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谢谢 ccmmjj !好题 !我来凑个热闹。
\(\sqrt{5}=\frac{5}{\sqrt{5}}=\frac{1+2^2}{\sqrt{1+2^2}}=>若\sin\theta+2\cos\theta=\sqrt{5},则:\tan\theta=\frac{1}{2}\)
\(\sqrt{10}=\frac{10}{\sqrt{10}}=\frac{1+3^2}{\sqrt{1+3^2}}=>若\sin\theta+3\cos\theta=\sqrt{10},则:\tan\theta=\frac{1}{3}\)
\(\sqrt{13}=\frac{13}{\sqrt{13}}=\frac{2^2+3^2}{\sqrt{2^2+3^2}}=>若2\sin\theta+3\cos\theta=\sqrt{13},则:\tan\theta=\frac{2}{3}\)
\(\sqrt{25}=\frac{25}{\sqrt{25}}=\frac{3^2+4^2}{\sqrt{3^2+4^2}}=>若3\sin\theta+4\cos\theta=\sqrt{25},则:\tan\theta=\frac{3}{4}\)
\(\sqrt{a^2+b^2}=\frac{a^2+b^2}{\sqrt{a^2+b^2}}=>若a\sin\theta+b\cos\theta=\sqrt{a^2+b^2},则:\tan\theta=\frac{a}{b}\) |
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