Trig Help!

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Question: Trying to find cos π/12, if cos π/6 = square root 3 over 2, how to find cos π/12 using DOUBLE angle formula?

This is what I got so far..
cos 2(π/6) = cos (π/6 + π/6)
= (cos π/6)(cos π/6) - (sin π/6)(sin π/6)
= cos^2 π/6 - sin^2 π/6

Is that right? Please and thank you.

  • Trig Help! -

    you got it backwards
    π/12 = (1/2) of π/6

    using cos 2A = cos^2 A - sin^2 A = 2cos^2 A - 1

    so cos π/6 = 2 cos^2 π/12 - 1
    √3/2 + 1 = 2cos^2 π/12
    cos^2 π/12 = √3/4 + 1/2 = (√3 + 2)/4
    cos π/12 = √(√3+2) /2

    check:
    by calculator, cos π/12 = .96592...
    √(√3+2)/2 = .96592...
    My answer is correct.

    Of course the answer I gave is not unique
    we could have done it this way
    cos π/12 = cos 15°
    = cos(45-30)°
    = cos45cos30 + sin45sin30
    = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4

    which when evaluated is also .96592...
    but it asked for a solution using the double angle formula

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