How do you evaluate the expression #sec3#?
1 Answer
Explanation:
I will assume that you mean
Let's work back from what we want to what we know...
#sec 3^@ = 1/cos 3^@#
#cos 3^@ = cos(18^@ - 15^@)#
#color(white)(cos 3^@) = cos 18^@ cos 15^@ + sin 18^@ sin 15^@#
#cos 15^@ = cos(45^@-30^@)#
#color(white)(cos 15^@) = cos 45^@ cos 30^@ + sin 45^@ sin 30^@#
#color(white)(cos 15^@) = 1/4(sqrt(6)+sqrt(2))#
#sin 15^@ = sin(45^@-30^@)#
#color(white)(sin 15^@) = sin 45^@ cos 30^@ - cos 45^@ sin 30^@#
#color(white)(sin 15^@) = 1/4(sqrt(6)-sqrt(2))#
#cos 18^@ = 1/4 sqrt(10+2sqrt(5))#
#sin 18^@ = 1/4 (sqrt(5)-1)#
So:
#sec 3^@ = 1/(cos 18^@ cos 15^@ + sin 18^@ sin 15^@)#
#color(white)(sec 3^@) = 1/((1/4 sqrt(10+2sqrt(5)))(1/4(sqrt(6)+sqrt(2))) + (1/4 (sqrt(5)-1))( 1/4(sqrt(6)-sqrt(2))))#
#color(white)(sec 3^@) = 16/( sqrt(10+2sqrt(5))(sqrt(6)+sqrt(2)) + (sqrt(5)-1)(sqrt(6)-sqrt(2)))#
It is possible, but somewhat tedious to rationalise the denominator, and results in a somewhat more complicated expression, so I'll stop here.