题目
Problem
f ( θ ) = 5 cos θ − 4 sin θ θ ∈ R f(\theta)=5\cos\theta-4\sin\theta\qquad \theta\in\mathbb{R} f ( θ ) = 5 cos θ − 4 sin θ θ ∈ R (a) Express f ( θ ) f(\theta) f ( θ ) R cos ( θ + α ) R\cos(\theta+\alpha) R cos ( θ + α ) R R R α \alpha α R > 0 R>0 R > 0 0 < α < π 2 0<\alpha<\dfrac{\pi}{2} 0 < α < 2 π
Give the exact value of R R R α \alpha α
(3)
The curve with equation y = cos θ y=\cos\theta y = cos θ y = f ( θ ) y=f(\theta) y = f ( θ )
Given that the first transformation is a stretch and the second a translation,
(b) (i) describe fully the transformation that is a stretch,
(ii) describe fully the transformation that is a translation.
(2)
Given
g ( θ ) = 90 4 + ( f ( θ ) ) 2 θ ∈ R g(\theta)=\frac{90}{4+(f(\theta))^2}\qquad \theta\in\mathbb{R} g ( θ ) = 4 + ( f ( θ ) ) 2 90 θ ∈ R (c) find the range of g g g
(2)
题目中文翻译
f ( θ ) = 5 cos θ − 4 sin θ θ ∈ R f(\theta)=5\cos\theta-4\sin\theta\qquad \theta\in\mathbb{R} f ( θ ) = 5 cos θ − 4 sin θ θ ∈ R (a) 将 f ( θ ) f(\theta) f ( θ ) R cos ( θ + α ) R\cos(\theta+\alpha) R cos ( θ + α ) R R R α \alpha α R > 0 , 0 < α < π 2 R>0,\ 0<\alpha<\dfrac{\pi}{2} R > 0 , 0 < α < 2 π
写出 R R R α \alpha α
曲线 y = cos θ y=\cos\theta y = cos θ y = f ( θ ) y=f(\theta) y = f ( θ )
已知第一次变换是伸缩，第二次变换是平移，
(b) (i) 完整描述这个伸缩变换；
(ii) 完整描述这个平移变换。
已知
g ( θ ) = 90 4 + ( f ( θ ) ) 2 θ ∈ R g(\theta)=\frac{90}{4+(f(\theta))^2}\qquad \theta\in\mathbb{R} g ( θ ) = 4 + ( f ( θ ) ) 2 90 θ ∈ R (c) 求 g g g
解答
(a)
We want
5 cos θ − 4 sin θ = R cos ( θ + α ) 5\cos\theta-4\sin\theta=R\cos(\theta+\alpha) 5 cos θ − 4 sin θ = R cos ( θ + α ) Expand the right hand side:
R cos ( θ + α ) = R cos θ cos α − R sin θ sin α R\cos(\theta+\alpha)=R\cos\theta\cos\alpha-R\sin\theta\sin\alpha R cos ( θ + α ) = R cos θ cos α − R sin θ sin α Compare coefficients of cos θ \cos\theta cos θ sin θ \sin\theta sin θ
R cos α = 5 , R sin α = 4 R\cos\alpha=5,\qquad R\sin\alpha=4 R cos α = 5 , R sin α = 4 Therefore
R 2 = 5 2 + 4 2 = 41 R^2=5^2+4^2=41 R 2 = 5 2 + 4 2 = 41 so
R = 41 R=\sqrt{41} R = 41 Also,
tan α = 4 5 \tan\alpha=\frac{4}{5} tan α = 5 4 Since 0 < α < π 2 0<\alpha<\dfrac{\pi}{2} 0 < α < 2 π
α = arctan 4 5 = 0.675 rad to 3 d.p. \alpha=\arctan\frac45=0.675\text{ rad}\quad \text{to 3 d.p.} α = arctan 5 4 = 0.675 rad to 3 d.p. So
f ( θ ) = 41 cos ( θ + 0.675 ) \boxed{f(\theta)=\sqrt{41}\cos(\theta+0.675)} f ( θ ) = 41 cos ( θ + 0.675 ) where the exact amplitude is
R = 41 \boxed{R=\sqrt{41}} R = 41 (b)
From part (a),
f ( θ ) = 41 cos ( θ + α ) f(\theta)=\sqrt{41}\cos(\theta+\alpha) f ( θ ) = 41 cos ( θ + α ) where
α = arctan 4 5 \alpha=\arctan\frac45 α = arctan 5 4 Starting from y = cos θ y=\cos\theta y = cos θ
(i)
Multiplying the output by 41 \sqrt{41} 41
y = 41 cos θ y=\sqrt{41}\cos\theta y = 41 cos θ So the stretch is:
a stretch parallel to the y -axis with scale factor 41 \boxed{\text{a stretch parallel to the }y\text{-axis with scale factor }\sqrt{41}} a stretch parallel to the y -axis with scale factor 41 (ii)
Replacing θ \theta θ θ + α \theta+\alpha θ + α
y = 41 cos ( θ + α ) y=\sqrt{41}\cos(\theta+\alpha) y = 41 cos ( θ + α ) This is a horizontal translation by α \alpha α
So the translation is:
a translation by − α in the θ -direction \boxed{\text{a translation by }-\alpha\text{ in the }\theta\text{-direction}} a translation by − α in the θ -direction where
α = arctan 4 5 \alpha=\arctan\frac45 α = arctan 5 4 Equivalently, the translation vector is
( − α 0 ) \boxed{\begin{pmatrix}-\alpha\\0\end{pmatrix}} ( − α 0 ) (c)
From part (a),
f ( θ ) = 41 cos ( θ + α ) f(\theta)=\sqrt{41}\cos(\theta+\alpha) f ( θ ) = 41 cos ( θ + α ) Since
− 1 ≤ cos ( θ + α ) ≤ 1 -1\leq \cos(\theta+\alpha)\leq 1 − 1 ≤ cos ( θ + α ) ≤ 1 we have
− 41 ≤ f ( θ ) ≤ 41 -\sqrt{41}\leq f(\theta)\leq \sqrt{41} − 41 ≤ f ( θ ) ≤ 41 So
0 ≤ ( f ( θ ) ) 2 ≤ 41 0\leq (f(\theta))^2\leq 41 0 ≤ ( f ( θ ) ) 2 ≤ 41 Then
4 ≤ 4 + ( f ( θ ) ) 2 ≤ 45 4\leq 4+(f(\theta))^2\leq 45 4 ≤ 4 + ( f ( θ ) ) 2 ≤ 45 Now
g ( θ ) = 90 4 + ( f ( θ ) ) 2 g(\theta)=\frac{90}{4+(f(\theta))^2} g ( θ ) = 4 + ( f ( θ ) ) 2 90 Because the denominator is positive, the largest value of g g g
Therefore
90 45 ≤ g ( θ ) ≤ 90 4 \frac{90}{45}\leq g(\theta)\leq \frac{90}{4} 45 90 ≤ g ( θ ) ≤ 4 90 So the range is
2 ≤ g ( θ ) ≤ 45 2 \boxed{2\leq g(\theta)\leq \frac{45}{2}} 2 ≤ g ( θ ) ≤ 2 45 