高専の数学解説

第2章 2.20

x=2a1+a2(−1≤a≤1)x = \frac{2a}{1+a^2} \quad (-1 \le a \le 1)

1+x−1−x1+x+1−x=1+2a1+a2−1−2a1+a21+2a1+a2+1−2a1+a2=1+a2+2a1+a2−1+a2−2a1+a21+a2+2a1+a2+1+a2−2a1+a2=(a+1)2−(a−1)2(a+1)2+(a−1)2\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} = \frac{\sqrt{1+\dfrac{2a}{1+a^2}}-\sqrt{1-\dfrac{2a}{1+a^2}}}{\sqrt{1+\dfrac{2a}{1+a^2}}+\sqrt{1-\dfrac{2a}{1+a^2}}} = \frac{\sqrt{\dfrac{1+a^2+2a}{1+a^2}}-\sqrt{\dfrac{1+a^2-2a}{1+a^2}}}{\sqrt{\dfrac{1+a^2+2a}{1+a^2}}+\sqrt{\dfrac{1+a^2-2a}{1+a^2}}} = \frac{\sqrt{(a+1)^2}-\sqrt{(a-1)^2}}{\sqrt{(a+1)^2}+\sqrt{(a-1)^2}}

=a+1−1+aa+1+1−a=a= \frac{a+1-1+a}{a+1+1-a} = a

解説: r31bn1z