Question

Use a sum or difference formula to find the exact value of the trigonometric function. sec− π 12

Answer 1

`\sec\left(-\frac\pi{12}\right)`

First use the fact that
`\sec x` is an even function, i.e.
`\sec(-x)=\sec x`. So this is the same as


`\sec\frac\pi{12}`

Now, note that
`\frac\pi{12}=\frac\pi3-\frac\pi4`, so


`\sec\frac\pi{12}=\sec\left(\frac\pi3-\frac\pi4\right)=\frac1{\cos\frac\pi3\cos\frac\pi4+\sin\frac\pi3\sin\frac\pi4}`

which comes from the sum identity for cosine,


`\cos(x\pm y)=\cos x\cos y\mp\sin x\sin y`

Now,


`\sec\frac\pi{12}=\frac1{\frac12\frac1{\sqrt2}+\frac{\sqrt3}2\frac1{\sqrt2}}=\sqrt6-\sqrt2`