Question

已知sin(30°+α)=3/5,0<α<π,求cosα的值

Answer 1

已知sin(30°+α)=3/5,0<α<π,求cosα的值
解：sin(30°+α)=sin30°cosα+cos30°sinα=(1/2)cosα+(√3/2)sinα=3/5
cosα+(√3)sinα=6/5
故sinα=[(6/5)-cosα]/(√3)
cos²α+sin²α=cos²α+[(6/5)-cosα]²/3=1
4cos²α-(12/5)cosα-39/25=0
100cos²α-60cosα-39=0
故cosα=(60±√19200)/200=(60±80√3)/200=(3±4√3)/10

Answer 2

sin(π/4 -α)=3/5推出sin(α-π/4 )=-3/5推出cos(π/4 α）=3/5 你的题目是错误的，无法解答！！！π/4

Answer 3

=sin30°cosa+cos30°sina
=(cosa+√3sina)/2=3/5
sina=(6/5-cosa)/√3
cos²a+sin²a=1
cos²a+(6/5-cosa)²/3=1
4cos²a-12cosa/5-3+36/25=0
100cos²a-60cosa-39=0
cosa=(3±4√3)/10