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- Question 2, Review Exercise
- 2 \theta&=\frac{144}{169} \\ \cos \phi&=\pm\frac{12}{13}\\ \end{align*} As $\phi$ is acute, so $\ph... This implies $\cos\pi >0$, thus $$\cos \phi=\frac{12}{13}$$ As, we have \begin{align*} \sin(\theta -\p... sin \phi\\ &=\left(\frac{3}{5}\right) \left(\frac{12}{13}\right)-\left(-\frac{4}{5}\right) \left(\frac... 2 \theta&=\frac{144}{169} \\ \cos \phi&=\pm\frac{12}{13}\\ \end{align*} As $\phi$ is acute, so $\ph
- Question 6 Exercise 8.2
- for the expression: $2 \cos ^{2}\left(\frac{\pi}{12}\right)-1$ ** Solution. ** We have a double-ang... alpha -1=\cos 2\alpha.$$ Put $\alpha= \dfrac{\pi}{12}$, we have \begin{align*} 2\cos^2 \left(\frac{\pi}{12}\right)-1&=\cos 2\left(\frac{\pi}{12}\right)\\ &=\cos \left(\frac{\pi}{6}\right)\\ \end{align*} \begi
- Question 5 and 6, Exercise 8.1
- $\sin \alpha=\dfrac{4}{5}, \tan \beta=-\dfrac{5}{12}$ with terminal side of an angles in QII, find $\... }$, $\alpha$ is in QII and $\tan \beta=-\dfrac{5}{12}$, $\beta$ is in QII. We have an identity: $$\co... t{1+\tan^2\beta} \\ &=-\sqrt{1+{{\left(-\dfrac{5}{12} \right)}^{2}}}\\ &=-\sqrt{1+\dfrac{25}{144}} \\ & =-\sqrt{\dfrac{169}{144}} =-\dfrac{13}{12} \end{align*} \begin{align*} \Rightarrow \quad
- Question 13, Exercise 8.1
- Express in the form of $r \sin (\theta+\phi)$: $12 \sin \theta-5 \cos \theta$ ** Solution. ** Let $12=r\cos \varphi $ and $-5=r\sin \varphi$.\\ Squaring and adding \begin{align*} & (12)^2+(-5)^2=r^2 \cos^2\varphi+r^2 \sin^2 \varphi \\... }=13 \end{align*} Also \begin{align*} & \frac{-5}{12}=\frac{r\sin \varphi }{r\cos \varphi } \\ \implie
- Question 4 Exercise 8.2
- $\tan \frac{\theta}{2}$ when: $\tan \theta=\frac{12}{5}$ where $\pi<\theta<\frac{3 \pi}{2}$ ** Solution. ** Given: \(\tan \theta = \frac{12}{5}\) where \(\pi < \theta < \frac{3\pi}{2}\), i.... \pm \sqrt{1+tan^2 \theta}\\ &= \pm \sqrt{1+(\frac{12}{5})^2 }\\ &=\pm \frac{13}{5}\end{align*} \(\thet... 2 \theta)} = -\sqrt{(1-\frac{25}{169}) } = -\frac{12}{13}\] (a) \(\sin 2 \theta\) \begin{align*} \si
- Question 7, Exercise 8.1
- $\beta$ are acute angles with $\sin \alpha=\dfrac{12}{13}$ and $\tan \beta=\dfrac{4}{3}$ find \\ (i) $... a)$. ** Solution. ** Given: $\sin \alpha=\dfrac{12}{13}$, where $\alpha$ is acute angle, i.e. is in ... \sqrt{1-\sin^2\alpha} \\ &= \sqrt{1-{{\left(\frac{12}{13}\right)}^2}} \\ &= \sqrt{1-\dfrac{144}{169}} ... gin{align*} \sin(\alpha + \beta) & = \left(\dfrac{12}{13}\right) \left(\dfrac{3}{5}\right) + \left(\df
- Question 8, Exercise 8.1
- $0<\alpha<\dfrac{\pi}{2}$ and $\cos \beta=\dfrac{12}{13}$, where $\dfrac{3 \pi}{2}<\beta<2 \pi$ find:... & = \frac{4}{5}. \end{align*} $\cos \beta=\dfrac{12}{13}$, where $\dfrac{3 \pi}{2}<\beta<2 \pi$, i.e.... t{1 - \cos^2 \beta} \\ &= -\sqrt{1 - \left( \frac{12}{13} \right)^2} \\ &= -\sqrt{\frac{25}{169}}\\ \s... eta \\ &= \left( \frac{3}{5} \right) \left( \frac{12}{13} \right) + \left( \frac{4}{5} \right) \left(
- Question 12, Exercise 8.1
- ====== Question 12, Exercise 8.1 ====== Solutions of Question 12 of Exercise 8.1 of Unit 08: Fundamental of Trigonometry... ook Board, Islamabad, Pakistan. ===== Question 12(i)===== If $\alpha+\beta+\gamma=180^{\circ}$, pro... \end{align*} as required. GOOD ===== Question 12(ii)===== If $\alpha+\beta+\gamma=180^{\circ}$, pr
- Question 2(i, ii, iii, iv and v) Exercise 8.3
- product of two function: $\cos 58^{\circ} + \cos 12^{\circ}$ ** Solution. ** \begin{align*} &\quad \cos 58^{\circ} + \cos 12^{\circ} \\ & = 2 \cos \left(\frac{58 + 12}{2}\right) \cos \left(\frac{58 - 12}{2}\right) \\ & = 2 \cos \left(\frac{70}{2}\right) \cos \left(\fr
- Question 11, Exercise 8.1
- ext align="right"><btn type="success">[[math-11-nbf:sol:unit08:ex8-1-p11|Question 12 >]]</btn></text>