



 
Hi Diana, I don't memorize many trig formulas but I do know \[ \sin^2 x + \cos^2 x = 1\] and the double angle formulas for the sine and cosine, in particuar \[\cos 2x = \cos^2 x  \sin^2 x.\] Substituting from the first equation into the second I get \[\cos 2x = 1 + 2 \cos^2 x\] and hence \[\cos x = \sqrt{ \left( \frac{cos 2x  1}{2} \right)} .\] Since \[\sec x = \frac{1}{\cos x}\] you can use the expression for the $\cos x$ above to determine the uantity you want if you know the cosine of $\large \frac{\pi}{6}.$ Harley 



Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 