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- Question 4 Exercise 8.2 @math-11-nbf:sol:unit08
- \cos 2 \theta$ %%(c)%% $\tan 2 \theta$ (d) $\sin \frac{\theta}{2}$ (e) $\cos \frac{\theta}{2}$ (f) $\tan \frac{\theta}{2}$ when: $\cos \theta=\frac{3}{5}$ where $0<\theta<\frac{\pi}{2}$ ** Solution. ** Given: $\c
- Question 1, Exercise 8.1 @math-11-nbf:sol:unit08
- \sin 60 \\ \implies \cos (180+60) & = (-1)\left(\frac{1}{2}\right) - (0)\left(\frac{\sqrt{3}}{2}\right) \\ & = -\frac{1}{2} - 0 = -\frac{1}{2} \end{align*} \begin{align*} \cos (\alpha - \beta) & = \cos \alpha \cos \be
- Question 3, Exercise 2.5 @math-11-nbf:sol:unit02
- nd{array} \right] \quad \text{by } R2-R_3 \quad \frac{1}{2} R3\\ \sim&{\text{R}} \left[ \begin{array}{c... 1 \\ 0 & 1 & 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 1 & \frac{1}{2} & \frac{1}{2} \end{array} \right] \quad \text{by } R1 + R2 \text{ and } \frac{1}{2} R3\\ \sim&{\text{R}} \left[ \begin{array}{c
- Question 5 Exercise 8.2 @math-11-nbf:sol:unit08
- eta$ using the information given: $\sin 2 \theta=\frac{24}{25}, 2 \theta$ in QII ** Solution. ** Given... = - \sqrt{1-\sin^2 2\theta}\\ &=- \sqrt{1-\left(\frac{24}{25}\right)^2} \\ &=- \sqrt{\frac{49}{625}} = -\frac{7}{25} \end{align*} Also we have $$\sin\theta = \pm \sqrt{\frac{1-\cos 2\theta}{2}}
- Question 12, Exercise 8.1 @math-11-nbf:sol:unit08
- +\beta) = \tan(180^{\circ}-\gamma) \\ \implies & \frac{\tan\alpha + \tan\beta}{1-\tan\alpha \tan\beta} ... pha+\beta+\gamma=180^{\circ}$, prove that: $\cot \frac{\alpha}{2}+\cot \frac{\beta}{2}+\cot \frac{\gamma}{2}=\cot \frac{\alpha}{2} \cot \frac{\beta}{2} \cot \frac{\gamma}{2}$ ** S
- Question 7 and 8, Exercise 4.8 @math-11-nbf:sol:unit04
- 7===== Find the sum of $n$ term of the series: $$\frac{1}{1 \times 4}+\frac{1}{4 \times 7}+\frac{1}{7 \times 10}+\ldots$$ ** Solution. ** Given: $$\frac{1}{1 \times 4}+\frac{1}{4 \times 7}+\frac{1}{7 \t
- Question 2, Review Exercise @math-11-nbf:sol:unit08
- n*} \cos^2 \theta &= 1-\sin^2\theta\\ &= 1-\left(\frac{3}{5}\right)^2\\ & =1-\frac{9}{25} \\ \implies \cos^2 \theta&=\frac{16}{25} \\ \cos \theta&=\pm\frac{4}{5}\\ \end{align*} As $\theta$ is obtuse, so $\theta$ lies in II
- Question 4, Exercise 2.6 @math-11-nbf:sol:unit02
- nd{bmatrix}\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} & \vert & 1 \\ 3 & -4 & 3 & \vert & 7 \\ 4 & 2 & -5 & \vert & 10 \end{bmatrix}\qua... dfrac{1}{2}\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} & : & 1 \\ 0 & -\frac{5}{2} & \frac{9}{2} & : & 4 \\ 4 & 2 & -5 & : & 10 \end{bmat
- Question 25 and 26, Exercise 4.7 @math-11-nbf:sol:unit04
- of the series (arithmetico-geometric series): $1+\frac{4}{7}+\frac{7}{7^{2}}+\frac{10}{7^{3}}+\ldots$ ** Solution. ** The given arithmetic-geometric series is: \[ 1 + \frac{4}{7} + \frac{7}{7^2} + \frac{10}{7^3} + \ldots \
- Question 11 and 12, Exercise 4.8 @math-11-nbf:sol:unit04
- Evaluate the sum of the series: $\sum_{k=1}^{n} \frac{1}{k(k+2)}$ ** Solution. ** Let $T_k$ represent... h 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
- Question 9, Exercise 1.2 @math-11-nbf:sol:unit01
- e the following formulas: \[\text{Re}(z^{-2}) = \frac{(\text{Re}(z))^2 - (\text{Im}(z))^2}{|z|^4},\] \[\text{Im}(z^{-2}) = \frac{-2 \text{Re}(z) \text{Im}(z)}{|z|^4}. \] First, n... bove formulas \begin{align} Re((3 - 2i)^{-2}) &= \frac{(3)^2 - (-2)^2}{(\sqrt{13})^4} \\ &= \frac{9 - 4}{169} = \frac{5}{169}. \end{align} \begin{align} Im((3
- Question 6, Exercise 2.3 @math-11-nbf:sol:unit02
- & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & -\frac{1}{9} & 0 & \frac{1}{9} \end{array}\right] \quad \frac{1}{9} R_3\\ &=\left[\begin{array}{ccc|ccc} 2 & 1 & 0 & 1 - 3 \left(\frac{1}{9}\right) & 0 & 3 \left(\frac{1}{9}\right) \\
- Question 7, Exercise 1.4 @math-11-nbf:sol:unit01
- frac{\pi}{3}$ ** Solution. ** \begin{align*} &-\frac{\pi}{3} \leq \arg (z-4) \leq \frac{\pi}{3}\\ \implies & -\frac{\pi}{3} \leq \arg(x+iy-4) \leq \frac{\pi}{3} \\ \implies & -\frac{\pi}{3} \leq \arg(x-4+iy) \leq \frac{
- Question 2, Exercise 2.5 @math-11-nbf:sol:unit02
- ]\\ \sim & \text{R}\left[ \begin{array}{ccc} 1 & \frac{9}{5} & \frac{3}{5} \\ 3 & -5 & 6 \\ 2 & 10 & 6 \end{array} \right]\quad \frac{1}{5} R1\\ \sim & \text{R}\left[ \begin{array}{ccc} 1 & \frac{9}{5} & \frac{3}{5} \\ 0 & -\frac{52}{5} & \frac
- Question 9, Exercise 8.1 @math-11-nbf:sol:unit08
- \cos \beta + \cos \alpha \sin \beta \\ &= \left( \frac{1}{\sqrt{2}} \right) \left( -\frac{3}{5} \right) + \left( -\frac{1}{\sqrt{2}} \right) \left( \frac{4}{5} \right) \\ &= -\frac{3}{5\sqrt{2}} - \frac{4}{5\sqrt{2}} \\ &=