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tAn5°

tan25°tan35° =tan(30°-5°)tan(30°+5°) =(tan30°-tan5°)/(1+tan30°tan5°) ·(tan30°+tan5°)/(1-tan30°tan5°) =(1/3-tan²5°)/(1-1/3·tan²5°) =(1-3tan²5°)/(3-tan²5°) tan10°=2tan5°/(1-tan²5°) 所以, tan15° =tan(10°+...

(1+cos20°)/2sin20°]-sin10°(1/tan5°-tan5°) =[(1+cos20°)/2sin20°]-sin10°(cot5°-tan5°) =[(1+cos20°)/4sin10°cos10°]-sin10°(cot5°-tan5°) =(2cos10°/4sin10°)-2sin5°cos5°(cot5°-tan5°) =(cos10°/2sin10°)-2((cos5°)^2-(sin5°)^2) =(cos10...

[(1+cos20°)/2sin20°]-sin10°(1/tan5°-tan5°) =[(1+cos20°)/2sin20°]-sin10°(cot5°-tan5°) =[(1+cos20°)/4sin10°cos10°]-sin10°(cot5°-tan5°) =(2cos10°/4sin10°)-2sin5°cos5°(cot5°-tan5°) =(cos10°/2sin10°)-2((cos5°)^2-(sin5°)^2) =(cos1...

0.087488663525924005222018669434961

反正切函数arctan 5 ≈ 78.6900675259798°

tan(5/4π)=tan(π+1/4π)=tan(1/4π)=tan45°=1

[(1+cos20°)/2sin20°]-sin10°(1/tan5°-tan5°) =[(1+cos20°)/2sin20°]-sin10°(cot5°-tan5°) =[(1+cos20°)/4sin10°cos10°]-sin10°(cot5°-tan5°) =(2cos10°/4sin10°)-2sin5°cos5°(cot5°-tan5°) =(cos10°/2sin10°)-2((cos5°)^2-(sin5°)^2) =(cos1...

(I)观察①、②,可得:若锐角α,β,γ满足α+β+γ=90°,则tanαtanβ+tanβtanγ+tanαtanγ=1.(II)对称轴x=a,当a<0时,[0,1]是f(x)的递减区间,f(x)max=f(0)=1-a=2∴a=-1;当a>1时,[0,1]是f(x)的递增区间,f(x)max=f(1)=a=2∴a=2;当...

可以观察到:10°+20°+60°=90°,5°+15°+70°=90°,故可以猜想此推广式为:若α+β+γ=π2,且α,β,γ都不等于kπ+π2(k∈Z),则有tanα?tanβ+tanβ?tanγ+tanγ?tanα=1. 证明:∵α+β+γ=π2,∴α+β=π2-γ,∴tan(α+β)=tan(π2-γ)=cotγ,∴tanα+tanβ=cotγ(1-ta...

原式=(sin5/cos5-cos5/sin5)cos70/(1+sin70) =[(sin5)^2-(cos5)^2]/sin5/cos5*cos70/(1+sin70) =-2cos10/sin10*sin20/(1+sin70) =-2cos10/sin10*sin10*cos10/(1+sin70) =-2*(1+cos20)/(1+sin70) =-2

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