prove that 2cosA=√(2+√(2+√((2)+√((2)+2cos16A))))

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2cos16A=2(2cos^2(8A)-1)=4cos^2(8A)-2; add 2: 4cos^2(8A), the square root of which is 2cos8A.

2cos8A+2=4cos^2(4A), square root is 2cos4A.

2cos4A+2=4cos^2(2A), square root is 2cos2A.

2cos2A+2=4cos^2A, square root is 2cosA QED.
by Top Rated User (1.1m points)

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