表达式 8sin⁡θ−3cos⁡θ8 \sin \theta - 3 \cos \theta 的定义域为 0∘≤θ≤180∘0^\circ \le \theta \le 180^\circ。
通过使用 Rsin⁡(θ−α)R \sin(\theta - \alpha)(其中 R>0R > 0 且 0∘<α<90∘0^\circ < \alpha < 90^\circ),

  1. (i)

    解方程 8sin⁡θ−3cos⁡θ=58 \sin \theta - 3 \cos \theta = 5,

  2. (ii)

    写出 1(8sin⁡θ−3cos⁡θ)2\frac{1}{(8 \sin \theta - 3 \cos \theta)^2} 的最小值,并求出当 0∘≤θ≤180∘0^\circ \le \theta \le 180^\circ 时对应的 θ\theta 值。

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完整解法

2(i)

1. Express $8 \sin \theta - 3 \cos \theta$ in the form $R \sin(\theta - \alpha)$

Let 8sin⁡θ−3cos⁡θ=Rsin⁡(θ−α)=R(sin⁡θcos⁡α−cos⁡θsin⁡α)8 \sin \theta - 3 \cos \theta = R \sin(\theta - \alpha) = R(\sin \theta \cos \alpha - \cos \theta \sin \alpha).
Comparing coefficients:
Rcos⁡α=8R \cos \alpha = 8
Rsin⁡α=3R \sin \alpha = 3
R=82+32=73R = \sqrt{8^2 + 3^2} = \sqrt{73}
tan⁡α=38  ⟹  α=tan⁡−1(38)≈20.556∘\tan \alpha = \frac{3}{8} \implies \alpha = \tan^{-1}\left(\frac{3}{8}\right) \approx 20.556^\circ
So, 8sin⁡θ−3cos⁡θ=73sin⁡(θ−20.556∘)8 \sin \theta - 3 \cos \theta = \sqrt{73} \sin(\theta - 20.556^\circ).

2. Set up the equation using the R-formula expression

73sin⁡(θ−20.556∘)=5\sqrt{73} \sin(\theta - 20.556^\circ) = 5
sin⁡(θ−20.556∘)=573\sin(\theta - 20.556^\circ) = \frac{5}{\sqrt{73}}

3. Find the basic angle and solve for $\theta$ in the given range

Basic angle =sin⁡−1(573)≈35.831∘= \sin^{-1}\left(\frac{5}{\sqrt{73}}\right) \approx 35.831^\circ
Since 0∘≤θ≤180∘0^\circ \le \theta \le 180^\circ, the range for (θ−20.556∘)(\theta - 20.556^\circ) is −20.556∘≤θ−20.556∘≤159.444∘-20.556^\circ \le \theta - 20.556^\circ \le 159.444^\circ.
θ−20.556∘=35.831∘\theta - 20.556^\circ = 35.831^\circ or 180∘−35.831∘=144.169∘180^\circ - 35.831^\circ = 144.169^\circ
θ=35.831∘+20.556∘=56.387∘≈56.4∘\theta = 35.831^\circ + 20.556^\circ = 56.387^\circ \approx 56.4^\circ
θ=144.169∘+20.556∘=164.725∘≈164.7∘\theta = 144.169^\circ + 20.556^\circ = 164.725^\circ \approx 164.7^\circ

答案

θ=56.1∘,165.0∘\theta = 56.1^\circ, 165.0^\circ

2(ii)

1. Express the denominator using the R-formula result from part (i)

From part (i), 8sin⁡θ−3cos⁡θ=73sin⁡(θ−20.556∘)8 \sin \theta - 3 \cos \theta = \sqrt{73} \sin(\theta - 20.556^\circ).

The expression is 1(73sin⁡(θ−20.556∘))2=173sin⁡2(θ−20.556∘)\frac{1}{(\sqrt{73} \sin(\theta - 20.556^\circ))^2} = \frac{1}{73 \sin^2(\theta - 20.556^\circ)}.

2. Determine the smallest value of the expression

For the expression to be as small as possible, the denominator must be as large as possible.
The maximum value of sin⁡2(θ−20.556∘)\sin^2(\theta - 20.556^\circ) is 11.
Smallest value = 173(1)=173\frac{1}{73(1)} = \frac{1}{73}.

3. Find the corresponding value of theta

This minimum occurs when sin⁡2(θ−20.556∘)=1\sin^2(\theta - 20.556^\circ) = 1.
Since 0∘≤θ≤180∘0^\circ \le \theta \le 180^\circ, we have −20.556∘≤θ−20.556∘≤159.444∘-20.556^\circ \le \theta - 20.556^\circ \le 159.444^\circ.
θ−20.556∘=90∘\theta - 20.556^\circ = 90^\circ
θ=110.556∘≈110.6∘\theta = 110.556^\circ \approx 110.6^\circ (to 1 d.p.)

答案

Smallest value is 173\frac{1}{73}, corresponding θ=110.6∘\theta = 110.6^\circ

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