作业帮已知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﹚]](/uploads/image/z/7158228-60-8.jpg?t=%E5%B7%B2%E7%9F%A5tan2%CE%B1%3D%26frac34%3B%EF%B9%99%CF%80%EF%BC%8F2%EF%BC%9C%CE%B1%EF%BC%9C%CF%80%EF%B9%9A%2C%E6%B1%82%5B2cos%E7%9A%84%E5%B9%B3%E6%96%B9%CE%B1%EF%BC%8F2%EF%BC%8Bsin%CE%B1-1%5D%EF%BC%8F%5B%E2%88%9A2cos%EF%B9%99%CE%B1%EF%BC%8B%CF%80%EF%BC%8F4%EF%B9%9A%5D)
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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