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- Question 4 Exercise 8.2
- \theta$ lies in QI, $\sin$ is positive in QI, so \begin{align*} \sin\theta & = \sqrt{1-\left(\frac{3}{5}... = \frac{4}{5} \end{align*} (a) $\sin 2 \theta$ \begin{align*} \sin 2\theta & = 2 \sin\theta \cos\theta ... \right) \left(\frac{3}{5} \right)\\ \end{align*} \begin{align*} \implies \boxed{\sin 2\theta = \frac{24}{25}}. \end{align*} (b) $\cos 2 \theta$ \begin{align*} \cos 2\theta & = 1-2\sin^2\theta \\ &= 1
- Question 1, Exercise 8.1
- iven: $\alpha=180^{\circ}$, $\beta=60^{\circ}$. \begin{align*} \cos (\alpha + \beta) & = \cos \alpha \c... -\frac{1}{2} - 0 = -\frac{1}{2} \end{align*} \begin{align*} \cos (\alpha - \beta) & = \cos \alpha \c... = -\frac{1}{2} + 0 = -\frac{1}{2} \end{align*} \begin{align*} \sin (\alpha + \beta) & = \sin \alpha \c... {3}}{2} + 0 = -\frac{\sqrt{3}}{2} \end{align*} \begin{align*} \sin (\alpha - \beta) & = \sin \alpha \c
- Question 5 Exercise 8.2
- $ in QII, therefore $\cos 2\theta$ is negative. \begin{align*}\cos 2\theta & = - \sqrt{1-\sin^2 2\theta}... $\theta$ lies in QI and $\sin\theta > 0$. Thus \begin{align*} \sin\theta & = \sqrt{\frac{1-\cos 2\theta... 32}{25}}{2}} = \sqrt{\frac{16}{25}} \end{align*} \begin{align*} \implies \boxed{\sin\theta = \frac{4}{5}}... eta$ lies in QI, therefore $\cos\theta >0$, thus \begin{align*} \cos\theta & = \sqrt{1-\sin\theta} \\ &=\
- Question 5 and 6, Exercise 8.1
- $\alpha$ lies in QII and $\cos$ is -ive in QII, \begin{align*}\cos\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=-... s $\beta$ lies in QII and $\sec$ is -ive in QII, \begin{align*}\sec \beta & =-\sqrt{1+\tan^2\beta} \\ &=-... dfrac{169}{144}} =-\dfrac{13}{12} \end{align*} \begin{align*} \Rightarrow \quad \cos \beta=\frac{1}{\sec \beta}=-\frac{12}{13}.\end{align*} This gives \begin{align*} \frac{\sin\beta}{\cos\beta} & = \tan\beta
- Question 9, Exercise 8.1
- $\alpha$ lies in QII and $\cos$ is -ive in QII, \begin{align*} \cos\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=... s $\beta$ lies in QII and $\sin$ is +ive in QII, \begin{align*} \cos \beta & =\sqrt{1-\cos^2\beta} \\ &=\... sqrt{\dfrac{16}{25}} =\dfrac{4}{5} \end{align*} \begin{align*} \sin (\alpha + \beta) &= \sin \alpha \cos... qrt{2}} \\ &= -\frac{7}{5\sqrt{2}}. \end{align*} \begin{align*} \sin (\alpha - \beta) &= \sin \alpha \cos
- Question 2, Review Exercise
- where $\theta$ is obtuse and $\phi$ is acute. As\begin{align*} \cos^2 \theta &= 1-\sin^2\theta\\ &= 1-\l... theta <0$, thus $$\cos \theta=-\frac{4}{5}$$ Now \begin{align*} \cos^2 \phi &= 1-\sin^2\phi\\ &= 1-\left(... 0$, thus $$\cos \phi=\frac{12}{13}$$ As, we have \begin{align*} \sin(\theta -\phi)&=\sin \theta \cos \phi... where $\theta$ is obtuse and $\phi$ is acute. As\begin{align*} \cos^2 \theta &= 1-\sin^2\theta\\ &= 1-\l
- Question 6 Exercise 8.2
- ac{1}{2}\sin 2\theta$$ Put $\theta = 15^{\circ}$ \begin{align*} \sin 15^{\circ} \cos 15^{\circ} & = \frac... c} = \frac{1}{2} \times \frac{1}{2} \end{align*} \begin{align*} \implies \boxed{\sin 15^{\circ} \cos 15^{... ta= \cos 2\theta = $$ Put $\theta = 15^{\circ}$ \begin{align*} \cos^2 15^\circ - \sin^2 15^\circ & = \co... 30^\circ \\ & = \frac{\sqrt{3}}{2} \end{align*} \begin{align*} \implies \boxed{\cos^2 15^\circ - \sin^2
- Question 2, Exercise 8.1
- 45^{\circ}-30^{\circ}\right)$. ** Solution. ** \begin{align*} \cos 15^{\circ} & = \cos \left(45^{\circ}... 80^{\circ}-15^{\circ}\right)$. ** Solution. ** \begin{align*} \cos 165^{\circ} & = \cos \left(180^{\cir... ive Method (if $\cos 15^{\circ}$ is not given)** \begin{align*} \cos 165^{\circ} & = \cos \left(180^{\cir... }{2\sqrt{2}}$. Now, let's find $\cos 345^\circ$. \begin{align*} \cos 345^\circ & = \cos \left(360^\circ -
- Question 3, Exercise 8.1
- 90^{\circ}+30^{\circ}\right)$. ** Solution. ** \begin{align*} \cos 120^{\circ} & = \cos \left(180^{\cir... ^{\circ}\\ &= -\dfrac{1}{2}. \end{align*} Also \begin{align*} \cos 120^{\circ} & = \cos \left(90^{\circ... $ and then $\tan 120^{\circ}$. ** Solution. ** \begin{align*} \sin 120^{\circ} & = \sin \left(180^{\cir... \dfrac{\sqrt{3}}{2}. \end{align*} Also, we have \begin{align*} \cos 120^{\circ} & = \cos \left(180^{\cir
- Question 11, Exercise 8.1
- 270^{\circ}-\lambda\right)}=1$ ** Solution. ** \begin{align*} L.H.S & = \dfrac{\sin \left(180^{\circ}+\... (90^{\circ}+\alpha\right)}=-1$ ** Solution. ** \begin{align*} L.H.S & = \frac{\sin \left(90^{\circ}+\al... eta)}{\cos \alpha \cos \beta}$ ** Solution. ** \begin{align*} L.H.S & = \tan \alpha+\tan \beta \\ & = \... n ^{2} \alpha-\sin ^{2} \beta$ ** Solution. ** \begin{align*} L.H.S & = \sin (\alpha + \beta) \sin (\al
- Question 13, Exercise 8.1
- $ and $-5=r\sin \varphi$.\\ Squaring and adding \begin{align*} & (12)^2+(-5)^2=r^2 \cos^2\varphi+r^2 \si... \\ \implies & r=\sqrt{169}=13 \end{align*} Also \begin{align*} & \frac{-5}{12}=\frac{r\sin \varphi }{r\c... -1}\left(-\frac{5}{12}\right) \end{align*} Now \begin{align*} & 12\sin \theta +5\cos \theta \\ =& r\cos... \( 4 = r \sin \varphi \).\\ Squaring and adding: \begin{align*} & (3)^2 + (4)^2 = r^2 \cos^2 \varphi + r^
- Question 10, Exercise 8.1
- 2}-\alpha\right)=\cos \alpha$ ** Solution. ** \begin{align*} L.H.S & = \sin \left(\frac{\pi}{2}-\alpha... cos (\pi-\alpha)=-\cos \alpha$ ** Solution. ** \begin{align*} L.H.S & = \cos(\pi - \alpha) \\ & = \cos ... }}(\cos \alpha-\sin \alpha)$ ** Solution. ** \begin{align*} L.H.S &= \cos \left(\alpha+\frac{\pi}{4}\... }{2}(\cos \beta+\sin \beta)$ ** Solution. ** \begin{align*} L.H.S & = \cos \left(\alpha + \frac{\pi}{
- Question 1, 2 and 3 Exercise 8.2
- -nbf-ex8-2-q1.svg |}} Given: $x=-3$ and $y=4$. \begin{align*} r&= \sqrt{(-3)^2+4^2} \\ &=\sqrt{25} = 5.... 5} \text{ and } \cos\theta = -\frac{3}{5}.$$ Now \begin{align*} \sin2\theta&= 2\sin\theta \cos\theta\\ &=... \right) \\ & = -\frac{24}{25}. \end{align*} and \begin{align*} \cos2\theta&= 1-2\sin^2\theta\\ &= 1-2\le... in QII and $\cos$ is negative in QII, therefore \begin{align*} \cos \alpha & = - \sqrt{1-\sin^2 \alpha}
- Question 7, Exercise 8.1
- pha\) lies in QI and \(\cos\) is positive in QI, \begin{align*} \cos \alpha & = \sqrt{1-\sin^2\alpha} \\ ... eta\) lies in QI and \(\sec\) is positive in QI, \begin{align*} \sec \beta & = \sqrt{1+\tan^2\beta} \\ &=... frac{25}{9}} = \dfrac{5}{3}. \end{align*} Thus, \begin{align*} \cos \beta & = \frac{1}{\sec \beta} = \frac{3}{5}. \end{align*} As \begin{align*} \frac{\sin\beta}{\cos\beta} & = \tan\beta
- Question 8, Exercise 8.1
- pha<\dfrac{\pi}{2}$, i.e. $\alpha$ lies in QI.\\ \begin{align*} \cos \alpha &= \sqrt{1 - \sin^2 \alpha} \... \pi}{2}<\beta<2 \pi$, i.e. $\beta$ lies in QIV. \begin{align*} \sin \beta &= -\sqrt{1 - \cos^2 \beta} \\... \ \sin \beta & = -\frac{5}{13}. \end{align*} (i) \begin{align*} \sin (\alpha + \beta) &= \sin \alpha \cos... \frac{20}{65} \\ &= \frac{16}{65}. \end{align*} \begin{align*} \csc (\alpha + \beta) &= \frac{1}{\sin (\