题目
Problem
(a) Express 3 cos 2 x − 3 sin 2 x 3\cos2x-\sqrt3\sin2x 3 cos 2 x − 3 sin 2 x R cos ( 2 x + α ) R\cos(2x+\alpha) R cos ( 2 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)
(b) Hence find the exact value of
∫ 0 1 12 π 3 ( 3 cos 2 x − 3 sin 2 x ) 2 d x , \int_0^{\frac1{12}\pi}\frac{3}{(3\cos2x-\sqrt3\sin2x)^2}\,\mathrm{d}x, ∫ 0 12 1 π ( 3 cos 2 x − 3 sin 2 x ) 2 3 d x , simplifying your answer.
(5)
题目中文翻译
(a) 将 3 cos 2 x − 3 sin 2 x 3\cos2x-\sqrt3\sin2x 3 cos 2 x − 3 sin 2 x R cos ( 2 x + α ) R\cos(2x+\alpha) R cos ( 2 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)
(b) 由此求
∫ 0 1 12 π 3 ( 3 cos 2 x − 3 sin 2 x ) 2 d x \int_0^{\frac1{12}\pi}\frac{3}{(3\cos2x-\sqrt3\sin2x)^2}\,\mathrm{d}x ∫ 0 12 1 π ( 3 cos 2 x − 3 sin 2 x ) 2 3 d x 的精确值，并化简答案。
(5)
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 ( 2 x + α ) R\cos(2x+\alpha) R cos ( 2 x + α )
答题过程
展开
Expanding,
R cos ( 2 x + α ) = R cos α cos 2 x − R sin α sin 2 x . R\cos(2x+\alpha)
=R\cos\alpha\cos2x-R\sin\alpha\sin2x. R cos ( 2 x + α ) = R cos α cos 2 x − R sin α sin 2 x . Comparing coefficients,
R cos α = 3 , R sin α = 3 . \begin{align*}
R\cos\alpha=&\,3,\\[4mm]
R\sin\alpha=&\,\sqrt3.
\end{align*} R cos α = R sin α = 3 , 3 . Therefore
R 2 = 9 + 3 = 12 , R = 2 3 . R^2=9+3=12,\qquad R=2\sqrt3. R 2 = 9 + 3 = 12 , R = 2 3 . Also tan α = 3 / 3 \tan\alpha=\sqrt3/3 tan α = 3 /3 0 < α < π / 2 0<\alpha<\pi/2 0 < α < π /2 α = π / 6 \alpha=\pi/6 α = π /6
3 cos 2 x − 3 sin 2 x = 2 3 cos ( 2 x + π 6 ) . \boxed{3\cos2x-\sqrt3\sin2x
=2\sqrt3\cos\bigg(2x+\frac\pi6\bigg)}. 3 cos 2 x − 3 sin 2 x = 2 3 cos ( 2 x + 6 π ) .
(b)
解法一
思路
展开
承接(a)，把分母的平方化成 12 cos 2 ( 2 x + π / 6 ) 12\cos^2(2x+\pi/6) 12 cos 2 ( 2 x + π /6 )
答题过程
展开
Using part (a),
I = ∫ 0 π / 12 3 ( 3 cos 2 x − 3 sin 2 x ) 2 d x = 1 4 ∫ 0 π / 12 sec 2 ( 2 x + π 6 ) d x = 1 8 [ tan ( 2 x + π 6 ) ] 0 π / 12 = 1 8 ( tan π 3 − tan π 6 ) = 1 8 ( 3 − 1 3 ) . \begin{align*}
I=&\,\int_0^{\pi/12}
\frac{3}{\big(3\cos2x-\sqrt3\sin2x\big)^2}
\,\mathrm{d}x\\[4mm]
=&\,\frac14\int_0^{\pi/12}
\sec^2\bigg(2x+\frac\pi6\bigg)\,\mathrm{d}x\\[4mm]
=&\,\frac18\bigg[\tan\bigg(2x+\frac\pi6\bigg)\bigg]_0^{\pi/12}\\[4mm]
=&\,\frac18\bigg(\tan\frac\pi3-\tan\frac\pi6\bigg)\\[4mm]
=&\,\frac18\bigg(\sqrt3-\frac1{\sqrt3}\bigg).
\end{align*} I = = = = = ∫ 0 π /12 ( 3 cos 2 x − 3 sin 2 x ) 2 3 d x 4 1 ∫ 0 π /12 sec 2 ( 2 x + 6 π ) d x 8 1 [ tan ( 2 x + 6 π ) ] 0 π /12 8 1 ( tan 3 π − tan 6 π ) 8 1 ( 3 − 3 1 ) . Therefore I = 3 / 12 \boxed{I=\sqrt3/12} I = 3 /12
