Maths Question (sin cos tan)

As follows:


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(i) sin 2x = 2 sin x cos x
cos 2x = (cos x)^2 - (sin x)^2
cot 2x = cos 2x/ sin 2x
= [(cos x)^2 - (sin x)^2] / (2 sin x cos x)
= [(cot x)^2 -1] / 2 (cos x/ sin x) _______(分子分母同除以(sin x)^2 )
=( (cot x)^2 -1 ) / (2 cot x)

(ii)
( (cot x)^2 -1 ) / (2 cot x) = cot 2x
(cot x)^2 -1 = 2 cot 2x cot x
(cot x)^2 - (2cot 2x) cot x -1 =0
Put x =22.5
(cot 22.5)^2 - (2cot 45) cot 22.5 -1 =0
(cot 22.5)^2 - 2cot 22.5 -1 =0
cot 22.5 = [2+/- √(4 -4(1)(-1)) ]/2 = (2+/- 2√2) /2 = 1 +/- √2
Since cot 22.5>0 , cot 22.5 = 1+√2

(sin x)^2 + (cos x) ^2 =1
divide both sides by (sin x)^2,
1 + (cot x)^2 = (csc x)^2
Put x = 22.5,
1 + (cot 22.5)^2 = (csc 22.5)^2
(csc 22.5)^2 = 1+ (1+√2) ^2 = 1 + 1 +2√2 +2 = 4 +2√2
i.e. (csc 22.5)^2 = 4 +2√2 #