Answered: "Find a double angle formula for sec⁡(2x) in..."

nemired9

Answered question

2021-12-13

Find a double angle formula for

\sec (2 x)

in terms of only

\csc (x)

and

\sec (x)

Natalie Yamamoto

Beginner2021-12-14Added 22 answers

\sec (2 x) = \frac{1}{\cos (2 x)}

= \frac{1}{\cos (x + x)}

= \frac{1}{\cos (x) \times \cos (x) + \sin (x) \times \sin (x)}

= \frac{1}{\frac{1}{\sec (x)} \times \frac{1}{\sec (x)} + \frac{1}{\csc (x)} \times \frac{1}{\csc (x)}}

= \frac{1}{\frac{\csc^{2} (x) + \sec^{2} (x)}{\sec^{2} (x) \times \sin^{2} (x)}}

\frac{\sec^{2} (x) \csc^{2} (x)}{\csc^{2} (x) + \sec^{2} (x)}

rodclassique4r

Beginner2021-12-15Added 37 answers

\sec (2 x) = \frac{1}{\cos (2 x)}

Using the double formula:

= \frac{1}{\cos^{2} (x) - \sin^{2} (x)}

= \frac{1}{\frac{1}{\sec^{2} (x)} - \frac{1}{\csc^{2} (x)}}

= \frac{1}{\frac{\csc^{2} (x) - \sec^{2} (x)}{\sec^{2} (x) \csc^{2} (x)}}

\frac{\sec^{2} (x) \csc^{2} (x)}{\csc^{2} (x) - \sec^{2} (x)}

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