(2cos^2-1)/2tan(π/4-α)sin^2(π/4+α)
人气:377 ℃ 时间:2019-12-24 10:10:25
解答
(2cos^2α-1)/2tan(π/4-α)sin^2(π/4+α)
=(1+cos2α-1)/[2tan(π/4-α)sin^2(π/4+α)]
=cos2x/[2tan(π/4-α)sin^2(π/4+α)]
2tan(π/4-α)sin^2(π/4+α)=[2(sin(π/2-2α))/(cos(π/2-2α)+1)]x[(1-cos(2α+π/2))/2]
=[2(cos2α/(sin2α+1)]*[(1+sin2α)/2]
=cos2x
推荐
猜你喜欢
- what time do they come我这样写对吗,为什么
- 一个长6dm,宽4dm,高5dm的长方体盒子,里面最多能放( )个棱长为2dm的正方体木块. A.12 B.13 C.14 D.15
- I am going to stay with my uncle this weekend .(对my uncle 提问)
- 三角形ABC内接于圆0,角BAC=120度,AB=AC,BD为直径,AD=6,求AC的长
- 如图,△ABC为等腰三角形,∠ACB=90°,延长AB至F,使∠ECF=135°.求证:AE:EC=BA:CF.
- 十六年前的回忆主要内容
- 甲乙两人以不变的速度在环形路上跑,相向而行2分钟相遇,同向而6分钟相遇,甲比乙快,甲乙两每分钟各跑几圈?求分析及解,
- 在△ABC中,a=2,cosB=3/5, (1)若b=4,求sinA的值; (2)若△ABC的面积S△ABC=4,求b和c的值.