a+weekend+picnic+作文

巫姚侮2020f(a)=∫[pi/2 0] |cos(x+a)|dx 求f(a)+f(a+pi/2)的值 以及f(a)的最大最小值 -
狄支斌13398596512 ______ 令t=x+a,则f(a)=∫[pi/2 0] |cos(x+a)|dx=∫[a+pi/2 a] |cost|dt,f(a+pi/2)=∫[a+pi a+pi/2] |cost|dt,所以f(a)+f(a+pi/2)=∫[a+pi a] |cost|dt因为被积函数f(t)=|cost|是周期为pi的周期函数,所以f(a)+f(a+pi/2)=∫[pi 0] |cost|dt=2f'(a)=|cos(a+pi/2)|-|cos(a)|令f'(a)=0...

巫姚侮2020sina+cosa<0,则a在第几象限 -
狄支斌13398596512 ______ sina+cosa= sqrt(2)sin(a+ pi/4)a+ pi/4 在3,4象限 a 在2,3,4象限都有可能

巫姚侮2020They are planing where they will go for a picinic this wekend改为同义句 -
狄支斌13398596512 ______ They are planing where they will go for a picinic this weekend 同义句:They are planing where to go for a picinic this weekend

巫姚侮2020sin a = 3/5 , a E (＂pi/2＂,pie) , sin (a+＂pi/3＂) = ? 是怎么做的? -
狄支斌13398596512 ______ sin(a+b)=sinacosb+sinbcosa.cosa=+-4/5.原式=sinacosp/3+sinp/3cosa=3/5*1/2+#3/2*(+-4/5)(#是根号)=3/10+-2#3/5

巫姚侮2020sec(a+pi/2)=? -
狄支斌13398596512 ______ 由secx=1/cosx.知sec(a+pi/2)=1/cos(a+pi/2)=1/-sina=-csca

巫姚侮2020求Y=sin(a)+cos(a)的最大值 -
狄支斌13398596512 ______ y=sina+cosa=1/2sqr2[sinacos45^0+cosasin45^0]=1/2sqr2sin(a+45^0) 由于sin(a+45^0)的最大值为1,故 y(max)=1/2sqr2(二分之根号二)

巫姚侮2020已知sina=1/3,sin(a+b)=1,求sin(a+2b) -
狄支斌13398596512 ______[答案] 因为sin(a+b)=1 所以a+b=2k*pi+pi/2 所以b=2k*pi+pi/2-a 所以 sin(a+2b) =sin(a+b+b) =sin(2k*pi+pi/2+2k*pi+pi/2-a) =sin(4k*pi+pi-a) =sin(pi-a) =sin(a) =1/3

巫姚侮2020化简tan(α+180度)+cos( - α)+cos(α+180度)分之sin(180度+α) - tan( - α) - tan(360度+α) -
狄支斌13398596512 ______[答案] (sin(a+pi)-tan(-a)-tan(a+2pi))/(tan(a+pi)+cos(-a)+cos(a+pi)) =(-sin a+tan a-tan a)/(tan a+cos a -cos a) =-sin a / tan a =-cos a 答案是负cos a

巫姚侮2020已知5+pi其中(p<0)是实数系一元二次方程x2+qx+26=0的一个根,则p=______,q=______. -
狄支斌13398596512 ______[答案] 设实数系一元二次方程x2+qx+26=0的另一个根为a+bi 则a+bi+5+pi=-q,(a+bi)•(5+pi)=26 则b=-p>0, 解得:a+bi=5+i ∴p=-1,q=-10 故答案为:-1,-10

巫姚侮2020(全题)已知sin(α+3pi/4)=5/13,cos(pi/4 - β)=3/5,且 - 4/pi小于α小于pi/4,pi/4小于β小于3pi/4,求cos2(α - β)? -
狄支斌13398596512 ______[答案] ∵sin(α+3π/4)=cos(α+π/4)=5/13 cos(π/4-β)=sin(β+π/4)=3/5 又 -π/4