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已知sin(-π/2 -α)×(乘号)cos(-5π/2 -α)=60/169,且π/4<α<π/2,求sinα与cos

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已知sin(-π/2 -α)×(乘号)cos(-5π/2 -α)=60/169,且π/4<α<π/2,求sinα与cosα的值.
已知sin(-π/2 -α)×(乘号)cos(-5π/2 -α)=60/169,且π/4<α<π/2,求sinα与cos
sin(-π/2 -α)= -sin(π/2 +α)=cosα;
cos(-5π/2 -α)=cos(5π/2 +α)=sinα;
所以原式=cosα*sinα=60/169
又因为(cosα)^2+(sinα)^2=1
所以(sinα-cosα)^2=(cosα)^2+(sinα)^2-2*cosα*sinα=49/169;
(sinα+cosα)^2=(cosα)^2+(sinα)^2+2*cosα*sinα=289/169;
开跟,所以sinα-cosα=7/13,sinα+cosα=17/13;
解得sinα=12/13;cosα=5/13