sin(pi/4-x)=5/13 0

sin(pi/4-x)=5/13 0 Matlab u(x,t)=sin(5*pi*x)cos(5*pi*t)+2sin(7*pi*x)cos(7*pi*t) 其中0 f(x)h(x)=√2/2sin(x+pi/4)cos(x+5pi/4) =-√2/2sin(x+pi/4)cos(x+pi/4)√2/2sin(x+pi/4)cos(x+5pi/4) =-√2/2sin(x+pi/4)cos(x+pi/4)这个是怎么化出来的? 已知sin(x+pi/6)=1/4,求sin（5pi/6-x)+sin^2(pi/3-x)的值 If sin(x) = 4/5,pi/2 < x < pi,find the value of sin(2x) 已知Sin(2x+pi/3)=1/3,求Sin(5Pi/6-4x),用三角变换 y=0.75sin(x+Pi/4) ( x属于[-pi,pi])的减区间 a,b∈（3pi/4,pi）,sin(a+b)=-3/5,sin(b-pi/4)=12/13,则cos（a+pi/4）= 已知a,b,属于（3pi/4,pi),sin(a+b)=-3/5,sin(b-pi/4)=12/13,则cos(a+pi/4)=? 已知sin(pi/4-x)=3/13(0ans=24/13 已知cos(x)=sqrt(10)/10,x属于(0,pi/2),求sin(pi/4+2x)的值 matlab提示Error:Incomplete or misformed expression or statement.x=0:2*pi:100;s=20*(4*x/pi-(1/(2*pi))*sin(8*pi*x/pi);plot(x,s);求指导. a*sin(pi/4 - x) - b*cos(pi/4 - x) = a*sin(pi/4 + x) - b*cos(pi/4 + x) 怎么得到（a + b）* sinx = 0 化简sin(2x-pi)sin[x+(3/2)pi]/cos[x+(5/2)pi][1-sin(pi+x)] cos(pi/4-a)=3/5 sin(3pi/4+b)=5/13 求sin(a+b) 已知x属于[0,pi],则函数y=2sin(pi/6-2x)为增函数的区间是a.[0,pi/3] b.[5pi/6,pi] f(x)=sin(Pi/4-x)+cos(Pi/4+x)的周期 2(sin^4 x + cos^4 x) =1 [ - pi ≤ x ≤ pi ]