第 4 題
題目:若 \(\displaystyle a,b,c\) 為相異正實數,則 \(\displaystyle a^{log\frac{b}{c}}\cdot b^{log\frac{c}{a}} \cdot c^{log\frac{a}{b}}=\)______________。
解答:
\(\displaystyle \log\left(a^{log\frac{b}{c}}\cdot b^{log\frac{c}{a}} \cdot c^{log\frac{a}{b}}\right)\)
\(\displaystyle =log\frac{b}{c}\cdot\log a + log\frac{c}{a}\cdot \log b + log\frac{a}{b}\cdot \log c\)
\(\displaystyle =\left(\log b - \log c\right)\log a +\left(\log c - \log a\right)\log b +\left(\log a - \log b\right)\log c\)
\(\displaystyle =0\)
故,\(\displaystyle a^{log\frac{b}{c}}\cdot b^{log\frac{c}{a}} \cdot c^{log\frac{a}{b}} = 1\)
第 6 題
題目: \(\displaystyle \left|\vec{a}\right|=\left|\vec{b}\right|\neq0\),若 \(\displaystyle \left|\vec{a}+\vec{b}\right|-\left|\vec{a}-\vec{b}\right|=\sqrt{2}\left|\vec{a}\right|\),則 \(\displaystyle \vec{a}\) 與 \(\displaystyle \vec{b}\) 的夾角為____________。
解答:
令 \(\displaystyle \vec{a}\) 與 \(\displaystyle \vec{b}\) 的夾角為 \(\displaystyle \theta\),\(\displaystyle \left|\vec{a}\right|=k>0\),
則
\(\displaystyle \left|\vec{a}+\vec{b}\right|=\sqrt{\left(\vec{a}+\vec{b}\right)\cdot\left(\vec{a}+\vec{b}\right)}=k\sqrt{2\left(1+\cos\theta\right)}\)
\(\displaystyle \left|\vec{a}-\vec{b}\right|=\sqrt{\left(\vec{a}-\vec{b}\right)\cdot\left(\vec{a}-\vec{b}\right)}=k\sqrt{2\left(1-\cos\theta\right)}\)
因此,可得 \(\displaystyle \sqrt{2\left(1+\cos\theta\right)} - \sqrt{2\left(1-\cos\theta\right)} = \sqrt{2}\)
\(\displaystyle \Rightarrow \sqrt{1+\cos\theta} - \sqrt{1-\cos\theta} = 1\)
\(\displaystyle \Rightarrow \sqrt{1+\left(2\cos^2\frac{\theta}{2}-1\right)} - \sqrt{1-\left(1-2\sin^2\frac{\theta}{2}\right)} = 1\)
\(\displaystyle \Rightarrow \left|\cos\frac{\theta}{2}\right|-\left|\sin\frac{\theta}{2}\right|=\frac{1}{\sqrt{2}}\)
因為 \(0^\circ\leq\theta\leq180^\circ\),所以 \(\displaystyle \cos\frac{\theta}{2}-\sin\frac{\theta}{2}=\frac{1}{\sqrt{2}}\)
\(\displaystyle \Rightarrow \sqrt{2}\cos\left(\frac{\theta}{2}+45^\circ\right)=\frac{1}{\sqrt{2}}\)
\(\displaystyle \Rightarrow \cos\left(\frac{\theta}{2}+45^\circ\right)=\frac{1}{2}\)
\(\displaystyle \Rightarrow \frac{\theta}{2}+45^\circ=\pm 60^\circ + 360^\circ\cdot k,\,k\in\mathbb{Z}\)
因為 \(0^\circ\leq\theta\leq180^\circ\),所以 \(\theta=30^\circ.\)