题目
Problem
(a) Express 3 2 sin ( x + 45 ∘ ) + cos x 3\sqrt2\sin(x+45^\circ)+\cos x 3 2 sin ( x + 4 5 ∘ ) + cos x R cos ( x − α ) R\cos(x-\alpha) R cos ( x − α ) R > 0 R>0 R > 0 0 ∘ < α < 90 ∘ 0^\circ<\alpha<90^\circ 0 ∘ < α < 9 0 ∘
(4)
(b) Hence solve the equation 3 2 sin ( 3 θ + 45 ∘ ) + cos 3 θ = − 4 3\sqrt2\sin(3\theta+45^\circ)+\cos3\theta=-4 3 2 sin ( 3 θ + 4 5 ∘ ) + cos 3 θ = − 4 0 ∘ < θ < 180 ∘ 0^\circ<\theta<180^\circ 0 ∘ < θ < 18 0 ∘
(4)
题目中文翻译
(a) 将 3 2 sin ( x + 45 ∘ ) + cos x 3\sqrt2\sin(x+45^\circ)+\cos x 3 2 sin ( x + 4 5 ∘ ) + cos x R cos ( x − α ) R\cos(x-\alpha) R cos ( x − α ) R > 0 R>0 R > 0 0 ∘ < α < 90 ∘ 0^\circ<\alpha<90^\circ 0 ∘ < α < 9 0 ∘
(4)
(b) 由此解方程 3 2 sin ( 3 θ + 45 ∘ ) + cos 3 θ = − 4 3\sqrt2\sin(3\theta+45^\circ)+\cos3\theta=-4 3 2 sin ( 3 θ + 4 5 ∘ ) + cos 3 θ = − 4 0 ∘ < θ < 180 ∘ 0^\circ<\theta<180^\circ 0 ∘ < θ < 18 0 ∘
(4)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
展开
展开和角正弦，合并成正弦与余弦的线性组合，再与目标形式比较系数。
答题过程
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Using the addition formula,
3 2 sin ( x + 45 ∘ ) + cos x = 3 sin x + 4 cos x . 3\sqrt2\sin(x+45^\circ)+\cos x=3\sin x+4\cos x. 3 2 sin ( x + 4 5 ∘ ) + cos x = 3 sin x + 4 cos x . Also,
R cos ( x − α ) = R cos α cos x + R sin α sin x . R\cos(x-\alpha)=R\cos\alpha\cos x+R\sin\alpha\sin x. R cos ( x − α ) = R cos α cos x + R sin α sin x . Comparing coefficients,
R cos α = 4 , R sin α = 3. R\cos\alpha=4,\qquad R\sin\alpha=3. R cos α = 4 , R sin α = 3. Thus
R = 4 2 + 3 2 = 5 , α = tan − 1 3 4 = 36.869897 … ∘ . \begin{align*}
R=&\,\sqrt{4^2+3^2}=5,\\[4mm]
\alpha=&\,\tan^{-1}\frac34=36.869897\ldots^\circ.
\end{align*} R = α = 4 2 + 3 2 = 5 , tan − 1 4 3 = 36.869897 … ∘ . Hence the required form is
5 cos ( x − 36.9 ∘ ) . \boxed{5\cos(x-36.9^\circ)}. 5 cos ( x − 36. 9 ∘ ) .
(b)
解法一
思路
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沿用(a)，把 x x x 3 θ 3\theta 3 θ α \alpha α
答题过程
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Using (a),
5 cos ( 3 θ − α ) = − 4 , α = tan − 1 ( 3 / 4 ) . 5\cos(3\theta-\alpha)=-4,
\qquad \alpha=\tan^{-1}(3/4). 5 cos ( 3 θ − α ) = − 4 , α = tan − 1 ( 3/4 ) . Since 0 ∘ < θ < 180 ∘ 0^\circ<\theta<180^\circ 0 ∘ < θ < 18 0 ∘
− α < 3 θ − α < 540 ∘ − α . -\alpha<3\theta-\alpha<540^\circ-\alpha. − α < 3 θ − α < 54 0 ∘ − α . Let β = cos − 1 ( − 4 / 5 ) = 143.130102 … ∘ \beta=\cos^{-1}(-4/5)=143.130102\ldots^\circ β = cos − 1 ( − 4/5 ) = 143.130102 … ∘ 3 θ − α 3\theta-\alpha 3 θ − α
β , 360 ∘ − β . \beta,\qquad 360^\circ-\beta. β , 36 0 ∘ − β . The next positive solution, 360 ∘ + β 360^\circ+\beta 36 0 ∘ + β 540 ∘ − α 540^\circ-\alpha 54 0 ∘ − α β = 180 ∘ − α \beta=180^\circ-\alpha β = 18 0 ∘ − α θ = 180 ∘ \theta=180^\circ θ = 18 0 ∘
θ = β + α 3 = 60 ∘ \theta=\frac{\beta+\alpha}{3}=60^\circ θ = 3 β + α = 6 0 ∘ or
θ = 360 ∘ − β + α 3 = 84.579931 … ∘ . \theta=\frac{360^\circ-\beta+\alpha}{3}
=84.579931\ldots^\circ. θ = 3 36 0 ∘ − β + α = 84.579931 … ∘ . Therefore θ = 60.0 ∘ , 84.6 ∘ \boxed{\theta=60.0^\circ,\ 84.6^\circ} θ = 60. 0 ∘ , 84. 6 ∘
