题目
Problem
(a) Express 5 sin θ + 12 cos θ 5\sin\theta+12\cos\theta 5 sin θ + 12 cos θ R cos ( θ − α ) R\cos(\theta-\alpha) R cos ( θ − α ) R > 0 R>0 R > 0 0 < α < 1 2 π 0<\alpha<\frac12\pi 0 < α < 2 1 π
(3)
(b) Hence solve the equation 5 sin 2 x + 12 cos 2 x = 6 5\sin2x+12\cos2x=6 5 sin 2 x + 12 cos 2 x = 6 0 ≤ x ≤ π 0\le x\le\pi 0 ≤ x ≤ π
(4)
题目中文翻译
(a) 将 5 sin θ + 12 cos θ 5\sin\theta+12\cos\theta 5 sin θ + 12 cos θ R cos ( θ − α ) R\cos(\theta-\alpha) R cos ( θ − α ) R > 0 R>0 R > 0 0 < α < 1 2 π 0<\alpha<\frac12\pi 0 < α < 2 1 π
(3)
(b) 由此解方程 5 sin 2 x + 12 cos 2 x = 6 5\sin2x+12\cos2x=6 5 sin 2 x + 12 cos 2 x = 6 0 ≤ x ≤ π 0\le x\le\pi 0 ≤ x ≤ π
(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 R cos ( θ − α ) R\cos(\theta-\alpha) R cos ( θ − α ) α \alpha α
答题过程
展开
Expanding,
R cos ( θ − α ) = R cos α cos θ + R sin α sin θ . R\cos(\theta-\alpha)
=R\cos\alpha\cos\theta+R\sin\alpha\sin\theta. R cos ( θ − α ) = R cos α cos θ + R sin α sin θ . Comparing coefficients,
R cos α = 12 , R sin α = 5. \begin{align*}
R\cos\alpha=&\,12,\\[4mm]
R\sin\alpha=&\,5.
\end{align*} R cos α = R sin α = 12 , 5. Hence
R = 12 2 + 5 2 = 13 , α = tan − 1 5 12 = 0.394791 … . \begin{align*}
R=&\,\sqrt{12^2+5^2}=13,\\[4mm]
\alpha=&\,\tan^{-1}\frac5{12}=0.394791\ldots.
\end{align*} R = α = 1 2 2 + 5 2 = 13 , tan − 1 12 5 = 0.394791 … . Therefore
5 sin θ + 12 cos θ = 13 cos ( θ − α ) , \boxed{5\sin\theta+12\cos\theta
=13\cos(\theta-\alpha)}, 5 sin θ + 12 cos θ = 13 cos ( θ − α ) , where α = tan − 1 ( 5 / 12 ) ≈ 0.395 \alpha=\tan^{-1}(5/12)\approx0.395 α = tan − 1 ( 5/12 ) ≈ 0.395
(b)
解法一
思路
展开
承接(a)，将角换成 2 x 2x 2 x 2 2 2
答题过程
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Using part (a), with the unrounded α = tan − 1 ( 5 / 12 ) \alpha=\tan^{-1}(5/12) α = tan − 1 ( 5/12 )
13 cos ( 2 x − α ) = 6. 13\cos(2x-\alpha)=6. 13 cos ( 2 x − α ) = 6. Let β = cos − 1 ( 6 / 13 ) \beta=\cos^{-1}(6/13) β = cos − 1 ( 6/13 ) 0 ≤ 2 x ≤ 2 π 0\le2x\le2\pi 0 ≤ 2 x ≤ 2 π
2 x = α + β or 2 x = α + 2 π − β . \begin{align*}
2x=&\,\alpha+\beta\\[4mm]
\text{or}\quad2x=&\,\alpha+2\pi-\beta.
\end{align*} 2 x = or 2 x = α + β α + 2 π − β . Thus
x = 0.743 or x = 2.79 \boxed{x=0.743\quad\text{or}\quad x=2.79} x = 0.743 or x = 2.79 in radians, to 3 significant figures.
