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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: $\c
- Question 1, Exercise 8.1 @math-11-nbf:sol:unit08
- \sin 60 \\ \implies \cos (180+60) & = (-1)\left(\frac{1}{2}\right) - (0)\left(\frac{\sqrt{3}}{2}\right) \\ & = -\frac{1}{2} - 0 = -\frac{1}{2} \end{align*} \begin{align*} \cos (\alpha - \beta) & = \cos \alpha \cos \be
- Question 4, Exercise 1.3 @math-11-nbf:sol:unit01
- -i \end{align} \begin{align} \implies \omega & =\dfrac{1-i}{9+5i}\\ &=\dfrac{1-i}{9+5i}\times\dfrac{9-5i}{9-5i}\\ &=\dfrac{9-5-5i-9i}{81+25}\\ &=\dfrac{4-14i}{106}\\ &=\dfrac{2}{53}-\dfrac{7}{53}i\end{al
- Question 5 Exercise 8.2 @math-11-nbf:sol:unit08
- eta$ using the information given: $\sin 2 \theta=\frac{24}{25}, 2 \theta$ in QII ** Solution. ** Given: $\sin 2\theta=\dfrac{24}{25}$, $2\theta$ in QII. We have $$\cos 2\the... = - \sqrt{1-\sin^2 2\theta}\\ &=- \sqrt{1-\left(\frac{24}{25}\right)^2} \\ &=- \sqrt{\frac{49}{625}} = -\frac{7}{25} \end{align*} Also we have $$\sin\theta
- Question 7, Exercise 1.4 @math-11-nbf:sol:unit01
- lowing equation in Cartesian form: $\arg (z-1)=-\dfrac{\pi}{4}$ ** Solution. ** Suppose $z=x+iy$, as \begin{align*} &\arg (z-1)=-\dfrac{\pi}{4} \\ \implies & \arg(x+iy-1) = -\dfrac{\pi}{4} \\ \implies & \arg(x-1+iy) = -\dfrac{\pi}{4} \\ \implies & \tan^{-1}\left(\dfrac{y}{x-1}\right)
- 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 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 9, Exercise 8.1 @math-11-nbf:sol:unit08
- and $\beta$ are obtuse angles with $\sin \alpha=\dfrac{1}{\sqrt{2}}$ and $\cos \beta=-\dfrac{3}{5}$ find: $\sin (\alpha \pm \beta)$ ** Solution. ** Given: $\sin \alpha=\dfrac{1}{\sqrt{2}}$, $\alpha$ is obtuse angle, i.e. it is in QII.\\ $\cos \beta=-\dfrac{3}{5}$, $\beta$ is obtuse angle, i.e. it is in QI
- Question 2, Review Exercise @math-11-nbf:sol:unit08
- =====Question 2(i)===== Given that $\sin \theta=\dfrac{3}{5}, \sin \phi=\dfrac{5}{13}$ where $\theta$ is obtuse and $\phi$ is acute. Find the values of $\sin ... ta-\phi)$. ** Solution. ** Given: $\sin \theta=\dfrac{3}{5}$ and $\sin \phi=\dfrac{5}{13}$, where $\theta$ is obtuse and $\phi$ is acute. As\begin{align*} \c
- 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: $$\frac{1}{1 \times 4}+\frac{1}{4 \times 7}+\frac{1}{7 \t
- Question 9, Exercise 1.2 @math-11-nbf:sol:unit01
- . \begin{align} Re(2+4i)^{-1} & = Re(z^{-1}) = \dfrac{Re(z)}{|z|^2} \\ & =\dfrac{2}{2^2+4^2} = \dfrac{2}{20}\\ &= \dfrac{1}{10}. \end{align} \begin{align} Im(2+4i)^{-1} & = Im(z^{-1}) = -\dfrac{Im(z)}{|z
- Question 2, Exercise 1.4 @math-11-nbf:sol:unit01
- 2(i)===== Write the complex number $\left(\cos \dfrac{\pi}{6}+i \sin \dfrac{\pi}{6}\right)\left(\cos \dfrac{\pi}{3}+i \sin \dfrac{\pi}{3}\right)$ in rectangular form. ** Solution. ** Let $z_1=\cos \dfrac{\pi}{
- Question 5 and 6, Exercise 8.1 @math-11-nbf:sol:unit08
- stan. ===== Question 5===== For $\sin \alpha=\dfrac{4}{5}, \tan \beta=-\dfrac{5}{12}$ with terminal side of an angles in QII, find $\cos (\alpha+\beta)$ an... -\beta)$. ** Solution. ** Given: $\sin \alpha=\dfrac{4}{5}$, $\alpha$ is in QII and $\tan \beta=-\dfrac{5}{12}$, $\beta$ is in QII. We have an identity: $$
- 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}\quad \dfrac{1}{2}\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} & : & 1 \\ 0 & -\frac{5}{2} & \fr
- Question 1, Exercise 9.1 @math-11-nbf:sol:unit09
- inimum values of the trigonometric function: $y=\dfrac{2}{3}-\dfrac{1}{2} \operatorname{Sin} \theta$ ** Solution. ** We know \begin{align*} -1 \leq \operato... } \theta \leq 1 \end{align*} Multiplying with $-\dfrac{1}{2}$ \begin{align*} & \dfrac{1}{2} \geq -\dfrac{1}{2} \operatorname{Sin} \theta \geq -\dfrac{1}{2} \e