 2{{\cos }^{2}}x \right).1+1 \right]-1 \\\ & =2\left[ {{\left( 2{{\cos }^{2}}x \right)}^{2}}-2\left( 2{{\cos }^{2}}x \right).1+1 \right]-1 \\\ & =2\left[ 4{{\cos }^{2}}x\left( {{\cos }^{2}}x-1 \right)+1 \right]-1 \\\ \end{aligned}$$ Since we know that $${{\cos }^{2}}x-1=-{{\sin }^{2}}x$$ $$\begin{aligned} & \Rightarrow \cos 4x=2\left[ 4{{\cos }^{2}}x\left( -{{\sin }^{2}}x \right)+1 \right]-1 \\\ & =2\left[ -4{{\cos }^{2}}x {{\sin }^{2}}x+1 \right]-1 \\\ & =-8{{\cos }^{2}}x {{\sin }^{2}}x+2-1 \\\ & =-8{{\cos }^{2}}x {{\sin }^{2}}x+1 \\\ \end{aligned}$$ $$\Rightarrow \cos 4x=1-8{{\cos }^{2}}x {{\sin }^{2}}x=$$ right hand side Hence the left hand side is equal to the right hand side. Therefore the given expression is proved. Note: Be careful of the sign while doing calculation. Also don’t substitute the value of $$\left( {{\cos }^{2}}x-1 \right)$$ is $${{\sin }^{2}}x$$ by mistake since we know that $$\left( {{\cos }^{2}}x-1 \right)$$ is equal to $$\left( -{{\sin }^{2}}x \right)$$ so take care of it. We can also prove the given trigonometric expression by considering the right hand side of the equation.