题目
Problem
(a) Given that
sin ( x + 1 6 π ) − sin ( x − 1 6 π ) = cos ( x + 1 3 π ) − cos ( x − 1 3 π ) , \begin{align*}
&\,\sin\big(x+\tfrac16\pi\big)-\sin\big(x-\tfrac16\pi\big)\\
=&\,\cos\big(x+\tfrac13\pi\big)-\cos\big(x-\tfrac13\pi\big),
\end{align*} = sin ( x + 6 1 π ) − sin ( x − 6 1 π ) cos ( x + 3 1 π ) − cos ( x − 3 1 π ) , find the exact value of tan x \tan x tan x
(4)
(b) Hence find the exact roots of the equation
sin ( x + 1 6 π ) − sin ( x − 1 6 π ) = cos ( x + 1 3 π ) − cos ( x − 1 3 π ) \begin{align*}
&\,\sin\big(x+\tfrac16\pi\big)-\sin\big(x-\tfrac16\pi\big)\\
=&\,\cos\big(x+\tfrac13\pi\big)-\cos\big(x-\tfrac13\pi\big)
\end{align*} = sin ( x + 6 1 π ) − sin ( x − 6 1 π ) cos ( x + 3 1 π ) − cos ( x − 3 1 π ) for 0 ≤ x ≤ 2 π 0\le x\le2\pi 0 ≤ x ≤ 2 π
(2)
题目中文翻译
(a) 已知
sin ( x + 1 6 π ) − sin ( x − 1 6 π ) = cos ( x + 1 3 π ) − cos ( x − 1 3 π ) , \begin{align*}
&\,\sin\big(x+\tfrac16\pi\big)-\sin\big(x-\tfrac16\pi\big)\\
=&\,\cos\big(x+\tfrac13\pi\big)-\cos\big(x-\tfrac13\pi\big),
\end{align*} = sin ( x + 6 1 π ) − sin ( x − 6 1 π ) cos ( x + 3 1 π ) − cos ( x − 3 1 π ) , 求 tan x \tan x tan x
(4)
(b) 由此求方程
sin ( x + 1 6 π ) − sin ( x − 1 6 π ) = cos ( x + 1 3 π ) − cos ( x − 1 3 π ) \begin{align*}
&\,\sin\big(x+\tfrac16\pi\big)-\sin\big(x-\tfrac16\pi\big)\\
=&\,\cos\big(x+\tfrac13\pi\big)-\cos\big(x-\tfrac13\pi\big)
\end{align*} = sin ( x + 6 1 π ) − sin ( x − 6 1 π ) cos ( x + 3 1 π ) − cos ( x − 3 1 π ) 在 0 ≤ x ≤ 2 π 0\le x\le2\pi 0 ≤ x ≤ 2 π
(2)
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
分别展开四个和差角，先让重复项相消，再代入特殊角的精确三角比。
答题过程
展开
Using the addition formulae,
sin ( x + π 6 ) − sin ( x − π 6 ) = ( 3 2 sin x + 1 2 cos x ) − ( 3 2 sin x − 1 2 cos x ) = 2 cos x sin π 6 = cos x , \begin{align*}
&\,\sin\bigg(x+\frac\pi6\bigg)-\sin\bigg(x-\frac\pi6\bigg)\\[4mm]
=&\,\bigg(\frac{\sqrt3}{2}\sin x+\frac12\cos x\bigg)\\[4mm]
&\,\hspace{2pt}-\bigg(\frac{\sqrt3}{2}\sin x-\frac12\cos x\bigg)\\[4mm]
=&\,2\cos x\sin\frac\pi6\\[4mm]
=&\,\cos x,
\end{align*} = = = sin ( x + 6 π ) − sin ( x − 6 π ) ( 2 3 sin x + 2 1 cos x ) − ( 2 3 sin x − 2 1 cos x ) 2 cos x sin 6 π cos x , and
cos ( x + π 3 ) − cos ( x − π 3 ) = − 2 sin x sin π 3 = − 3 sin x . \begin{align*}
&\,\cos\bigg(x+\frac\pi3\bigg)-\cos\bigg(x-\frac\pi3\bigg)\\[4mm]
=&\,-2\sin x\sin\frac\pi3\\[4mm]
=&\,-\sqrt3\sin x.
\end{align*} = = cos ( x + 3 π ) − cos ( x − 3 π ) − 2 sin x sin 3 π − 3 sin x . The given equation therefore becomes
cos x = − 3 sin x . \cos x=-\sqrt3\sin x. cos x = − 3 sin x . Here cos x \cos x cos x sin x \sin x sin x cos x \cos x cos x
tan x = − 1 3 . \boxed{\tan x=-\frac1{\sqrt3}}. tan x = − 3 1 .
(b)
解法一
思路
展开
利用(a)得到的正切值。参考角为 π / 6 \pi/6 π /6
答题过程
展开
From part (a),
tan x = − 1 3 . \tan x=-\frac1{\sqrt3}. tan x = − 3 1 . The reference angle is π / 6 \pi/6 π /6 0 ≤ x ≤ 2 π 0\le x\le2\pi 0 ≤ x ≤ 2 π
x = π − π 6 = 5 π 6 or x = 2 π − π 6 = 11 π 6 . \boxed{x=\pi-\frac\pi6=\frac{5\pi}{6}
\quad\text{or}\quad
x=2\pi-\frac\pi6=\frac{11\pi}{6}}. x = π − 6 π = 6 5 π or x = 2 π − 6 π = 6 11 π . 