如何推导-√2<sin∠α+cos∠α<√2
问题描述:
如何推导-√2<sin∠α+cos∠α<√2
答
sin∠α+cos∠α=√2sin(∠α+π/3)
-1=
-√2=
答
sinα+cosα=√2(sina √2/2+cosα2/2)=√2sin(α+π/4)
-1=
答
(sin∠α - cos∠α)^2 >= 0展开移项得:(sin∠α)^2 + (cos∠α)^2 >= 2 * sin∠α * cos∠α不等式两边同时加上(sin∠α)^2 + (cos∠α)^2 ,不等号不变,得:2 * [(sin∠α)^2 + (cos∠α)^2 ] >= (sin∠α + ...
答
sinα+cosα=√2*sin(α+π/4)
因为-1≤sin(α+π/4)≤1
所以-√2≤√2*sin(α+π/4)≤√2
即-√2≤sinα+cosα≤√2