How do you evaluate the expression sec3?

1 Answer
George C.
Jan 15, 2018

sec 3^@ = 16/( sqrt(10+2sqrt(5))(sqrt(6)+sqrt(2)) + (sqrt(5)-1)(sqrt(6)-sqrt(2)))

Explanation:

I will assume that you mean sec 3^@, since that has an interesting answer.

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)

(See https://socratic.org/s/aMzLvAhu)

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.

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