sin(α-5π/4)=sinα cos5π/4 - cosα sin5π/4 = -√2/2sinα + √2/2cosα =-√2/2(sinα -cosα)因为sinα+cosα=√6/2所以(sinα+cosα)^2 =1+2sinαcosα =3/2所以 sinαcosα =1/4所以(sinα -cosα)^2 =...
一道高一数学题,急死了已知sinα cosα=√6/2,α∈(0,π/4),则sin(α-5π/4)=?是sinα+co
5个回答
相关问题
-
已知sin(2π-α)=[4/5],α∈(3π2,2π),则[sinα+cosα/sinα-cosα]等于( )
-
已知α∈﹙0,π/2),且2Sinα-SinαCosα-3Cosα=0.求[Sin(α+π/4)]/(Sin2α+Cos
-
已知sin(7π-α)-3cos(3π/2+α)=2,则(sin(π-α)+cos(π+α))/(sinα+cos(-α
-
已知tanα=2,sinα+cosα<0,求[sin(2π-α)*sin(π+α)*cos(-π+α)]/[sin(3π
-
已知sin(α+π)=4/5,且sinα*cosα<0,求【2sin(α-π)+3tan(3π -α)】除以【4cos(
-
若sin(π-α)+cos(2π-α)=1 ,则sin4α+cos4α+sin(π+α)·cos(4π-α)的值为___
-
已知α∈(0,π/4),a=(sinα)^(cosα),b=(sinα)^(sinα),c=(cosα)^(sinα),
-
(2008•和平区三模)已知sin(2π-α)=[4/5],α∈(3π2,2π),则[sinα+cosα/sinα-co
-
已知α∈(π/4,π/2),sinα+cosα=√2/2,求sinα-cosα
-
已知方程sin(α-3π)=2cos(α-4π),求sin(π−α)+5cos(2π−α)2sin(3π2−α)−sin