题目
Problem
(a) (i) By writing cos ( 1 12 π ) \cos\big(\frac1{12}\pi\big) cos ( 12 1 π ) cos ( 1 4 π − 1 6 π ) \cos\big(\frac14\pi-\frac16\pi\big) cos ( 4 1 π − 6 1 π ) cos ( 1 12 π ) = 1 4 ( 6 + 2 ) \cos\big(\frac1{12}\pi\big)=\frac14(\sqrt6+\sqrt2) cos ( 12 1 π ) = 4 1 ( 6 + 2 )
(1)
(ii) Show also that sin ( 1 12 π ) = 1 4 ( 6 − 2 ) \sin\big(\frac1{12}\pi\big)=\frac14(\sqrt6-\sqrt2) sin ( 12 1 π ) = 4 1 ( 6 − 2 )
(1)
The matrix M \mathbf{M} M
M = 1 4 ( 6 + 2 2 − 6 6 − 2 6 + 2 ) ( 1 0 0 − 1 ) . \mathbf{M}=\frac14\begin{pmatrix}\sqrt6+\sqrt2&\sqrt2-\sqrt6\\\sqrt6-\sqrt2&\sqrt6+\sqrt2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}. M = 4 1 ( 6 + 2 6 − 2 2 − 6 6 + 2 ) ( 1 0 0 − 1 ) . (b) The matrix M \mathbf{M} M x x x y y y
Give full details of each transformation, and make clear the order in which they are applied.
(4)
(c) Write M − 1 \mathbf{M}^{-1} M − 1 I \mathbf{I} I
(2)
(d) Given that y = m x y=mx y = m x M \mathbf{M} M
m 2 sin ( 1 12 π ) + 2 m cos ( 1 12 π ) − sin ( 1 12 π ) = 0 m^2\sin\big(\tfrac1{12}\pi\big)+2m\cos\big(\tfrac1{12}\pi\big)-\sin\big(\tfrac1{12}\pi\big)=0 m 2 sin ( 12 1 π ) + 2 m cos ( 12 1 π ) − sin ( 12 1 π ) = 0 and find the values of m m m a cot ( 1 12 π ) + b cosec ( 1 12 π ) a\cot\big(\frac1{12}\pi\big)+b\operatorname{cosec}\big(\frac1{12}\pi\big) a cot ( 12 1 π ) + b cosec ( 12 1 π ) a a a b b b
(6)
题目中文翻译
(a) (i) 将 cos ( 1 12 π ) \cos\big(\frac1{12}\pi\big) cos ( 12 1 π ) cos ( 1 4 π − 1 6 π ) \cos\big(\frac14\pi-\frac16\pi\big) cos ( 4 1 π − 6 1 π ) cos ( 1 12 π ) = 1 4 ( 6 + 2 ) \cos\big(\frac1{12}\pi\big)=\frac14(\sqrt6+\sqrt2) cos ( 12 1 π ) = 4 1 ( 6 + 2 )
(1)
(ii) 再证明 sin ( 1 12 π ) = 1 4 ( 6 − 2 ) \sin\big(\frac1{12}\pi\big)=\frac14(\sqrt6-\sqrt2) sin ( 12 1 π ) = 4 1 ( 6 − 2 )
(1)
矩阵 M \mathbf{M} M
M = 1 4 ( 6 + 2 2 − 6 6 − 2 6 + 2 ) ( 1 0 0 − 1 ) . \mathbf{M}=\frac14\begin{pmatrix}\sqrt6+\sqrt2&\sqrt2-\sqrt6\\\sqrt6-\sqrt2&\sqrt6+\sqrt2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}. M = 4 1 ( 6 + 2 6 − 2 2 − 6 6 + 2 ) ( 1 0 0 − 1 ) . (b) 矩阵 M \mathbf{M} M x x x y y y
完整描述每个变换，并明确说明它们的作用顺序。
(4)
(c) 将 M − 1 \mathbf{M}^{-1} M − 1 I \mathbf{I} I
(2)
(d) 已知 y = m x y=mx y = m x M \mathbf{M} M
m 2 sin ( 1 12 π ) + 2 m cos ( 1 12 π ) − sin ( 1 12 π ) = 0 , m^2\sin\big(\tfrac1{12}\pi\big)+2m\cos\big(\tfrac1{12}\pi\big)-\sin\big(\tfrac1{12}\pi\big)=0, m 2 sin ( 12 1 π ) + 2 m cos ( 12 1 π ) − sin ( 12 1 π ) = 0 , 并求 m m m a cot ( 1 12 π ) + b cosec ( 1 12 π ) a\cot\big(\frac1{12}\pi\big)+b\operatorname{cosec}\big(\frac1{12}\pi\big) a cot ( 12 1 π ) + b cosec ( 12 1 π ) a a a b b b
(6)
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)(i)
解法一
思路
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按指定角差使用余弦差角公式，代入两个特殊角的精确值。
答题过程
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Using the angle-difference formula,
cos π 12 = cos π 4 cos π 6 + sin π 4 sin π 6 = 2 2 3 2 + 2 2 1 2 = 6 + 2 4 , \begin{align*}
\cos\frac\pi{12}
=&\,\cos\frac\pi4\cos\frac\pi6+\sin\frac\pi4\sin\frac\pi6 \\[4mm]
=&\,\frac{\sqrt2}{2}\frac{\sqrt3}{2}
+\frac{\sqrt2}{2}\frac12 \\[4mm]
=&\,\boxed{\frac{\sqrt6+\sqrt2}{4}},
\end{align*} cos 12 π = = = cos 4 π cos 6 π + sin 4 π sin 6 π 2 2 2 3 + 2 2 2 1 4 6 + 2 , as required.
(a)(ii)
解法一
思路
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同样用正弦差角公式，注意第二项为减号。
答题过程
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Similarly,
sin π 12 = sin π 4 cos π 6 − cos π 4 sin π 6 = 2 2 3 2 − 2 2 1 2 = 6 − 2 4 , \begin{align*}
\sin\frac\pi{12}
=&\,\sin\frac\pi4\cos\frac\pi6-\cos\frac\pi4\sin\frac\pi6 \\[4mm]
=&\,\frac{\sqrt2}{2}\frac{\sqrt3}{2}
-\frac{\sqrt2}{2}\frac12 \\[4mm]
=&\,\boxed{\frac{\sqrt6-\sqrt2}{4}},
\end{align*} sin 12 π = = = sin 4 π cos 6 π − cos 4 π sin 6 π 2 2 2 3 − 2 2 2 1 4 6 − 2 , as required.
(b)
解法一
思路
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用(a)识别左边为旋转矩阵；右边的反射先做，左边的旋转后做。
答题过程
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By (a), the left matrix is
R = ( cos ( π / 12 ) − sin ( π / 12 ) sin ( π / 12 ) cos ( π / 12 ) ) . \mathbf R=\begin{pmatrix}
\cos(\pi/12)&-\sin(\pi/12)\\
\sin(\pi/12)&\cos(\pi/12)
\end{pmatrix}. R = ( cos ( π /12 ) sin ( π /12 ) − sin ( π /12 ) cos ( π /12 ) ) . The right matrix S = ( 1 0 0 − 1 ) \mathbf S=\begin{pmatrix}1&0\\0&-1\end{pmatrix} S = ( 1 0 0 − 1 ) x x x M = R S \mathbf M=\mathbf R\mathbf S M = RS
Reflection in the x x x
Anticlockwise rotation through π / 12 \pi/12 π /12
(c)
解法一
思路
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复合变换取逆要反转次序；反射是自身的逆，旋转取相反角度。
答题过程
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Since M = R S \mathbf M=\mathbf R\mathbf S M = RS
M − 1 = S − 1 R − 1 = S R − 1 . \mathbf M^{-1}=\mathbf S^{-1}\mathbf R^{-1}=\mathbf S\mathbf R^{-1}. M − 1 = S − 1 R − 1 = S R − 1 . Thus the required product is
M − 1 = ( 1 0 0 − 1 ) ( cos ( π / 12 ) sin ( π / 12 ) − sin ( π / 12 ) cos ( π / 12 ) ) . \boxed{\mathbf M^{-1}=
\begin{pmatrix}1&0\\0&-1\end{pmatrix}
\begin{pmatrix}
\cos(\pi/12)&\sin(\pi/12)\\
-\sin(\pi/12)&\cos(\pi/12)
\end{pmatrix}}. M − 1 = ( 1 0 0 − 1 ) ( cos ( π /12 ) − sin ( π /12 ) sin ( π /12 ) cos ( π /12 ) ) . Neither matrix is the identity.
(d)
解法一
思路
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先将M乘开，再设y=mx的点变换后仍满足Y=mX，得到指定二次式。求根时用正弦余弦平方和化简。
答题过程
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Put ϕ = π / 12 \phi=\pi/12 ϕ = π /12
M = ( cos ϕ sin ϕ sin ϕ − cos ϕ ) . \mathbf M=\begin{pmatrix}\cos\phi&\sin\phi\\\sin\phi&-\cos\phi\end{pmatrix}. M = ( cos ϕ sin ϕ sin ϕ − cos ϕ ) . The image of ( x , m x ) (x,mx) ( x , m x )
X = x cos ϕ + m x sin ϕ , Y = x sin ϕ − m x cos ϕ . \begin{align*}
X=&\,x\cos\phi+mx\sin\phi,\\[4mm]
Y=&\,x\sin\phi-mx\cos\phi.
\end{align*} X = Y = x cos ϕ + m x sin ϕ , x sin ϕ − m x cos ϕ . For an invariant line, Y = m X Y=mX Y = m X
sin ϕ − m cos ϕ = m cos ϕ + m 2 sin ϕ . \sin\phi-m\cos\phi=m\cos\phi+m^2\sin\phi. sin ϕ − m cos ϕ = m cos ϕ + m 2 sin ϕ . Hence
m 2 sin ϕ + 2 m cos ϕ − sin ϕ = 0 , \boxed{m^2\sin\phi+2m\cos\phi-\sin\phi=0}, m 2 sin ϕ + 2 m cos ϕ − sin ϕ = 0 , as required. The quadratic formula gives
2 m sin ϕ = − 2 cos ϕ ± 4 cos 2 ϕ + 4 sin 2 ϕ . 2m\sin\phi=-2\cos\phi\pm\sqrt{4\cos^2\phi+4\sin^2\phi}. 2 m sin ϕ = − 2 cos ϕ ± 4 cos 2 ϕ + 4 sin 2 ϕ . Using cos 2 ϕ + sin 2 ϕ = 1 \cos^2\phi+\sin^2\phi=1 cos 2 ϕ + sin 2 ϕ = 1 sin ϕ > 0 \sin\phi>0 sin ϕ > 0
m = − cos ϕ ± 1 sin ϕ = − cot π 12 ± cosec π 12 . \begin{align*}
m=&\,\frac{-\cos\phi\pm1}{\sin\phi} \\[4mm]
=&\,\boxed{-\cot\frac\pi{12}\pm\operatorname{cosec}\frac\pi{12}}.
\end{align*} m = = sin ϕ − cos ϕ ± 1 − cot 12 π ± cosec 12 π . Thus a = − 1 a=-1 a = − 1 b = 1 b=1 b = 1 − 1 -1 − 1
