题目
Problem
(a) Express 3 sin x + 2 2 cos ( x + 1 4 π ) 3\sin x+2\sqrt2\cos\big(x+\frac14\pi\big) 3 sin x + 2 2 cos ( x + 4 1 π ) R sin ( x + α ) R\sin(x+\alpha) R sin ( 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 3 3
(4)
(b) Hence solve the equation
6 sin 1 2 θ + 4 2 cos ( 1 2 θ + 1 4 π ) = 3 6\sin\frac12\theta+4\sqrt2\cos\bigg(\frac12\theta+\frac14\pi\bigg)=3 6 sin 2 1 θ + 4 2 cos ( 2 1 θ + 4 1 π ) = 3 for − 4 π < θ < 4 π -4\pi<\theta<4\pi − 4 π < θ < 4 π
(5)
题目中文翻译
(a) 将 3 sin x + 2 2 cos ( x + 1 4 π ) 3\sin x+2\sqrt2\cos\big(x+\frac14\pi\big) 3 sin x + 2 2 cos ( x + 4 1 π ) R sin ( x + α ) R\sin(x+\alpha) R sin ( 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 3 3
(4)
(b) 由此解方程
6 sin 1 2 θ + 4 2 cos ( 1 2 θ + 1 4 π ) = 3 , 6\sin\frac12\theta+4\sqrt2\cos\bigg(\frac12\theta+\frac14\pi\bigg)=3, 6 sin 2 1 θ + 4 2 cos ( 2 1 θ + 4 1 π ) = 3 , 其中 − 4 π < θ < 4 π -4\pi<\theta<4\pi − 4 π < θ < 4 π
(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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先展开平移角的余弦并合并同类项，再与 R sin ( x + α ) R\sin(x+\alpha) R sin ( x + α )
答题过程
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Using the addition formula,
3 sin x + 2 2 cos ( x + π 4 ) = 3 sin x + 2 2 ( cos x 2 − sin x 2 ) = sin x + 2 cos x . \begin{align*}
&\,3\sin x+2\sqrt2\cos\bigg(x+\frac\pi4\bigg)\\[4mm]
=&\,3\sin x+2\sqrt2\bigg(
\frac{\cos x}{\sqrt2}-\frac{\sin x}{\sqrt2}\bigg)\\[4mm]
=&\,\sin x+2\cos x.
\end{align*} = = 3 sin x + 2 2 cos ( x + 4 π ) 3 sin x + 2 2 ( 2 cos x − 2 sin x ) sin x + 2 cos x . Also,
R sin ( x + α ) = R cos α sin x + R sin α cos x . R\sin(x+\alpha)=R\cos\alpha\sin x+R\sin\alpha\cos x. R sin ( x + α ) = R cos α sin x + R sin α cos x . Thus R cos α = 1 R\cos\alpha=1 R cos α = 1 R sin α = 2 R\sin\alpha=2 R sin α = 2
R = 1 2 + 2 2 = 5 , α = tan − 1 2 = 1.107148 … . \begin{align*}
R=&\,\sqrt{1^2+2^2}=\sqrt5,\\[4mm]
\alpha=&\,\tan^{-1}2=1.107148\ldots.
\end{align*} R = α = 1 2 + 2 2 = 5 , tan − 1 2 = 1.107148 … . Therefore
R = 5 , α = 1.107 radians \boxed{R=\sqrt5,\qquad\alpha=1.107\text{ radians}} R = 5 , α = 1.107 radians with α \alpha α
(b)
解法一
思路
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原式是(a)表达式的两倍，并把 x x x θ / 2 \theta/2 θ /2
答题过程
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Using part (a), with the unrounded value α = tan − 1 2 \alpha=\tan^{-1}2 α = tan − 1 2
2 5 sin ( θ 2 + α ) = 3. 2\sqrt5\sin\bigg(\frac{\theta}{2}+\alpha\bigg)=3. 2 5 sin ( 2 θ + α ) = 3. Let ϕ = θ / 2 + α \phi=\theta/2+\alpha ϕ = θ /2 + α
β = sin − 1 ( 3 2 5 ) . \beta=\sin^{-1}\bigg(\frac3{2\sqrt5}\bigg). β = sin − 1 ( 2 5 3 ) . Since − 4 π < θ < 4 π -4\pi<\theta<4\pi − 4 π < θ < 4 π
− 2 π + α < ϕ < 2 π + α . -2\pi+\alpha<\phi<2\pi+\alpha. − 2 π + α < ϕ < 2 π + α . The solutions in this interval are
ϕ = π − β − 2 π , β , π − β , β + 2 π . \phi=\pi-\beta-2\pi,\quad
\beta,\quad \pi-\beta,\quad \beta+2\pi. ϕ = π − β − 2 π , β , π − β , β + 2 π . Using θ = 2 ( ϕ − α ) \theta=2(\phi-\alpha) θ = 2 ( ϕ − α )
θ = − 9.96811 … , − 0.743668 … , 2.598258 … , 11.82270 … . \theta=-9.96811\ldots,\ -0.743668\ldots,\
2.598258\ldots,\ 11.82270\ldots. θ = − 9.96811 … , − 0.743668 … , 2.598258 … , 11.82270 … . Hence
θ = − 9.97 , − 0.744 , 2.60 , 11.8 \boxed{\theta=-9.97,\ -0.744,\ 2.60,\ 11.8} θ = − 9.97 , − 0.744 , 2.60 , 11.8 in radians, to 3 significant figures.
