题目
Problem
(a) Prove the identity sin 4 x ≡ 4 sin x ( 2 cos 3 x − cos x ) \sin4x\equiv4\sin x(2\cos^3x-\cos x) sin 4 x ≡ 4 sin x ( 2 cos 3 x − cos x ) .
(3)
(b) Hence find the exact value of ∫ 0 1 4 π cos 3 x sin 4 x d x \displaystyle\int_0^{\frac14\pi}\cos^3x\sin4x\,\mathrm{d}x ∫ 0 4 1 π cos 3 x sin 4 x d x .
(5)
题目中文翻译
(a) 证明恒等式 sin 4 x ≡ 4 sin x ( 2 cos 3 x − cos x ) \sin4x\equiv4\sin x(2\cos^3x-\cos x) sin 4 x ≡ 4 sin x ( 2 cos 3 x − cos x ) 。
(3)
(b) 由此求 ∫ 0 1 4 π cos 3 x sin 4 x d x \displaystyle\int_0^{\frac14\pi}\cos^3x\sin4x\,\mathrm{d}x ∫ 0 4 1 π cos 3 x sin 4 x d x 的精确值。
(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
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连续用两次二倍角公式,再把 cos 2 x \cos2x cos 2 x 写成仅含余弦的形式。
答题过程
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Using the double-angle identities,
sin 4 x ≡ 2 sin 2 x cos 2 x ≡ 2 ( 2 sin x cos x ) ( 2 cos 2 x − 1 ) ≡ 4 sin x ( 2 cos 3 x − cos x ) , \begin{align*}
\sin4x
\equiv&\,2\sin2x\cos2x\\
\equiv&\,2(2\sin x\cos x)(2\cos^2x-1)\\
\equiv&\,4\sin x(2\cos^3x-\cos x),
\end{align*} sin 4 x ≡ ≡ ≡ 2 sin 2 x cos 2 x 2 ( 2 sin x cos x ) ( 2 cos 2 x − 1 ) 4 sin x ( 2 cos 3 x − cos x ) ,
as required.
(b)
解法一
思路
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承接(a),让被积函数出现 sin x \sin x sin x 乘余弦的幂,然后用链式法则的逆过程积分。代入端点时注意根号的幂次。
答题过程
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Using the identity from (a),
cos 3 x sin 4 x = 8 sin x cos 6 x − 4 sin x cos 4 x . \cos^3x\sin4x
=8\sin x\cos^6x-4\sin x\cos^4x. cos 3 x sin 4 x = 8 sin x cos 6 x − 4 sin x cos 4 x .
With u = cos x u=\cos x u = cos x and d u = − sin x d x \mathrm{d}u=-\sin x\,\mathrm{d}x d u = − sin x d x , the limits become u = 1 u=1 u = 1 and u = 2 / 2 u=\sqrt2/2 u = 2 /2 . Thus
I = ∫ 2 / 2 1 ( 8 u 6 − 4 u 4 ) d u . I=\int_{\sqrt2/2}^{1}(8u^6-4u^4)\,\mathrm{d}u. I = ∫ 2 /2 1 ( 8 u 6 − 4 u 4 ) d u .
Equivalently, integrating in terms of cos x \cos x cos x ,
I = [ − 8 7 cos 7 x + 4 5 cos 5 x ] 0 π / 4 . I=\bigg[-\frac87\cos^7x+\frac45\cos^5x\bigg]_0^{\pi/4}. I = [ − 7 8 cos 7 x + 5 4 cos 5 x ] 0 π /4 .
At x = π / 4 x=\pi/4 x = π /4 ,
− 8 7 ( 2 2 ) 7 + 4 5 ( 2 2 ) 5 = − 2 14 + 2 10 = 2 35 . \begin{align*}
&\,-\frac87\bigg(\frac{\sqrt2}{2}\bigg)^7
+\frac45\bigg(\frac{\sqrt2}{2}\bigg)^5\\[4mm]
=&\,-\frac{\sqrt2}{14}+\frac{\sqrt2}{10}\\[4mm]
=&\,\frac{\sqrt2}{35}.
\end{align*} = = − 7 8 ( 2 2 ) 7 + 5 4 ( 2 2 ) 5 − 14 2 + 10 2 35 2 .
At x = 0 x=0 x = 0 , the bracket is − 8 / 7 + 4 / 5 = − 12 / 35 -8/7+4/5=-12/35 − 8/7 + 4/5 = − 12/35 . Therefore
I = 12 + 2 35 . \boxed{I=\frac{12+\sqrt2}{35}}. I = 35 12 + 2 .