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- Question 4 Exercise 8.2 @math-11-nbf:sol:unit08
- ta<\dfrac{\pi}{2}$, i.e. $\theta$ lies in QI. We have $$\sin\theta = \pm \sqrt{1-\cos^2}.$$ Since $\the... }. \end{align*} (d) $\sin \dfrac{\theta}{2}$ We have $$\sin\left(\frac{\theta}{2} \right) = \pm \sqrt{... }} \end{align*} (e) $\cos \dfrac{\theta}{2}$ We have $$\cos\left(\frac{\theta}{2} \right) = \pm \sqrt{... ac{3\pi}{2}\), i.e., \(\theta\) lies in QIII. We have: \begin{align*} \sec \theta &= \pm \sqrt{1+tan^2
- Question 6, Exercise 1.3 @math-11-kpk:sol:unit01
- $ $$(z^2+z+1 )=0$$ By using quadratic formula, we have $$z=\dfrac{-1\pm \sqrt{1-4}}{2}$$ $$z=\dfrac{-1\p... $ $$(z^2-z+1 )=0$$ By using quadratic formula, we have $$z=\dfrac{1\pm \sqrt{1-4}}{2}$$ $$z=\dfrac{1\pm ... }i}{2}$$ The value of $z$ from both equations, we have $$z=\pm \dfrac{1}{2}\pm \dfrac{\sqrt{3}}{2}i$$ ... 2-2z+4=0$$ According to the quadratic formula, we have $a=1$, $b=-2$ and $c=4$ Thus, we have \begin{alig
- Question 4, Exercise 1.3 @math-11-nbf:sol:unit01
- {53}i\end{align} Put value of $\omega$ in (1), we have \begin{align} &(1-i) z+(1+i)\left(\dfrac{2}{53}-\... c{155}{106}+\dfrac{145}{106}i\end{align} Thus, we have $$z=\dfrac{155}{106}+\dfrac{145}{106}i, \omega=\d... -12+10i \quad \cdots(4) \end{align} $(3)-(4)$,we have \begin{align} &(-1-8i+4-6i)\omega=3-i+12-10i\\ \i... c{36}{205}+\dfrac{373}{205}i \end{align} Thus, we have: $$z = \dfrac{36}{205} + \dfrac{373}{205}i;\quad
- Question 2, Exercise 2.6 @math-11-nbf:sol:unit02
- ich the system of homogeneous linear equation may have non-trivial solution. Also solve the system for v... }+4 x_{3}=0\cdots(vi)\\ \end{align*} (iv)-(v), we have\\ \begin{align*} &\begin{array}{cccc} 2x_1&+\frac... \end{align*} Put the value of $x_2$ in (vi), we have \begin{align*} &3 x_{1}-2(\frac{11}{13})x_{3}+4 x... ich the system of homogeneous linear equation may have non-trivial solution. Also solve the system for v
- Formatting Syntax @wiki
- syntax you may use when editing the pages. Simply have a look at the source of this page by pressing "Ed... ls of headlines to structure your content. If you have more than three headlines, a table of contents is... before the video has started. That image needs to have the same filename as the video and be either a jp... This is a list * The second item * You may have different levels * Another item - The same l
- Question 5 Exercise 8.2 @math-11-nbf:sol:unit08
- in 2\theta=\dfrac{24}{25}$, $2\theta$ in QII. We have $$\cos 2\theta = \pm \sqrt{1-\sin^2 2\theta}$$ S... c{49}{625}} = -\frac{7}{25} \end{align*} Also we have $$\sin\theta = \pm \sqrt{\frac{1-\cos 2\theta}{2}... dfrac{7}{25}\) and \(2\theta\) lies in QIII. We have: \[\sin 2\theta = \pm \sqrt{1 - \cos^2 2\theta}\... 76}{625}} = -\frac{24}{25}. \end{align*} Also, we have: \[ \sin\theta = \pm \sqrt{\frac{1 - \cos 2\the
- Question 2 Exercise 4.3 @math-11-kpk:sol:unit04
- on==== Given: $a_1=-40$ and $S_{21}=210$.\\ So we have $n=21$ and we have to find $a_{21}$ and $d$. As \begin{align}&S_{21}=\dfrac{21}{2}(a_1+a_{21}) \\ \impl... ===Solution==== Given: $a_1=-7, d=8, S_n=225$, we have to find $n$ and $a_n$. We know that $$S_n=\dfrac{n}{2}[2 a_1+(n-1) d].$$ Thus, we have \begin{align} & 225=\dfrac{n}{2}[2 \cdot(-7)+(n-1
- Question 10, Exercise 1.2 @math-11-nbf:sol:unit01
- -- (4) \end{align} From (1), (2), (3) and (4), we have: $$\left| z_1 \right| = \left| -z_1 \right| = \le... + \frac{7}{10}i. \,\, -- (i) \end{align} Now, we have \begin{align} \overline{z_1} = -3 - 2i, \quad \ov... frac{7}{10}i.\,\, -- (ii)$$ From (i) and (ii), we have \[ \overline{\left( \frac{z_1}{z_2} \right)} = \f... 11i. -- (ii) \end{align} From (i) and (ii), we have \[ \overline{z_1 z_2} = \overline{z_1} \overline
- Question 6 Exercise 8.2 @math-11-nbf:sol:unit08
- 15^{\circ} \cos 15^{\circ}$ ** Solution. ** We have double-angle identity: $$\sin 2 \theta = 2\sin\th... \circ}-\sin ^{2} 15^{\circ}$ ** Solution. ** We have double-angle identity: $$\cos^2\theta -\sin^2\the... 2}\left(\frac{\pi}{8}\right)$ ** Solution. ** We have a double-angle identity: $$\cos 2\alpha = 1-2\sin... =\cos 2\alpha.$$ Put $\alpha= \dfrac{\pi}{8}$, we have \begin{align*} 1-2\sin^2 \left(\frac{\pi}{8}\righ
- Question 2, Exercise 10.3 @fsc-part1-kpk:sol:unit10
- c }}+\sin {{43}^{\circ }}.$$ ====Solution==== We have an identity: $$\sin \alpha +\sin \beta =2\sin \le... ht)\end{align} Since $cos(-\theta)=cos\theta$, we have $$\sin {{37}^{\circ }}+\sin {{43}^{\circ }}=2\sin... rc }}-\cos {{82}^{\circ }}$. ====Solution==== We have an identity: $$\cos \alpha -\cos \beta =-2\sin \l... &=-2\sin(59^\circ)\sin(-23^\circ) \end{align} We have $\sin(-\theta)=-\sin\theta$, therefore $$\cos {{3
- Question 5, Exercise 1.3 @math-11-kpk:sol:unit01
- }+z+3=0.$$ According to the quadratic formula, we have\\ $a=1$, $b=1$ and $c=3$.\\ Thus we have \begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ &=\... }}-z-1=0$$ According to the quadratic formula, we have\\ $a=1$, $b=-1$ and $c=-1$\\ So we have \begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ &=\d
- Question 9 Exercise 3.4 @math-11-kpk:sol:unit03
- }+2 \hat{k}\end{align} From $\triangle A E B$, we have\\ \begin{align}\vec{c}&=\overrightarrow{A E}+\ove... s \ldots(1)\end{align} From $\triangle A E D$. we have\\ \begin{align}\vec{d}&=\overrightarrow{A E}+\ove... {. }....(2) \end{align} By using (1) and (2), we have,\\ \begin{align}\vec{c} \times \vec{d}&=\left| \b... at{k}\end{align} Which from $\triangle A E B$, we have\\ \begin{align}\vec{c} & =\overrightarrow{A E}+\v
- Question 2, Exercise 10.3 @math-11-kpk:sol:unit10
- c }}+\sin {{43}^{\circ }}.$$ ====Solution==== We have an identity: $$\sin \alpha +\sin \beta =2\sin \le... ht)\end{align} Since $cos(-\theta)=cos\theta$, we have $$\sin {{37}^{\circ }}+\sin {{43}^{\circ }}=2\sin... rc }}-\cos {{82}^{\circ }}$. ====Solution==== We have an identity: $$\cos \alpha -\cos \beta =-2\sin \l... &=-2\sin(59^\circ)\sin(-23^\circ) \end{align} We have $\sin(-\theta)=-\sin\theta$, therefore $$\cos {{3
- Question 8, Exercise 1.2 @math-11-nbf:sol:unit01
- ution.** Given: $$|2z-i|=4.$$ Put $z=x+i y$, we have \begin{align} & |2(x+iy)-i|=4 \\ \implies & |2x+i... * Given: $$|z-1|=|\bar{z}+i|.$$ Put $z=x+iy$, we have \begin{align} & |(x+iy)-1| = |(x-iy)+i| \\ \impli... n: $$|z-4i| + |z+4i| = 10.$$ Put $z = x + iy$, we have \begin{align} & |(x + iy) - 4i| + |(x + iy) + 4i|... Put $z = x + i y$, then $\bar{z} = x - i y$. We have \begin{align} & \dfrac{1}{2}Re(i(x-iy)) = 4 \\ \i
- Question 20 and 21, Exercise 4.4 @math-11-nbf:sol:unit04
- _\_ , \_\_\_ , \_\_\_ , 48$$ ** Solution. ** We have given $a_1=3$ and $a_5=48$. Assume $r$ be common... ference, then by general formula for nth term, we have $$ a_n=ar^{n-1}. $$ This gives \begin{align*} &a_... ="true"> **The good solution is as follows:** We have given $a_1=3$ and $a_5=48$. Assume $r$ be common... ference, then by general formula for nth term, we have $$ a_n=ar^{n-1}. $$ This gives \begin{align*} &a_
- How to prepare admission test (A short guide) @papers:old_admission_test_of_assms_for_ph.d._mathematics