Question 1, Exercise 9.1

Solutions of Question 1 of Exercise 9.1 of Unit 09: Trigonometric Functions. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Find the maximum and minimum values of the trigonometric function: $y=2-2 \operatorname{Cos} \theta$

Solution.

We know \begin{align*} -1 \leq \operatorname{Cos} \theta \leq 1 \end{align*} Multiplying with $-2$ \begin{align*} & 2 \geq -2 \operatorname{Cos} \theta \geq -2 \end{align*} Adding $2$ \begin{align*} & 4 \geq 2-2 \operatorname{Cos} \theta \geq 0 \\ \implies & 0 \leq 2-2 \operatorname{Cos} \theta \leq 4 \\ \end{align*} Hence maximum value $(M) = 4$
and minimum value $(m) = 0$.

Alternative Method:

Given: $$y=2-2 \operatorname{Cos} \theta$$ Consider $$y=a+b \operatorname{Cos} \theta$$ Comparing the coeffients: $$a=2, \quad b=-2$$ \begin{align*} \text{Maximum value (M)} & = a+|b|\\ & = 2+|-2| \\ & = 2+2 = 4 \end{align*} \begin{align*} \text{Minimum value (m)} & = a-|b|\\ & = 2-|-2| \\ & = 2-2 = 0 \end{align*} Hence Maximum value (M) = 4
and mimimum value (m) = 0.

Find the maximum and minimum values of the trigonometric function: $y=\dfrac{2}{3}-\dfrac{1}{2} \operatorname{Sin} \theta$

Solution.

We know \begin{align*} -1 \leq \operatorname{Sin} \theta \leq 1 \end{align*} Multiplying with $-\dfrac{1}{2}$ \begin{align*} & \dfrac{1}{2} \geq -\dfrac{1}{2} \operatorname{Sin} \theta \geq -\dfrac{1}{2} \end{align*} Adding $\dfrac{2}{3}$ \begin{align*} & \dfrac{2}{3}+\dfrac{1}{2} \geq \dfrac{2}{3}-\dfrac{1}{2} \operatorname{Sin} \theta \geq \dfrac{2}{3}-\dfrac{1}{2} \\ \implies & \dfrac{7}{6} \geq \dfrac{2}{3}-\dfrac{1}{2} \operatorname{Sin} \theta \geq \dfrac{1}{6} \\ \implies & \dfrac{1}{6} \leq \dfrac{2}{3}-\dfrac{1}{2} \operatorname{Sin} \theta\leq \dfrac{7}{6} \\ \end{align*} Hence maximum value $(M) = \dfrac{7}{6}$
and minimum value $(m) = \dfrac{1}{6}$.

Alternative Method:

Given: $$y=\dfrac{2}{3}-\dfrac{1}{2} \operatorname{Sin} \theta \theta$$ Consider $$y=a+b \operatorname{Sin} \theta$$ Comparing the coeffients: $$a=\dfrac{2}{3}, \quad b=-\dfrac{1}{2}$$ \begin{align*} \text{Maximum value (M)} & = a+|b|\\ & = \dfrac{2}{3}+\left|-\dfrac{1}{2} \right| \\ & = \dfrac{2}{3}+\dfrac{1}{2} = \dfrac{7}{6} \end{align*} \begin{align*} \text{Minimum value (m)} & = a-|b|\\ & = \dfrac{2}{3}-\left|-\dfrac{1}{2} \right| \\ & = \dfrac{2}{3}-\dfrac{1}{2} = \dfrac{1}{6} \end{align*} Hence Maximum value $(M) = \dfrac{7}{6}$
and mimimum value $(m) = \dfrac{1}{6}$.

Find the maximum and minimum values of the trigonometric function: $y=\dfrac{1}{5}-2 \operatorname{Sin}(3 \theta-7)$

Solution.

We know \begin{align*} -1 \leq \operatorname{Sin}(3 \theta-7) \leq 1 \end{align*} Multiplying with $-2$ \begin{align*} & 2 \geq -2 \operatorname{Sin}(3 \theta-7) \geq -2 \end{align*} Adding $2$ \begin{align*} & \dfrac{1}{5}+2 \geq \dfrac{1}{5}-2 \operatorname{Sin}(3 \theta-7) \geq \dfrac{1}{5}-2 \\ \implies & \dfrac{11}{5} \geq \dfrac{1}{5}-2 \operatorname{Sin}(3 \theta-7) \geq -\dfrac{9}{5} \\ \implies & -\dfrac{9}{5} \leq \dfrac{1}{5}-2 \operatorname{Sin}(3 \theta-7) \leq \dfrac{11}{5} \\ \end{align*} Hence maximum value $(M) = \dfrac{11}{5}$
and minimum value $(m) = -\dfrac{9}{5}$.

Find the maximum and minimum values of the trigonometric function: $\mathrm{y}=7+\frac{3}{5} \operatorname{Cos}(2 \theta-1)$

Solution.

Given \[-1 \leq \operatorname{Cos} \theta \leq 1\] Multiplying by \(\frac{3}{5}\): \[-\frac{3}{5} \leq \frac{3}{5} \operatorname{Cos} \theta \leq \frac{3}{5}\] Adding 7: \[7 - \frac{3}{5} \leq 7 + \frac{3}{5} \operatorname{Cos} \theta \leq 7 + \frac{3}{5}\] Now replacing \(\mathrm{y} = 7 + \frac{3}{5} \operatorname{Cos}(2\theta - 1)\):
Since the range of \(\operatorname{Cos}(2\theta - 1)\) is the same as that of \(\operatorname{Cos} \theta\),
we can write: \[7 - \frac{3}{5} \leq \mathrm{y} \leq 7 + \frac{3}{5}\]

Hence,
\begin{align*} \text{Maximum value } (M)& = 7 + \dfrac{3}{5} =\dfrac{38}{5},\\ \text{Minimum value } (m)& = 7 - \frac{3}{5} = \frac{32}{5}. \end{align*}

Question 2 >