已知tan2α=¾﹙π/2<α<π﹚,求[2cos的平方α/2+sinα-1]/[√2cos﹙α+π/4﹚]

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已知tan2α=¾﹙π/2<α<π﹚,求[2cos的平方α/2+sinα-1]/[√2cos﹙α+π/4﹚]
已知tan2α=¾﹙π/2<α<π﹚,求[2cos的平方α/2+sinα-1]/[√2cos﹙α+π/4﹚]

已知tan2α=¾﹙π/2<α<π﹚,求[2cos的平方α/2+sinα-1]/[√2cos﹙α+π/4﹚]
tan2α=3/4
2tanα/(1-tan^2α) = 3/4
8tanα = 3 - 3tan^2α
3tan^2α + 8tanα - 3 = 0
(tanα+3)(3tanα-1) = 0
∵π/2<α<π
∴tanα<0
∴3tanα-1<0
∴tanα+3 = 0
∴tanα = -3
[2cos^2(α/2)+sinα-1]/[√2cos(α+π/4)]
= [(cosα + 1)+sinα-1]/[√2(cosαcosπ/4-sinαsinπ/4)]
= (cosα + sinα)/(cosα - sinα)
= (1+tanα)/(1-tanα)
= (1-3)/(1+3)
= -1/2

tan2α = 3/4
2tanα/(1 - tan²α) = 3/4
3tan²α + 8)tanα - 3 = 0
(3tanα - 1)(tanα + 3) = 0
∵π/2 < α < π
∴tanα < 0
即 tanα = -3
[2cos²(α/2) + sinα - 1]/[√2cos(α + π...

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tan2α = 3/4
2tanα/(1 - tan²α) = 3/4
3tan²α + 8)tanα - 3 = 0
(3tanα - 1)(tanα + 3) = 0
∵π/2 < α < π
∴tanα < 0
即 tanα = -3
[2cos²(α/2) + sinα - 1]/[√2cos(α + π/4)]
= (cosα + 1 + sinα - 1)/[√2cos(α + π/4)]
= √2sin(α + π/4)/[√2cos(α + π/4)]
= tan(α + π/4)
= (tanα + 1)/(1 - tanα)
= -2/4
= -1/2

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