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为什么%tAn11/4π=tAnπ/4

先运用诱导公式化简: tan(-11π/5)=tan(-2π -π/5)=tan(-π/5)=-tan(π/5) tan(4π/5)=-tan(π- 4π/5)=-tan(π/5) tan(-11π/5)=tan(4π/5)

=sin(-4π+5π/6)+cos(-8π+π/3)-tan(2π+3/4π) =sin(5π/6)+cos(π/3)-tan(3/4π) =1/2+1/2-1 =0 希望对你有所帮助

cos 9 4 π+tan(- 11 6 π) =cos π 4 +tan π 6 = 2 2 + 3 3 = 3 2 +2 3 6 故答案为: 3 2 +2 3 6

sin13π/3 +cos( -25π/6)+tan11π/4) =sin(4π+π/3)+cos(4π+π/6)+tan(3π-π/4) =v3/2 +v3/2 -1 =v3-1 您的满意是我继续的动力!

你好!!! 依题意有,用韦达定律 tanα+tan(π/4-α)=-P,tanαtan(π/4-α)=q. ∴tan[α+(π/4-α)]=[tanα+tan(π/4-α)]/[1-tanαtan(π/4-α)] →tan(π/4)=-p/(1-q) →p-q=-1 …… (1) 而p+q=11 …… (2) 解(1)、(2)得, p=5,q=6. 希望能够帮助你!!!

arctan(2/3)+kπ. (k=0,1,2...).正切值定义...4、单调性:在区间(-π/2+kπ,π/2+kπ),k...(对做题有帮助):当A+B=π/4时候,必有(1+tanA...

10. 令x=tanu x:1→√3,则u:π/4→π/3 ∫[1:√3]dx/[x²√(1+x²)] =∫[π/4:π/3]d(tanu)/[tan²u√(1+tan²u)] =∫[π/4:π/3] (cosu/sin²u)du =-cscu|[π/4:π/3] =-(2/√3 -2/√2) =(3√2-2√3)/3 11. 令x=√2sint x:0→1,则t:0→π/...

[4cos²(-15π/4)]/[tan(-11π/3)+√2sin(25π/4)] = [4cos²(-15π/4+4π)]/[tan(-11π/3+4π)+√2sin(25π/4-6π)] = [4cos²(π/4)]/[tan(π/3)+√2sin(π/4)] = [4(√2/2)²]/[√3+√2(√2/2)] = 2/[√3+1] = √3-1

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