题目
Problem
(a) Express 5 sin ( x + 1 6 π ) − 4 cos x 5\sin\big(x+\frac16\pi\big)-4\cos x 5 sin ( x + 6 1 π ) − 4 cos x R sin ( x − α ) R\sin(x-\alpha) R sin ( x − α ) R > 0 R>0 R > 0 0 < α < 1 2 π 0<\alpha<\frac12\pi 0 < α < 2 1 π R R R α \alpha α 3 3 3
(4)
(b) Hence solve the equation 5 sin ( 2 θ + 1 6 π ) − 4 cos 2 θ = 7 5\sin\big(2\theta+\frac16\pi\big)-4\cos2\theta=\sqrt7 5 sin ( 2 θ + 6 1 π ) − 4 cos 2 θ = 7 0 ≤ θ ≤ π 0\le\theta\le\pi 0 ≤ θ ≤ π 2 2 2
(4)
题目中文翻译
(a) 将 5 sin ( x + 1 6 π ) − 4 cos x 5\sin\big(x+\frac16\pi\big)-4\cos x 5 sin ( x + 6 1 π ) − 4 cos x R sin ( x − α ) R\sin(x-\alpha) R sin ( x − α ) R > 0 R>0 R > 0 0 < α < 1 2 π 0<\alpha<\frac12\pi 0 < α < 2 1 π R R R α \alpha α 3 3 3
(4)
(b) 由此解方程 5 sin ( 2 θ + 1 6 π ) − 4 cos 2 θ = 7 5\sin\big(2\theta+\frac16\pi\big)-4\cos2\theta=\sqrt7 5 sin ( 2 θ + 6 1 π ) − 4 cos 2 θ = 7 0 ≤ θ ≤ π 0\le\theta\le\pi 0 ≤ θ ≤ π 2 2 2
(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
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展开和角后，正弦系数为正、余弦系数为负，与 sin ( x − α ) \sin(x-\alpha) sin ( x − α )
答题过程
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Expanding,
5 sin ( x + π / 6 ) − 4 cos x = 5 3 2 sin x + 5 2 cos x − 4 cos x = 5 3 2 sin x − 3 2 cos x . \begin{align*}
5\sin(x+\pi/6)-4\cos x
=&\,\frac{5\sqrt3}{2}\sin x+\frac52\cos x\\[4mm]
&\,\hspace{2pt}-4\cos x\\[4mm]
=&\,\frac{5\sqrt3}{2}\sin x-\frac32\cos x.
\end{align*} 5 sin ( x + π /6 ) − 4 cos x = = 2 5 3 sin x + 2 5 cos x − 4 cos x 2 5 3 sin x − 2 3 cos x . Since
R sin ( x − α ) = R cos α sin x − R sin α cos x , R\sin(x-\alpha)=R\cos\alpha\sin x-R\sin\alpha\cos x, R sin ( x − α ) = R cos α sin x − R sin α cos x , we require
R cos α = 5 3 2 , R sin α = 3 2 . \begin{align*}
R\cos\alpha=&\,\frac{5\sqrt3}{2},\\[4mm]
R\sin\alpha=&\,\frac32.
\end{align*} R cos α = R sin α = 2 5 3 , 2 3 . Therefore
R = 75 4 + 9 4 = 21 , R=\sqrt{\frac{75}{4}+\frac94}=\sqrt{21}, R = 4 75 + 4 9 = 21 , and
α = tan − 1 3 5 3 = 0.333473 … = 0.333 . \alpha=\tan^{-1}\frac{3}{5\sqrt3}
=0.333473\ldots=\boxed{0.333}. α = tan − 1 5 3 3 = 0.333473 … = 0.333 . The required form is 21 sin ( x − α ) \boxed{\sqrt{21}\sin(x-\alpha)} 21 sin ( x − α )
(b)
解法一
思路
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承接(a)，把新角范围写成 − α -\alpha − α 2 π − α 2\pi-\alpha 2 π − α
答题过程
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Using (a),
sin ( 2 θ − α ) = 7 21 = 1 3 . \sin(2\theta-\alpha)=\frac{\sqrt7}{\sqrt{21}}
=\frac1{\sqrt3}. sin ( 2 θ − α ) = 21 7 = 3 1 . Let β = sin − 1 ( 1 / 3 ) \beta=\sin^{-1}(1/\sqrt3) β = sin − 1 ( 1/ 3 ) 0 ⩽ θ ⩽ π 0\leqslant\theta\leqslant\pi 0 ⩽ θ ⩽ π
− α ⩽ 2 θ − α ⩽ 2 π − α . -\alpha\leqslant2\theta-\alpha\leqslant2\pi-\alpha. − α ⩽ 2 θ − α ⩽ 2 π − α . The only admissible angles are β \beta β π − β \pi-\beta π − β
θ = β + α 2 = 0.474476 … \theta=\frac{\beta+\alpha}{2}=0.474476\ldots θ = 2 β + α = 0.474476 … or
θ = π − β + α 2 = 1.429793 … . \theta=\frac{\pi-\beta+\alpha}{2}=1.429793\ldots. θ = 2 π − β + α = 1.429793 … . Therefore, to 2 decimal places,
θ = 0.47 , 1.43 radians . \boxed{\theta=0.47,\ 1.43\text{ radians}}. θ = 0.47 , 1.43 radians . 