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- Question 2, Exercise 9.1
- atorname{Sin} \theta \leq 7 \\ \implies & 1 \geq \frac{1}{4+3 \operatorname{Sin}} \theta \geq \frac{1}{7} \\ \implies & \frac{1}{7} \leq \frac{1}{4+3 \operatorname{Sin}} \theta\leq 1 \\ \end{align*} Hence maximum value $(M) = 1$
- Question 1,Review Exercise
- hose the correct option.\\ i. If $\cos \theta=\frac{\sqrt{3}}{2}$ and the terminal arm of angle is in III quadrant. Then $\sin \theta=$\\ * $\frac{1}{2}$\\ * (b) $-\frac{1}{2}$\\ * %%(c)%% $\sqrt{3}$\\ * (d) $-\frac{2}{\sqrt{3}}$ \\ <btn type="link" collapse="a1">S
- Question 4(v-viii), Exercise 9.1
- x}{x+\tan x}$ ** Solution. ** Consider \[y = \frac{\sin^2 x}{x + \tan x}\] Take \begin{align*} y(-x) &= \frac{\big(-\sin x\big)^2}{-x - \tan x} \\ &= \frac{\sin^2 x}{-x - \tan x}\\ & = \frac{\sin^2 x}{-(x + \tan x)}\\ & = -\frac{\sin^2 x}{x + \tan x}\\ &=-y(x
- Question 1, Exercise 9.1
- ues of the trigonometric function: $\mathrm{y}=7+\frac{3}{5} \operatorname{Cos}(2 \theta-1)$ ** Solutio... atorname{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} \le
- Question 3, Exercise 9.1
- estion 3(ii)===== Find domain and range: $y=\cos \frac{x}{3}$ ** Solution. ** AS \begin{align*} & -1\leq \cos \frac{x}{3} \leq 1 \,\, \forall \,\, x\in \mathbb{R} \\... stion 3(iii)===== Find domain and range: $y=\sin \frac{2 x}{3}$ ** Solution. ** AS \begin{align*} & -1 \leq \sin \frac{2x}{3} \leq 1 \,\, \forall \,\, x \in \mathbb{R}
- Question 4(i-iv), Exercise 9.1
- x}{x+\sin x}$ ** Solution. ** Consider \[y = \frac{x^2 \cdot \tan x}{x + \sin x}.\] Take\\ \[y(-x) = \frac{(-x)^2 \cdot \tan(-x)}{-x + \sin(-x)}.\] \begin{align*} y(-x) &= \frac{(-x)^2 \cdot \tan(-x)}{-x + \sin(-x)} \\ &= \frac{x^2 \cdot (-\tan x)}{-x - \sin x}\\ &= \frac{-x^2 \cd
- Question 2 and 3, Review Exercise
- (\sqrt{2}+1)\sin \theta\\ \implies & \sin \theta=\frac{1}{\sqrt{2}+1}\cos \theta ... (1) \end{align*} No... = \cos \theta+ \sin \theta \\ & = \cos \theta+ \frac{1}{\sqrt{2}+1}\cos \theta \quad \text{from (1)}\\ & = \left(1+ \frac{1}{\sqrt{2}+1}\right)\cos \theta \\ & = \frac{\sqrt{2}+1+1}{\sqrt{2}+1} \cos \theta \\ & = \frac{2+\sqr
- Question 6, Exercise 9.1
- & = 6 \cos(5x+4+2\pi) \\ & = \cos\left(5\left(x+\frac{2\pi}{5}\right)+4\right) \end{align*} Hence perio... on 6(iii)===== Find the period: $y=\cot 4 x+\sin \frac{5 x}{2}$ ** Solution. ** FIXME(unable to solve) ... \sin(3x + 3 + 2\pi) \\ &= 7 \sin\left(3\left(x + \frac{2\pi}{3}\right) + 3\right). \end{align*} Hence, t
- Question 5(i-v), Exercise 9.1
- of the function: $\mathrm{y}=\operatorname{Cos} \frac{\mathrm{x}}{2}$ ** Solution. ** ====Go to ===
- Question 5(vi-x), Exercise 9.1
- h of each of the function: $y=\operatorname{Sin} \frac{x}{2}$ ** Solution. ** ====Go to ==== <text