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Question 1, Exercise 9.1

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=2-2 \operatorname{Cos} \theta$ ** Solution. ** We know \begin{align*} -1 \leq \operatorname{Cos} \t... }{2} \operatorname{Sin} \theta$ ** Solution. ** We know \begin{align*} -1 \leq \operatorname{Sin} \t... \operatorname{Sin}(3 \theta-7)$ ** Solution. ** We know \begin{align*} -1 \leq \operatorname{Sin}(3 ... same as that of \(\operatorname{Cos} \theta\),\\ we can write: \[7 - \frac{3}{5} \leq \mathrm{y} \l

Question 2, Exercise 9.1

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4+3 \operatorname{Sin} \theta}$ ** Solution. ** We know \begin{align*} -1 \leq \operatorname{Sin} \t... 1-q2_ii_.png?400 |Graph of y}} **From the graph, we see that given $y$ is not bounded and hence its m... -5)}$ ** Solution. ** **Same as Question 2(ii), we see that given $\dfrac{1}{\frac{1}{3}-4 \sin (2 \... frac{2}{5} \sin (5 \theta-7)}$ ** Solution. ** We know \begin{align*} -1 \leq \operatorname{Sin(5 \

Question 4(i-iv), Exercise 9.1

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= \sin (-x) + (-x)\cdot \cos (-x) \end{align*} As we know $\sin(-x)=-\sin x$ and $\cos (-x) = \cos x$,... } \cdot \sin (-x) \cdot \cos (-x) \end{align*} As we know $\sin(-x)=-\sin x$ and $\cos (-x) = \cos x$,