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- Question 4 Exercise 8.2
- $ 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
- $\alpha=180^{\circ}$, $\beta=60^{\circ}$. \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
- 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 $$\sin\theta = \pm \sqrt{\frac{1-... ta$ lies in QI and $\sin\theta > 0$. Thus \begin{align*} \sin\theta & = \sqrt{\frac{1-\cos 2\theta}{2}} ... ac{\frac{32}{25}}{2}} = \sqrt{\frac{16}{25}} \end{align*} \begin{align*} \implies \boxed{\sin\theta = \fr
- Question 5 and 6, Exercise 8.1
- ha$ lies in QII and $\cos$ is -ive in QII, \begin{align*}\cos\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=-\sqrt{... \Rightarrow \quad \cos \alpha&=-\dfrac{3}{5}.\end{align*} Also $$\sec \beta=\pm\sqrt{1+\tan^2\beta}.$$ As... ta$ lies in QII and $\sec$ is -ive in QII, \begin{align*}\sec \beta & =-\sqrt{1+\tan^2\beta} \\ &=-\sqrt{... =-\sqrt{\dfrac{169}{144}} =-\dfrac{13}{12} \end{align*} \begin{align*} \Rightarrow \quad \cos \beta=\f
- Question 9, Exercise 8.1
- ha$ lies in QII and $\cos$ is -ive in QII, \begin{align*} \cos\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=-\sqrt... \implies \cos \alpha&=-\dfrac{1}{\sqrt{2}}. \end{align*} Also $$\sin\beta=\pm\sqrt{1-\sin^2\beta}.$$ As... ta$ lies in QII and $\sin$ is +ive in QII, \begin{align*} \cos \beta & =\sqrt{1-\cos^2\beta} \\ &=\sqrt{1... } \\ & =\sqrt{\dfrac{16}{25}} =\dfrac{4}{5} \end{align*} \begin{align*} \sin (\alpha + \beta) &= \sin \a
- Question 2, Review Exercise
- $\theta$ is obtuse and $\phi$ is acute. As\begin{align*} \cos^2 \theta &= 1-\sin^2\theta\\ &= 1-\left(\f... {16}{25} \\ \cos \theta&=\pm\frac{4}{5}\\ \end{align*} As $\theta$ is obtuse, so $\theta$ lies in II Q... <0$, thus $$\cos \theta=-\frac{4}{5}$$ Now \begin{align*} \cos^2 \phi &= 1-\sin^2\phi\\ &= 1-\left(\frac{... 44}{169} \\ \cos \phi&=\pm\frac{12}{13}\\ \end{align*} As $\phi$ is acute, so $\phi$ lies in I Q. This
- Question 6 Exercise 8.2
- 2}\sin 2\theta$$ Put $\theta = 15^{\circ}$ \begin{align*} \sin 15^{\circ} \cos 15^{\circ} & = \frac{1}{2}... 30^{\circ} = \frac{1}{2} \times \frac{1}{2} \end{align*} \begin{align*} \implies \boxed{\sin 15^{\circ} \cos 15^{\circ} = \frac{1}{4}} \end{align*} GOOD =====Question 6(ii)===== Use a double-a
- Question 2, Exercise 8.1
- irc}-30^{\circ}\right)$. ** Solution. ** \begin{align*} \cos 15^{\circ} & = \cos \left(45^{\circ}-30^{\... rt{2}} \\ & = \dfrac{\sqrt{3}+1}{2\sqrt{2}}. \end{align*} GOOD ===== Question 2(b)===== Use the val... irc}-15^{\circ}\right)$. ** Solution. ** \begin{align*} \cos 165^{\circ} & = \cos \left(180^{\circ}-15^... theta)\\ & = -\dfrac{\sqrt{3}+1}{2\sqrt{2}}. \end{align*} GOOD **Alternative Method (if $\cos 15^{\circ}
- Question 3, Exercise 8.1
- irc}+30^{\circ}\right)$. ** Solution. ** \begin{align*} \cos 120^{\circ} & = \cos \left(180^{\circ}-60^... \\ &= - \cos 60 ^{\circ}\\ &= -\dfrac{1}{2}. \end{align*} Also \begin{align*} \cos 120^{\circ} & = \cos \left(90^{\circ}+30^{\circ}\right) \\ &= - \sin 30^{\circ}\\ &= -\dfrac{1}{2}. \end{align*} GOOD ===== Question 3(b)===== Find the exact va
- Question 11, Exercise 8.1
- circ}-\lambda\right)}=1$ ** Solution. ** \begin{align*} L.H.S & = \dfrac{\sin \left(180^{\circ}+\lambda... \sin\lambda)(-\sin\lambda)} \\ & = 1 = R.H.S \end{align*} GOOD ===== Question 11(ii)===== Show that: $\... circ}+\alpha\right)}=-1$ ** Solution. ** \begin{align*} L.H.S & = \frac{\sin \left(90^{\circ}+\alpha\ri... & = -1 \quad \text{(if $\tan \alpha = -1$)} \end{align*} ===== Question 11(iii)===== Show that: $\tan \
- Question 13, Exercise 8.1
- $-5=r\sin \varphi$.\\ Squaring and adding \begin{align*} & (12)^2+(-5)^2=r^2 \cos^2\varphi+r^2 \sin^2 \v... ft( 1 \right) \\ \implies & r=\sqrt{169}=13 \end{align*} Also \begin{align*} & \frac{-5}{12}=\frac{r\sin \varphi }{r\cos \varphi } \\ \implies & \frac{-5}{12}=... \varphi =\tan^{-1}\left(-\frac{5}{12}\right) \end{align*} Now \begin{align*} & 12\sin \theta +5\cos \th
- Question 10, Exercise 8.1
- pha\right)=\cos \alpha$ ** Solution. ** \begin{align*} L.H.S & = \sin \left(\frac{\pi}{2}-\alpha\right... \times \sin\alpha \\ & = \cos\alpha = R.H.S \end{align*} GOOD ===== Question 10(ii)===== Verify: $\cos... pi-\alpha)=-\cos \alpha$ ** Solution. ** \begin{align*} L.H.S & = \cos(\pi - \alpha) \\ & = \cos \pi \c... sin \alpha \\ & = -\cos \alpha \\ & = R.H.S. \end{align*} ===== Question 10(iii)===== Verify: $\cos \le
- Question 1, 2 and 3 Exercise 8.2
- x8-2-q1.svg |}} Given: $x=-3$ and $y=4$. \begin{align*} r&= \sqrt{(-3)^2+4^2} \\ &=\sqrt{25} = 5. \end{align*} Thus $$\sin\theta = \frac{4}{5} \text{ and } \cos\theta = -\frac{3}{5}.$$ Now \begin{align*} \sin2\theta&= 2\sin\theta \cos\theta\\ &= 2\lef... (-\frac{3}{5} \right) \\ & = -\frac{24}{25}. \end{align*} and \begin{align*} \cos2\theta&= 1-2\sin^2\the
- Question 7, Exercise 8.1
- lies in QI and \(\cos\) is positive in QI, \begin{align*} \cos \alpha & = \sqrt{1-\sin^2\alpha} \\ &= \sq... \ &= \sqrt{\dfrac{25}{169}} = \dfrac{5}{13}. \end{align*} Also, $$\sec \beta=\pm\sqrt{1+\tan^2\beta}.$$ ... lies in QI and \(\sec\) is positive in QI, \begin{align*} \sec \beta & = \sqrt{1+\tan^2\beta} \\ &= \sqrt... } \\ &= \sqrt{\dfrac{25}{9}} = \dfrac{5}{3}. \end{align*} Thus, \begin{align*} \cos \beta & = \frac{1}{\
- Question 8, Exercise 8.1
- frac{\pi}{2}$, i.e. $\alpha$ lies in QI.\\ \begin{align*} \cos \alpha &= \sqrt{1 - \sin^2 \alpha} \\ &= \... frac{16}{25}}\\ \cos \alpha & = \frac{4}{5}. \end{align*} $\cos \beta=\dfrac{12}{13}$, where $\dfrac{3 \p... 2}<\beta<2 \pi$, i.e. $\beta$ lies in QIV. \begin{align*} \sin \beta &= -\sqrt{1 - \cos^2 \beta} \\ &= -\... ac{25}{169}}\\ \sin \beta & = -\frac{5}{13}. \end{align*} (i) \begin{align*} \sin (\alpha + \beta) &= \si