题目
Problem
The diagram shows the curve y = sin x cos 2 x y=\sin x\cos2x y = sin x cos 2 x 0 ≤ x ≤ π 0\le x\le\pi 0 ≤ x ≤ π M M M x = a x=a x = a x x x R R R
(a) Find the value of a a a 2 2 2
(5)
(b) Find the exact area of the region R R R
(4)
题目中文翻译
图中显示了曲线 y = sin x cos 2 x y=\sin x\cos2x y = sin x cos 2 x 0 ≤ x ≤ π 0\le x\le\pi 0 ≤ x ≤ π M M M x x x a a a x x x R R R
(a) 求 a a a 2 2 2
(5)
(b) 求区域 R R R
(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
用乘积法则求导，再用二倍角公式整理因子。图中的 M M M
答题过程
展开
For y = sin x cos 2 x y=\sin x\cos2x y = sin x cos 2 x
d y d x = cos x cos 2 x − 2 sin x sin 2 x . \frac{\mathrm{d}y}{\mathrm{d}x}
=\cos x\cos2x-2\sin x\sin2x. d x d y = cos x cos 2 x − 2 sin x sin 2 x . Using double-angle formulae,
d y d x = cos x ( 1 − 2 sin 2 x ) − 4 sin 2 x cos x = cos x ( 1 − 6 sin 2 x ) . \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x}
=&\,\cos x(1-2\sin^2x)-4\sin^2x\cos x\\[4mm]
=&\,\cos x(1-6\sin^2x).
\end{align*} d x d y = = cos x ( 1 − 2 sin 2 x ) − 4 sin 2 x cos x cos x ( 1 − 6 sin 2 x ) . For the left-hand maximum M M M 0 < a < π / 4 0<a<\pi/4 0 < a < π /4 cos a > 0 \cos a>0 cos a > 0
1 − 6 sin 2 a = 0 , sin a = 1 6 . 1-6\sin^2a=0,\qquad \sin a=\frac1{\sqrt6}. 1 − 6 sin 2 a = 0 , sin a = 6 1 . The derivative changes from positive to negative there. Therefore
a = sin − 1 1 6 ≈ 0.42 a=\sin^{-1}\frac1{\sqrt6}\approx\boxed{0.42} a = sin − 1 6 1 ≈ 0.42 to 2 decimal places.
解法二
思路
展开
先用二倍角公式把原函数改成 sin x \sin x sin x
答题过程
展开
Using cos 2 x = 1 − 2 sin 2 x \cos2x=1-2\sin^2x cos 2 x = 1 − 2 sin 2 x
y = sin x − 2 sin 3 x . y=\sin x-2\sin^3x. y = sin x − 2 sin 3 x . Differentiating,
d y d x = cos x − 6 sin 2 x cos x = cos x ( 1 − 6 sin 2 x ) . \frac{\mathrm{d}y}{\mathrm{d}x}
=\cos x-6\sin^2x\cos x
=\cos x(1-6\sin^2x). d x d y = cos x − 6 sin 2 x cos x = cos x ( 1 − 6 sin 2 x ) . At the left-hand maximum M M M 0 < a < π / 4 0<a<\pi/4 0 < a < π /4
sin a = 1 6 . \sin a=\frac1{\sqrt6}. sin a = 6 1 . The factor 1 − 6 sin 2 x 1-6\sin^2x 1 − 6 sin 2 x
a = sin − 1 1 6 ≈ 0.42 a=\sin^{-1}\frac1{\sqrt6}\approx\boxed{0.42} a = sin − 1 6 1 ≈ 0.42 to 2 decimal places.
(b)
解法一
思路
展开
阴影在 π / 4 \pi/4 π /4 3 π / 4 3\pi/4 3 π /4 cos 2 x \cos2x cos 2 x 2 cos 2 x − 1 2\cos^2x-1 2 cos 2 x − 1
答题过程
展开
The shaded region is below the x x x x = π / 4 x=\pi/4 x = π /4 x = 3 π / 4 x=3\pi/4 x = 3 π /4
A = − ∫ π / 4 3 π / 4 sin x cos 2 x d x . A=-\int_{\pi/4}^{3\pi/4}\sin x\cos2x\,\mathrm{d}x. A = − ∫ π /4 3 π /4 sin x cos 2 x d x . Using cos 2 x = 2 cos 2 x − 1 \cos2x=2\cos^2x-1 cos 2 x = 2 cos 2 x − 1 u = cos x u=\cos x u = cos x d u = − sin x d x \mathrm{d}u=-\sin x\,\mathrm{d}x d u = − sin x d x
∫ sin x cos 2 x d x = − ∫ ( 2 u 2 − 1 ) d u = − 2 3 u 3 + u + C = − 2 3 cos 3 x + cos x + C , \begin{align*}
\int\sin x\cos2x\,\mathrm{d}x
=&\,-\int(2u^2-1)\,\mathrm{d}u\\[4mm]
=&\,-\frac23u^3+u+C\\[4mm]
=&\,-\frac23\cos^3x+\cos x+C,
\end{align*} ∫ sin x cos 2 x d x = = = − ∫ ( 2 u 2 − 1 ) d u − 3 2 u 3 + u + C − 3 2 cos 3 x + cos x + C , Consequently,
A = [ 2 3 cos 3 x − cos x ] π / 4 3 π / 4 . A=\bigg[\frac23\cos^3x-\cos x\bigg]_{\pi/4}^{3\pi/4}. A = [ 3 2 cos 3 x − cos x ] π /4 3 π /4 . At π / 4 \pi/4 π /4 − 2 / 3 -\sqrt2/3 − 2 /3 3 π / 4 3\pi/4 3 π /4 2 / 3 \sqrt2/3 2 /3
A = 2 2 3 . \boxed{A=\frac{2\sqrt2}{3}}. A = 3 2 2 .
解法二
思路
展开
用积化和差将乘积改成两个正弦函数，逐项积分，仍须注意阴影对应负积分。
答题过程
展开
By the product-to-sum formula,
sin x cos 2 x = 1 2 ( sin 3 x − sin x ) . \sin x\cos2x=\frac12(\sin3x-\sin x). sin x cos 2 x = 2 1 ( sin 3 x − sin x ) . Thus the area is
A = − 1 2 ∫ π / 4 3 π / 4 ( sin 3 x − sin x ) d x = [ 1 6 cos 3 x − 1 2 cos x ] π / 4 3 π / 4 . \begin{align*}
A
=&\,-\frac12\int_{\pi/4}^{3\pi/4}(\sin3x-\sin x)\,\mathrm{d}x\\[4mm]
=&\,\bigg[\frac16\cos3x-\frac12\cos x\bigg]_{\pi/4}^{3\pi/4}.
\end{align*} A = = − 2 1 ∫ π /4 3 π /4 ( sin 3 x − sin x ) d x [ 6 1 cos 3 x − 2 1 cos x ] π /4 3 π /4 . The upper value is 2 / 3 \sqrt2/3 2 /3 − 2 / 3 -\sqrt2/3 − 2 /3
A = 2 2 3 . \boxed{A=\frac{2\sqrt2}{3}}. A = 3 2 2 . 
