证明(要过程):1.(tan(α+β)-tanα)/(1+tanαtan(α+β))=sin2β/2cos平方β2.sinβ(1+cos2β)=2sin2βcosβ3.sin^4α+cos^4α=1-1/2sin平方2α
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/17 04:47:07
![证明(要过程):1.(tan(α+β)-tanα)/(1+tanαtan(α+β))=sin2β/2cos平方β2.sinβ(1+cos2β)=2sin2βcosβ3.sin^4α+cos^4α=1-1/2sin平方2α](/uploads/image/z/8642605-13-5.jpg?t=%E8%AF%81%E6%98%8E%28%E8%A6%81%E8%BF%87%E7%A8%8B%29%3A1.%28tan%28%CE%B1%2B%CE%B2%29-tan%CE%B1%29%2F%281%2Btan%CE%B1tan%28%CE%B1%2B%CE%B2%29%29%3Dsin2%CE%B2%2F2cos%E5%B9%B3%E6%96%B9%CE%B22.sin%CE%B2%281%2Bcos2%CE%B2%29%3D2sin2%CE%B2cos%CE%B23.sin%5E4%CE%B1%2Bcos%5E4%CE%B1%3D1-1%2F2sin%E5%B9%B3%E6%96%B92%CE%B1)
证明(要过程):1.(tan(α+β)-tanα)/(1+tanαtan(α+β))=sin2β/2cos平方β2.sinβ(1+cos2β)=2sin2βcosβ3.sin^4α+cos^4α=1-1/2sin平方2α
证明(要过程):1.(tan(α+β)-tanα)/(1+tanαtan(α+β))=sin2β/2cos平方β
2.sinβ(1+cos2β)=2sin2βcosβ
3.sin^4α+cos^4α=1-1/2sin平方2α
证明(要过程):1.(tan(α+β)-tanα)/(1+tanαtan(α+β))=sin2β/2cos平方β2.sinβ(1+cos2β)=2sin2βcosβ3.sin^4α+cos^4α=1-1/2sin平方2α
1.根据公式tan(α-β)=(tanα-tanβ)/(1+tanαtanβ)有
左边=.(tan(α+β)-tanα)/(1+tanαtancos2βtan[(α+β)-α]=tanβ
右边=sin2β/2cos²β=(2sinβcosβ)/2cos²β=sinβ/cosβ=tanβ
由左边=右边,得
(tan(α+β)-tanα)/(1+tanαtan(α+β))=sin2β/2cos平方β
2.sinβ(1+cos2β)=2sin2βcosβ
因为cos2β=cos(β+β)=cosβcosβ-sinβsinβ=2cos²β-1
sinβ(1+cos2β)
=sinβ[1+(2cos²β-1)]
=sinβ2cos²β
=(2sinβcosβ)cosβ
=sin2βcosβ
3.sin^4α+cos^4α=1-1/2sin平方2α
sin^4α+cos^4α
=(sin²α)²+(cos²α)²
=(sin²α+cos²α)²-2sin²αcos²α
=1²-2(sinαcosα)²
=1-2(1/2sin2α)²
=1-1/2sin²2α