====== 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. 



=====Question 1(i)=====
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 $(m) = 0$. GOOD

**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. GOOD



=====Question 1(ii)=====
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 $(m) = \dfrac{1}{6}$. GOOD

**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 $(m) = \dfrac{1}{6}$. GOOD
=====Question 1(iii)=====
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 $(m) = -\dfrac{9}{5}$. GOOD

=====Question 1(iv)=====
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*}

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