Question

已知sin(a+135°)=5/13,cos(45°-β ),且-45°<a<45°,45°<β <135°,求cos2(a-β)的值

cos(45°-β )=3/5

Answer 1

cos2(a-β)=2cos^2(a-β)-1
-45°<a<喊蔽45° ,90°<a+135°<180° sin(a+135°)=5/13
所以：cos(a+135°)=-√[1-sin^2(a+135°)]=-√枝渗凯(1-25/169)=-12/13
45°<β <135°, -45°>-β >-135°, -90°<45°-β<0° cos(45°-β )=3/5
所以猛唤：sin(45°-β)=-√[1-cos(45°-β)]=-√(1-9/25)=-4/5
cos(a-β)=-cos(a-β+180°)
=-cos[(a+135°)+(45°-β)]
=-cos(a+135°)cos(45°-β)+sin(a+135°)sin(45°-β)
=-（-12/13）*(3/5)+(5/13)*(-4/5)
=36/65-20/65
=16/65
所以：cos2(a-β)=2cos^2(a-β)-1=2*（16/65）^2-1