cos^2(π/4-α)+cos^2(π/4+α)
=sin^2[π/2-(π/4-α)]+cos^2(π/4+α)
=sin^2(π/4+α)+cos^2(π/4+α)
=1
式子cos^2(π/4-α)+cos^2(π/4+α)=?
1个回答
相关问题
-
sin(3π+α )=1/4,求cos(π+α)/cosα[cos(π+α)-1]+cos(α-2π)/cos(α+2π
-
cosα+sinα不等于 A.√2cos[α-(π/4)] B.√2cos[(π/4)-α] C.cos[α-(π/4)
-
若sin(π-α)+cos(2π-α)＝1 ,则sin4α+cos4α+sin(π+α)·cos(4π-α)的值为___
-
化简：（1）cos(α+π)sin(−α)cos(−3π−α)sin(−α−4π)（2）cos(α−π2)sin(5π2
-
sin(π/4 -α)=5/13,且α∈（0,π/2）,则cos2α / cos(π/4 +α）=?
-
化简2cos²α-1/2tan(π/4-α)cos²(π/4-α)
-
sin2α=sin^2(α+π/4)-cos^2(α+π/4)=2sin^2(a+π/4)-1=1-2cos^2(α+π
-
已知cos（α+π/4）=-3/5,π/4＜α＜π/2,（1）求cos（2α+π/4）；（2）求tanα/2
-
化简：[1+2sin(α+2π)*cos(α-2π)]/[sin(α+4π)+cos(α+8π)]
-
sin(π/2+α)·cos(π/2-α)/cos(π+α)+sin（π-α）·cos（π/2+α）/sin（π+α）=