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- Question 4 Exercise 8.2
- {9}{25}}\\ &= \sqrt{\frac{16}{25}} = \frac{4}{5} \end{align*} (a) $\sin 2 \theta$ \begin{align*} \sin ... (\frac{4}{5} \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 ... \right)^2\\ &= 1-2\left(\frac{16}{25} \right)\\ \end{align*} \begin{align*} \implies \boxed{\cos 2\the
- Question 1, Exercise 8.1
- \right) \\ & = -\frac{1}{2} - 0 = -\frac{1}{2} \end{align*} \begin{align*} \cos (\alpha - \beta) & ... }\right) \\ & = -\frac{1}{2} + 0 = -\frac{1}{2} \end{align*} \begin{align*} \sin (\alpha + \beta) & ... 0-\frac{\sqrt{3}}{2} + 0 = -\frac{\sqrt{3}}{2} \end{align*} \begin{align*} \sin (\alpha - \beta) & ... = 0+\frac{\sqrt{3}}{2} + 0 = \frac{\sqrt{3}}{2} \end{align*} \begin{align*} \tan (\alpha + \beta) &
- Question 5 Exercise 8.2
- ^2} \\ &=- \sqrt{\frac{49}{625}} = -\frac{7}{25} \end{align*} Also we have $$\sin\theta = \pm \sqrt{\f... {\frac{\frac{32}{25}}{2}} = \sqrt{\frac{16}{25}} \end{align*} \begin{align*} \implies \boxed{\sin\theta = \frac{4}{5}} \end{align*} Also $$\cos\theta = \pm\sqrt{1-\sin\theta... & = \sqrt{1-\frac{16}{25}} = \sqrt{\frac{9}{25}} \end{align*} \begin{align*} \implies \boxed{\cos\theta
- Question 5 and 6, Exercise 8.1
- }\\ \Rightarrow \quad \cos \alpha&=-\dfrac{3}{5}.\end{align*} Also $$\sec \beta=\pm\sqrt{1+\tan^2\beta}... \\ & =-\sqrt{\dfrac{169}{144}} =-\dfrac{13}{12} \end{align*} \begin{align*} \Rightarrow \quad \cos \beta=\frac{1}{\sec \beta}=-\frac{12}{13}.\end{align*} This gives \begin{align*} \frac{\sin\beta... \right) \\ \implies \sin\beta & = \frac{5}{13}. \end{align*} Now \begin{align*}\cos (\alpha+\beta )&=\
- Question 9, Exercise 8.1
- }}\\ \implies \cos \alpha&=-\dfrac{1}{\sqrt{2}}. \end{align*} Also $$\sin\beta=\pm\sqrt{1-\sin^2\beta}... {25}} \\ & =\sqrt{\dfrac{16}{25}} =\dfrac{4}{5} \end{align*} \begin{align*} \sin (\alpha + \beta) &= \... \frac{4}{5\sqrt{2}} \\ &= -\frac{7}{5\sqrt{2}}. \end{align*} \begin{align*} \sin (\alpha - \beta) &= \... + \frac{4}{5\sqrt{2}} \\ &= \frac{1}{5\sqrt{2}}. \end{align*} ===== Question 9(ii)===== Given $\alpha$
- Question 2, Review Exercise
- frac{16}{25} \\ \cos \theta&=\pm\frac{4}{5}\\ \end{align*} As $\theta$ is obtuse, so $\theta$ lies i... ac{144}{169} \\ \cos \phi&=\pm\frac{12}{13}\\ \end{align*} As $\phi$ is acute, so $\phi$ lies in I Q... 65}\\ \implies \sin(\theta -\phi)&=\frac{56}{65} \end{align*} =====Question 2(ii)===== Given that $\si... frac{16}{25} \\ \cos \theta&=\pm\frac{4}{5}\\ \end{align*} As $\theta$ is obtuse, so $\theta$ lies i
- Question 6 Exercise 8.2
- \sin 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 do... ) \\ & = \cos 30^\circ \\ & = \frac{\sqrt{3}}{2} \end{align*} \begin{align*} \implies \boxed{\cos^2 15^\circ - \sin^2 15^\circ = \frac{\sqrt{3}}{2}} \end{align*} =====Question 6(iii)===== Use a double-a
- Question 2, Exercise 8.1
- 2\sqrt{2}} \\ & = \dfrac{\sqrt{3}+1}{2\sqrt{2}}. \end{align*} GOOD ===== Question 2(b)===== Use t... cos\theta)\\ & = -\dfrac{\sqrt{3}+1}{2\sqrt{2}}. \end{align*} GOOD **Alternative Method (if $\cos 15^{... \right] \\ & = -\dfrac{\sqrt{3}+1}{2\sqrt{2}}. \end{align*} GOOD ===== Question 2(c)===== Use the va... os \theta) \\ & = \dfrac{\sqrt{3}+1}{2\sqrt{2}}. \end{align*} Thus, $\cos 345^\circ = \dfrac{\sqrt{3}+1
- Question 3, Exercise 8.1
- ht) \\ &= - \cos 60 ^{\circ}\\ &= -\dfrac{1}{2}. \end{align*} Also \begin{align*} \cos 120^{\circ} & =... ght) \\ &= - \sin 30^{\circ}\\ &= -\dfrac{1}{2}. \end{align*} GOOD ===== Question 3(b)===== Find the ex... \ &= \sin 60 ^{\circ}\\ &= \dfrac{\sqrt{3}}{2}. \end{align*} Also, we have \begin{align*} \cos 120^{\... ht) \\ &= - \cos 60 ^{\circ}\\ &= -\dfrac{1}{2}. \end{align*} Now \begin{align*} \tan 120^{\circ} & =
- Question 11, Exercise 8.1
- )}{(\sin\lambda)(-\sin\lambda)} \\ & = 1 = R.H.S \end{align*} GOOD ===== Question 11(ii)===== Show th... } \\ & = -1 \quad \text{(if $\tan \alpha = -1$)} \end{align*} ===== Question 11(iii)===== Show that: $... pha+\beta)}{\cos \alpha \cos \beta} \\ & = R.H.S \end{align*} GOOD ===== Question 11(iv)===== Show tha... \\ & = \left(\cos^2 \beta - \cos^2 \alpha\right) \end{align*} Now\begin{align*}\left(\cos^2 \beta - \co
- Question 13, Exercise 8.1
- }\left( 1 \right) \\ \implies & r=\sqrt{169}=13 \end{align*} Also \begin{align*} & \frac{-5}{12}=\frac... s & \varphi =\tan^{-1}\left(-\frac{5}{12}\right) \end{align*} Now \begin{align*} & 12\sin \theta +5\c... ht) \\ =& r\sin \left( \theta +\varphi \right), \end{align*} where $r=13$ and $\varphi =\tan^{-1}\left... left( 1 \right) \\ \implies & r = \sqrt{25} = 5. \end{align*} Also \begin{align*} & \frac{4}{3} = \frac
- Question 10, Exercise 8.1
- - 0 \times \sin\alpha \\ & = \cos\alpha = R.H.S \end{align*} GOOD ===== Question 10(ii)===== Verify:... ot \sin \alpha \\ & = -\cos \alpha \\ & = R.H.S. \end{align*} ===== Question 10(iii)===== Verify: $\c... }{\sqrt{2}}(\cos \alpha-\sin \alpha)\\ & = R.H.S \end{align*} GOOD ===== Question 10(iv)===== Verify:... \cos \alpha - \sin \alpha \right) \\ & = R.H.S. \end{align*} ===== Question 10(v)===== Verify: $\tan
- Question 1, 2 and 3 Exercise 8.2
- lign*} r&= \sqrt{(-3)^2+4^2} \\ &=\sqrt{25} = 5. \end{align*} Thus $$\sin\theta = \frac{4}{5} \text{ an... left(-\frac{3}{5} \right) \\ & = -\frac{24}{25}. \end{align*} and \begin{align*} \cos2\theta&= 1-2\sin... t)^2 \\ & = 1-\frac{32}{25} \\ & = -\frac{7}{25} \end{align*} Since $\sin2\theta$ and $\cos 2\theta$ bo... = - \sqrt{1-\sin^2 \alpha} \\ & =- \sqrt{1-y^2} \end{align*}. Now \begin{align*} & \sin 2\alpha = 2 \
- Question 7, Exercise 8.1
- }} \\ &= \sqrt{\dfrac{25}{169}} = \dfrac{5}{13}. \end{align*} Also, $$\sec \beta=\pm\sqrt{1+\tan^2\bet... }{9}} \\ &= \sqrt{\dfrac{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\... 5}\right) \\ \implies \sin\beta & = \frac{4}{5}. \end{align*} (i) $\sin(\alpha + \beta)$ \begin{align*
- Question 8, Exercise 8.1
- rt{\frac{16}{25}}\\ \cos \alpha & = \frac{4}{5}. \end{align*} $\cos \beta=\dfrac{12}{13}$, where $\dfra... {\frac{25}{169}}\\ \sin \beta & = -\frac{5}{13}. \end{align*} (i) \begin{align*} \sin (\alpha + \beta) ... rac{36}{65} - \frac{20}{65} \\ &= \frac{16}{65}. \end{align*} \begin{align*} \csc (\alpha + \beta) &= \... &= \frac{1}{\dfrac{16}{65}} \\ &= \frac{65}{16}. \end{align*} (ii) \begin{align*} \cos (\alpha + \beta