Search
You can find the results of your search below.
Fulltext results:
- Question 1, Exercise 9.1
- ratorname{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 \... $$ Comparing the coeffients: $$a=2, \quad b=-2$$ \begin{align*} \text{Maximum value (M)} & = a+|b|\\ & =
- Question 2, Exercise 9.1
- atorname{Sin} \theta}$ ** Solution. ** We know \begin{align*} -1 \leq \operatorname{Sin} \theta \leq 1 \end{align*} Multiplying with $3$ \begin{align*} -3 \leq 3 \operatorname{Sin} \theta \leq 3 \end{align*} Adding $4$ \begin{align*} & 1 \leq 4+3 \operatorname{Sin} \theta \... } \sin (5 \theta-7)}$ ** Solution. ** We know \begin{align*} -1 \leq \operatorname{Sin(5 \theta-7)} \t
- Question 3, Exercise 9.1
- in and range: $y=7 \cos 4x$ ** Solution. ** AS \begin{align*} & -1\leq \cos 4x \leq 1 \,\, \forall \,\... range: $y=\cos \frac{x}{3}$ ** Solution. ** AS \begin{align*} & -1\leq \cos \frac{x}{3} \leq 1 \,\, \f... nge: $y=\sin \frac{2 x}{3}$ ** Solution. ** AS \begin{align*} & -1 \leq \sin \frac{2x}{3} \leq 1 \,\, ... integer} \right\}$ Range of $y=\mathbb{R}$ As \begin{align*} & \theta \neq n\pi \\ \implies & \dfrac{
- Question 4(i-iv), Exercise 9.1
- * Consider $f(x)=\sin x+x \cdot \cos x$. Take \begin{align*} f(-x) = \sin (-x) + (-x)\cdot \cos (-x) ... $\sin(-x)=-\sin x$ and $\cos (-x) = \cos x$, so \begin{align*} f(x) & = -\sin x - x \cdot \cos x \\ & = ... r $f(x)=x^{3} \cdot \sin x \cdot \cos x$. Take \begin{align*} f(-x) = (-x)^{3} \cdot \sin (-x) \cdot \... $\sin(-x)=-\sin x$ and $\cos (-x) = \cos x$, so \begin{align*} f(-x) & = -x^3 (-\sin x) (\cos x) \\ & =
- Question 4(v-viii), Exercise 9.1
- nsider \[y = \frac{\sin^2 x}{x + \tan x}\] Take \begin{align*} y(-x) &= \frac{\big(-\sin x\big)^2}{-x - ... \[y = \frac{\tan x - \sin x}{\sin^3 x}\] Take \begin{align*} y(-x) &= \frac{-\tan x - (-\sin x)}{(-\si... \[y = \frac{\tan x - \sin x}{\sin^3 x}\] Take \begin{align*} y(-x) &= \frac{-\tan x - (-\sin x)}{(-\si... onsider \[y = x^2 \cdot \sin x - \cot x\] Take \begin{align*} y(-x) &= (-x)^2 \cdot (-\sin x) - (-\cot
- Question 6, Exercise 9.1
- Since period of the $\sec$ is $2\pi$, therefore \begin{align*} 6 \sec(2 x-3) & = 6 \sec(2 x-3+2\pi) \\ &... Since period of the $\cos$ is $2\pi$, therefore \begin{align*} \cos (5 x+4) & = 6 \cos(5x+4+2\pi) \\ & =... e period of \( \sin \) is \( 2\pi \), therefore: \begin{align*} 7 \sin(3x + 3) &= 7 \sin(3x + 3 + 2\pi) \... e period of \( \sin \) is \( 2\pi \), therefore: \begin{align*} 5 \sin(2x + 3) &= 5 \sin(2x + 3 + 2\pi) \
- Question 2 and 3, Review Exercise
- ta -\sin \theta=\sqrt{2}\sin \theta$$ This gives \begin{align*} & \cos \theta=\sqrt{2}\sin \theta + \sin ... {\sqrt{2}+1}\cos \theta ... (1) \end{align*} Now \begin{align*} LHS & = \cos \theta+ \sin \theta \\ & = ... \cos x} = \sec^2 x - \csc^2 x$ ** Solution. ** \begin{align*} LHS & = \dfrac{\tan x - \cot x}{\sin x \c... n^2 x} = \sec^2 x - \tan^2 x$ ** Solution. ** \begin{align*} LHS & = \dfrac{\sec^4 x - \tan^4 x}{\sec