Question

How do you find a double angle formula for $\sec \left( 2x \right)$ in terms of only $\csc x$ and $\sec x$?

Answer 1

We find the formulas for $\cos \left( 2x \right)$ where $\cos \left( 2x \right)=2{{\cos }^{2}}x-1=1-2{{\sin }^{2}}x$. We also use the inverse relations where $\csc x=\dfrac{1}{\sin x}$ and $\sec x=\dfrac{1}{\cos x}$. We put the values and find the relations between them.

Complete step by step answer:
We use the multiple angle formula for $\cos \left( 2x \right)$ where $\cos \left( 2x \right)=2{{\cos }^{2}}x-1=1-2{{\sin }^{2}}x$.
We also know the inverse relations where $\csc x=\dfrac{1}{\sin x}$ and $\sec x=\dfrac{1}{\cos x}$. The vice versa relations are also true.
So, $\sec 2x=\dfrac{1}{\cos 2x}$ and we put the value $\cos \left( 2x \right)=2{{\cos }^{2}}x-1$ to get
$\sec 2x=\dfrac{1}{\cos 2x}=\dfrac{1}{2{{\cos }^{2}}x-1}$.
We now need to find the relation in terms of $\sec x$ and we put $\sec x=\dfrac{1}{\cos x}$.
$\sec 2x=\dfrac{1}{2{{\cos }^{2}}x-1}=\dfrac{1}{2{{\left( \dfrac{1}{\sec x} \right)}^{2}}-1}$.
We multiply both numerator and denominator with ${{\sec }^{2}}x$ to get
$\sec 2x=\dfrac{1}{\dfrac{2}{{{\sec }^{2}}x}-1}=\dfrac{{{\sec }^{2}}x}{2-{{\sec }^{2}}x}$.
We put the value $\cos \left( 2x \right)=1-2{{\sin }^{2}}x$ to get
$\sec 2x=\dfrac{1}{\cos 2x}=\dfrac{1}{1-2{{\sin }^{2}}x}$.
We now need to find the relation in terms of $\csc x$ and we put $\csc x=\dfrac{1}{\sin x}$.
$\sec 2x=\dfrac{1}{1-2{{\sin }^{2}}x}=\dfrac{1}{1-2{{\left( \dfrac{1}{\csc x} \right)}^{2}}}$.
We multiply both numerator and denominator with ${{\csc }^{2}}x$ to get
$\sec 2x=\dfrac{1}{1-\dfrac{2}{{{\csc }^{2}}x}}=\dfrac{{{\csc }^{2}}x}{{{\csc }^{2}}x-2}$.
The double angle formulas for $\sec \left( 2x \right)$ in terms of only $\csc x$ and $\sec x$ are $\sec 2x=\dfrac{{{\csc }^{2}}x}{{{\csc }^{2}}x-2}$ and $\sec 2x=\dfrac{{{\sec }^{2}}x}{2-{{\sec }^{2}}x}$ respectively.

Note: The double-angle formulas can be quite useful when we need to simplify complicated trigonometric expressions later. With these formulas, it is better to remember where they come from, rather than trying to remember the actual formulas.