顯然2cos²α/2 -1=cosα
所以
原式=(cosα -sinα) / √2sin(α+π/4)
而
cosα -sinα
=√2*(cosα*√2/2 - sinα*√2/2)
=√2 *(cosα*cosπ/4 - sinα*sinπ/4)
=√2 *cos(α+π/4)
所以
原式
=(cosα -sinα) / √2sin(α+π/4)
=√2 *cos(α+π/4) / √2sin(α+π/4)
= cot(α+π/4)
= tan(π/4 -α)
= (tanπ/4 -tanα)/(1+tanπ/4* tanα)
而tan2α= -2√2 =2tanα/(1-tan²α)
即
2tan²α -√2tanα -2=0
解得tanα=√2或 -√2/2
那么
tanα=√2時
原式= (tanπ/4 -tanα)/(1+tanπ/4* tanα)
=(1 -tanα)/(1+ tanα)
=(1 -√2)/(1+√2)
= -3+2√2
tanα= -√2/2時
原式=(1+√2/2)/(1-√2/2)
= 3+2√2
所以解得
原式= -3+2√2或 3+2√2