cos+x+sin+x

厍董皇3778(sinx+cosx)/(sinx - cosx)=3 求sin^2(x)+sinxcosx -
仲具成18955721645 ______ sinx+cosx=3sinx-3cosx2sinx=4cosx tanx=2 cosx^2=1/(1+tanx^2)=1/5 sinx^2=1-cosx^2=4/5 sinxcosx=tanxcosx^2=2/5 sinx^2+sinxcosx=4/5+2/5=6/5

厍董皇3778sinx+cosx=1求证sin^2012x+cos2012x=1 -
仲具成18955721645 ______[答案] sinx+cosx=1 两边同时平方得:sin²x+cos²x+2sinxcosx=1 1+2sinxcosx=1 sinxcosx=0 则sinx=0或cosx=0 且sinx+cosx=1 所以sinx和cosx一个是0,一个是1 所以sin^2012x+cos^2012x=1+0=1

厍董皇3778(sinx+cosx)/(sinx - cosx)=2,求sinxcosx的值 -
仲具成18955721645 ______[答案] sinx+cosx=2sinx-2cosx sinx=3cosx sin^2x+cos^2x=1 10cos^2x=1 sinx*cosx=3cos^2x=3/10 谢了

厍董皇3778已知sinx+cosx=1/2,则sin三次方x - cos三次方x=? -
仲具成18955721645 ______[答案] 1、若是求sin³x-cos³x的值 sin³x-cos³x=(sinx-cosx)(sin²x+cos²x+sinxcosx) =(sinx-cosx)(1+sinxcosx) 因为 sinx+cosx=1/2,所以(sinx+cosx)²=1/4,1+2sinxcosx=1/4 sinxcosx=-3/8 而因为(sinx-cosx)²=(sinx+cosx)²-4sinxcosx=1/4-4*(-3/...

厍董皇3778为什么y=√3cosx+sinx=2sin(x+π/3) -
仲具成18955721645 ______[答案] y=2(√3/2cosx+1/2sinx) =2(sinπ/3cosx+cosπ/3sinx) =2sin(x+π/3)

厍董皇3778已知y=sinXcosX+sinX的平方.求y的值域和最小正周期!
仲具成18955721645 ______ y=sinxcosx+sin²x =1/2sin2x+1/2(1-cos2x) =1/2sin2x-1/2cos2x+1/2 =√2/2sin(2x-π/4)+1/2 ∴其最小值为-√2/2+1/2=(1-√2)/2 最大值为√2/2+1/2=(1+√2)/2 则其值域为[(1-√2)/2,(1+√2)/2] 最小正周期T=2π/2=π

厍董皇3778函数y=sin+cosx+sinxcosx(x∈R)的值域为 -
仲具成18955721645 ______ 设sin+cosx=t,t∈(-√2,√2)则sinxcosx=(t^2-1)/2y=t+(t^2-1)/2=1/2(t+1)^2-1∵t∈(-√2,√2)所以y∈[1,1/2(√2+1)^2-1]

厍董皇3778若x=sinx+cosx,y=sinxcosx,试求y=f(x)的解析式 -
仲具成18955721645 ______[答案] x=sinx+cosx 两边同时平方得 x²=sin²x+2sinxcosx+cos²x =1+2sinxcosx =1+2y 那么2y=x²-1→y=(x²-1)/2 所以y=f(x)=(x²-1)/2

厍董皇3778(sinx+cosx)/ (sinx - cosx)=2,则sinxcosx=? -
仲具成18955721645 ______[答案] sinx+cosx=2sinx-2cosx sinx=3cosx sin²x=9cos²x 因为sin²x+cos²x=1 所以cos²x=1/10 sinxcosx =(3cosx)cosx =3cos²x =3/10

厍董皇3778求y=sinx+2sinxcosx - cosx的化简 -
仲具成18955721645 ______[答案] y=sinx+2sinxcosx-cosx=(sinx-cosx)+[1-(sinx-cosx)²]/2=-(1/2)(sinx-cosx)²+(sinx-cosx)+1/2=-(1/2)[√2sin(x-π/4)]²+√2sin(x-π/4)+1/2=-sin²(x-π/4)+√2sin(x-π/4)+1/2