来自葛订记的问题
已知sinθ=(m-3)/(m+5),cosθ=(4-2m)/(m+5)(π/2
已知sinθ=(m-3)/(m+5),cosθ=(4-2m)/(m+5)(π/2
1回答
2020-11-01 14:43
已知sinθ=(m-3)/(m+5),cosθ=(4-2m)/(m+5)(π/2
已知sinθ=(m-3)/(m+5),cosθ=(4-2m)/(m+5)(π/2
∵sin²θ+cos²θ=1∴(m-3)²/(m+5)²+(4-2m)²/(m+5)²=1∴(m-3)²+(4-2m)²=(m+5)²即m²-6m+9+16-16m+4m²=m²+10m+25即25-22m+4m²=10m+25即-32m+4m²...