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Question 4(i-iv), Exercise 9.1
4 Hits, Last modified: 5 months ago
\cdot \cos x) \\ & = -f(x) \end{align*} Thus the given function is odd. =====Question 4(ii)===== Check ... x \cdot \cos x \\ & = f(x) \end{align*} Thus the given function is even. =====Question 4(iii)===== Chec... an x}{x + \sin x}\\ &=y(x) \end{align*} Thus, the given function is even. =====Question 4(iv)===== Check... \sin x \cos^2 x\\ &=-y(x) \end{align*} Hence, the given function is odd. ====Go to ==== <text align="l
Question 4(v-viii), Exercise 9.1
4 Hits, Last modified: 5 months ago
^2 x}{x + \tan x}\\ &=-y(x)\end{align*} Thus, the given function is odd. =====Question 4(vi)===== Check ... n x}{\sin^3 x}\\ & = -y(x) \end{align*} Thus, the given function is odd. =====Question 4(vii)===== Check... sin x}{\sin^3 x}\\ &= -y(x)\end{align*} Thus, the given function is odd. =====Question 4(viii)===== Chec... t \sin x + \cot x\\ &=-y(x)\end{align*} Thus, the given function is odd. ====Go to ==== <text align="le
Question 1, Exercise 9.1
3 Hits, Last modified: 5 months ago
m value $(m) = 0$. GOOD **Alternative Method:** Given: $$y=2-2 \operatorname{Cos} \theta$$ Consider $$y... ) = \dfrac{1}{6}$. GOOD **Alternative Method:** Given: $$y=\dfrac{2}{3}-\dfrac{1}{2} \operatorname{Sin}... \operatorname{Cos}(2 \theta-1)$ ** Solution. ** Given \[-1 \leq \operatorname{Cos} \theta \leq 1\] Mu
Question 2, Exercise 9.1
2 Hits, Last modified: 5 months ago
?400 |Graph of y}} **From the graph, we see that given $y$ is not bounded and hence its maximum and mini... lution. ** **Same as Question 2(ii), we see that given $\dfrac{1}{\frac{1}{3}-4 \sin (2 \theta-5)}$ is n
Question 1,Review Exercise
2 Hits, Last modified: 5 months ago
$ and $\theta=45^{\circ}$, then the value of the given trigonometric identity is:\\ * (a) $\frac{1}{\sq... sed="true">%%(c)%%: $-17$</collapse> xix. If the given figure represent the curve $y=3\sin x$, then $|a|
Question 2 and 3, Review Exercise
1 Hits, Last modified: 5 months ago
in \theta=\sqrt{2} \cos \theta$ ** Solution. ** Given $$\cos \theta -\sin \theta=\sqrt{2}\sin \theta$$