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- Question 4 Exercise 8.2 @math-11-nbf:sol:unit08
- $ lies in QI, $\sin$ is positive in QI, so \begin{align*} \sin\theta & = \sqrt{1-\left(\frac{3}{5} \righ... 25}}\\ &= \sqrt{\frac{16}{25}} = \frac{4}{5} \end{align*} (a) $\sin 2 \theta$ \begin{align*} \sin 2\theta & = 2 \sin\theta \cos\theta \\ &= 2\left(\frac{4}{5} \right) \left(\frac{3}{5} \right)\\ \end{align*} \begin{align*} \implies \boxed{\sin 2\theta =
- Question 1, Exercise 8.1 @math-11-nbf:sol:unit08
- $\alpha=180^{\circ}$, $\beta=60^{\circ}$. \begin{align*} \cos (\alpha + \beta) & = \cos \alpha \cos \be... ht) \\ & = -\frac{1}{2} - 0 = -\frac{1}{2} \end{align*} \begin{align*} \cos (\alpha - \beta) & = \cos \alpha \cos \beta + \sin \alpha \sin \beta \\ \implies... ght) \\ & = -\frac{1}{2} + 0 = -\frac{1}{2} \end{align*} \begin{align*} \sin (\alpha + \beta) & = \sin
- Question 5 Exercise 8.2 @math-11-nbf:sol:unit08
- II, therefore $\cos 2\theta$ is negative. \begin{align*}\cos 2\theta & = - \sqrt{1-\sin^2 2\theta}\\ &=-... \\ &=- \sqrt{\frac{49}{625}} = -\frac{7}{25} \end{align*} Also we have $$\sin\theta = \pm \sqrt{\frac{1-... ta$ lies in QI and $\sin\theta > 0$. Thus \begin{align*} \sin\theta & = \sqrt{\frac{1-\cos 2\theta}{2}} ... ac{\frac{32}{25}}{2}} = \sqrt{\frac{16}{25}} \end{align*} \begin{align*} \implies \boxed{\sin\theta = \fr
- Question 4, Exercise 1.3 @math-11-nbf:sol:unit01
- z-(2+5 i) \omega=2+3 i$. ** Solution. ** \begin{align} &(1-i) z+(1+i) \omega=3 \quad \cdots(1)\\ &2 z-(2+5 i) \omega=2+3i \quad\cdots(2) \end{align} Multiplying Eq. (1) by $2$: \begin{align} &(2-2i)z+(2+2i) \omega=6 \quad \cdots (3) \end{align} Multiplying Eq. (2) by $(1-i)$: \begin{align} &2
- Question 10, Exercise 1.2 @math-11-nbf:sol:unit01
- \overline{z_{!}}\right|.$$ **Solution.** \begin{align} |z_1| &= \sqrt{(-3)^2 + (2)^2} \\ &= \sqrt{9 + 4} = \sqrt{13} \,\, -- (1) \end{align} Now \begin{align} -z_1 &= -(-3 + 2i) = 3 - 2i\\ \implies |-z_1| &= \sqrt{(3)^2 + (-2)^2} \\ &= \sqrt{9 + 4} = \sqrt{13} \,\, -- (2) \end{align} Also \begin{align} \overline{z_1} &= -3 - 2i
- Question 11 and 12, Exercise 4.8 @math-11-nbf:sol:unit04
- present the $k$th term of the series. Then \begin{align*} T_k &= \frac{1}{k(k+2)}. \end{align*} Resolving it into partial fractions: \begin{align*} \frac{1}{k(k+2)} = \frac{A}{k} + \frac{B}{k+2} \ldots (1) \end{align*} Multiplying both sides by $k(k+2)$, we get \beg
- Question 2, Exercise 2.6 @math-11-nbf:sol:unit02
- _{1}-2 x_{2}+4 x_{3}=0$\\ ** Solution. ** \begin{align*} &2 x_{1}-\lambda x_{2}+x_{3}=0 \cdots(i)\\ &2 x... i)\\ &3 x_{1}-2 x_{2}+4 x_{3}=0\cdots(iii)\\ \end{align*} Homogenous system has non-trivial solution, if \begin{align*} &\left| \begin{array}{ccc} 2 & -\lambda & 1 \\... 0+11\lambda-13=0\\ &\lambda =-\frac{7}{11}\\ \end{align*} The system becomes \begin{align*} &2 x_{1}+ \f
- Question 1, Exercise 1.3 @math-11-nbf:sol:unit01
- ar functions: $z^{2}+169$. **Solution.** \begin{align} & z^{2} + 169 \\ = & z^{2} - (13i)^2 \\ = &(z + 13i)(z - 13i). \end{align} ====Question 1(ii)==== Factorize the polynomial ... r functions: $2 z^{2}+18$. **Solution.** \begin{align} & 2z^2 + 18 \\ = &2(z^2 - (3i)^2)\\ = &2(z + 3i)(z - 3i) \end{align} ====Question 1(iii)==== Factorize the polynomia
- Question 9, Exercise 1.2 @math-11-nbf:sol:unit01
- {-1}$. **Solution.** Suppose $z=2+4i$. \begin{align} Re(2+4i)^{-1} & = Re(z^{-1}) = \dfrac{Re(z)}{|z|... ^2+4^2} = \dfrac{2}{20}\\ &= \dfrac{1}{10}. \end{align} \begin{align} Im(2+4i)^{-1} & = Im(z^{-1}) = -\dfrac{Im(z)}{|z|^2} \\ & =-\dfrac{4}{2^2+4^2} = -\dfrac{4}{20}\\ &= \dfrac{1}{5}. \end{align} GOOD ====Question 9(ii)==== Find real and imagin
- Question 7 and 8, Exercise 4.8 @math-11-nbf:sol:unit04
- presents the kth term of the series. Then \begin{align*} T_k &=\frac{1}{(3k-2)(3k+1)}. \end{align*} Resolving it into partial fraction: \begin{align*} \frac{1}{(3k-2)(3k+1)} = \frac{A}{3k-2}+\frac{B}{3k+1} \ldots (1) \end{align*} Multiplying with $(3k-2)(3k+1)$ \begin{align*}
- Question 5 and 6, Exercise 8.1 @math-11-nbf:sol:unit08
- ha$ lies in QII and $\cos$ is -ive in QII, \begin{align*}\cos\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=-\sqrt{... \Rightarrow \quad \cos \alpha&=-\dfrac{3}{5}.\end{align*} Also $$\sec \beta=\pm\sqrt{1+\tan^2\beta}.$$ As... ta$ lies in QII and $\sec$ is -ive in QII, \begin{align*}\sec \beta & =-\sqrt{1+\tan^2\beta} \\ &=-\sqrt{... =-\sqrt{\dfrac{169}{144}} =-\dfrac{13}{12} \end{align*} \begin{align*} \Rightarrow \quad \cos \beta=\f
- Question 9, Exercise 8.1 @math-11-nbf:sol:unit08
- ha$ lies in QII and $\cos$ is -ive in QII, \begin{align*} \cos\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=-\sqrt... \implies \cos \alpha&=-\dfrac{1}{\sqrt{2}}. \end{align*} Also $$\sin\beta=\pm\sqrt{1-\sin^2\beta}.$$ As... ta$ lies in QII and $\sin$ is +ive in QII, \begin{align*} \cos \beta & =\sqrt{1-\cos^2\beta} \\ &=\sqrt{1... } \\ & =\sqrt{\dfrac{16}{25}} =\dfrac{4}{5} \end{align*} \begin{align*} \sin (\alpha + \beta) &= \sin \a
- Question 1, Exercise 1.4 @math-11-nbf:sol:unit01
- ** Let $z=x+iy=2 + i 2 \sqrt{3}$. We have \begin{align} r & = \sqrt{x^2 + y^2} = \sqrt{2^2 + (2\sqrt{3})^2} \\ & = \sqrt{4 + 12} = \sqrt{16} = 4. \end{align} and \begin{align} \alpha & = \tan^{-1}\left|\frac{y}{x}\right| = \tan^{-1}\left|\frac{2\sqrt{3}}{2}\rig... \\ & = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}. \end{align} Since the complex number \( 2 + i 2 \sqrt{3} \)
- Question 1, Exercise 9.1 @math-11-nbf:sol:unit09
- ame{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 \operat
- Question 3, Exercise 2.6 @math-11-nbf:sol:unit02
- olution. ** Given the system of equations: \begin{align*} \begin{aligned} 2x + 3y + 4z &= 2 \\ 2x + y + z &= 5 \\ 3x - 2y + z &= -3 \end{aligned}\end{align*} The associated augmented matrix is: \begin{align*} A_{b} &=\quad \left[\begin{array}{cc