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- Question 1, Exercise 8.1
- ign*} \begin{align*} \tan (\alpha + \beta) & = \dfrac{\tan\alpha + \tan\beta}{1-\tan\alpha \tan\beta} \\ \implies \tan (180 + 60) & = \dfrac{\tan 180 + \tan 60}{1-\tan 180 \tan 60} \\ \implies \tan (180 + 60) & = \dfrac{0 + \sqrt{3}}{1-(0)(\sqrt{3})} \\ & = \dfrac{\sqrt{3}}{1-0} = \sqrt{3} \end{align*} \begin{align*} \t
- Question 2, Exercise 8.1
- ht)\\ &= \cos 45 \cos 30 + \sin 45 \sin 30 \\ &= \dfrac{1}{\sqrt{2}}\cdot \dfrac{\sqrt{3}}{2} + \dfrac{1}{\sqrt{2}}\cdot \dfrac{1}{2} \\ & = \dfrac{\sqrt{3}}{2\sqrt{2}}+\dfrac{1}{2\sqrt{2}} \\ & = \dfrac{\sq
- Question 9, Exercise 8.1
- 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 4 Exercise 8.2
- ac{\pi}{2}$ ** Solution. ** Given: $\cos\theta=\dfrac{3}{5}$ where $0<\theta<\dfrac{\pi}{2}$, i.e. $\theta$ lies in QI. We have $$\sin\theta = \pm \sqrt{1-\c... theta = -\frac{24}{7}}. \end{align*} (d) $\sin \dfrac{\theta}{2}$ We have $$\sin\left(\frac{\theta}{2}... pm \sqrt{\frac{1-\cos\theta}{2}}$$ As $0<\theta<\dfrac{\pi}{2}$ implies $0<\dfrac{\theta}{2}<\dfrac{\pi}
- Question 7, Exercise 8.1
- $ and $\beta$ are acute angles with $\sin \alpha=\dfrac{12}{13}$ and $\tan \beta=\dfrac{4}{3}$ find \\ (i) $\sin(\alpha+\beta)$ (ii) $\cos(\alpha+\beta)$ (iii) ... a+\beta)$. ** Solution. ** Given: $\sin \alpha=\dfrac{12}{13}$, where $\alpha$ is acute angle, i.e. is in QI.\\ $\tan \beta=\dfrac{4}{3}$, where $\beta$ is acute angle, i.e. is in
- Question 5 and 6, Exercise 8.1
- istan. ===== Question 5===== For $\sin \alpha=\dfrac{4}{5}, \tan \beta=-\dfrac{5}{12}$ with terminal side of an angles in QII, find $\cos (\alpha+\beta)$ and... a-\beta)$. ** Solution. ** Given: $\sin \alpha=\dfrac{4}{5}$, $\alpha$ is in QII and $\tan \beta=-\dfrac{5}{12}$, $\beta$ is in QII. We have an identity: $$\
- Question 8(xvi, xvii & xviii) Exercise 8.2
- ====Question 8(xvi)===== Verify the identities: $\dfrac{1-\cos ^{2} \beta}{2-2 \cos \beta}=\cos ^{2} \dfrac{\beta}{2}$ ** Solution. ** \begin{align*} LHS &= \dfrac{1-\cos ^{2} \beta}{2-2 \cos \beta}\\ &= \dfrac{\sin ^{2} \beta}{2-2 \cos \beta}\\ &=\dfrac{4\sin ^{2} \f
- Question 3(vi, vii, viii, ix & x) Exercise 8.3
- OOD =====Questio 3(vii)===== Prove the identity $\dfrac{ \sin 6 \beta + \sin 4 \beta}{\sin 6 \beta - \sin... \beta$ ** Solution. ** \begin{align*} LHS & = \dfrac{ \sin 6 \beta + \sin 4 \beta}{\sin 6 \beta - \sin 4 \beta} \\ & = \dfrac{2 \sin \left( \frac{6\beta + 4\beta}{2} \right) \... \left( \frac{6\beta - 4\beta}{2} \right)} \\ & = \dfrac{\sin 5\beta \cos 5\beta}{\sin \beta \cos \beta} \
- Question 8(xiii, xiv & xv) Exercise 8.2
- ====Question 8(xiv)===== Verify the identities: $\dfrac{\cos 3 x-\sin 3 x}{\cos x-\sin x}=\dfrac{2+\sin 2 x}{2}$ ** Solution. ** \begin{align*} LHS & = \dfrac{\cos 3 x-\sin 3 x}{\cos x-\sin x} \\ & = \dfrac{(\cos 3 x-\sin 3 x)(\cos x + \sin x)}{(\cos x-\sin x)((\
- Question 3(i, ii, iii, iv & v) Exercise 8.3
- an. =====Questio 3(i)===== Prove the identity $\dfrac{\cos (\alpha + \beta)}{\cos(\alpha - \beta)}=\dfrac{1- \tan \alpha \tan \beta}{1+ \tan \alpha \tan \beta}$ ** Solution. ** \begin{align*} RHS & = \dfrac{1- \tan \alpha \tan \beta}{1+ \tan \alpha \tan \beta} \\ & = \dfrac{1- \frac{\sin\alpha}{\cos\alpha} \frac{\sin\beta}
- Question 3, Exercise 8.1
- 60^{\circ}\right) \\ &= - \cos 60 ^{\circ}\\ &= -\dfrac{1}{2}. \end{align*} Also \begin{align*} \cos 120... +30^{\circ}\right) \\ &= - \sin 30^{\circ}\\ &= -\dfrac{1}{2}. \end{align*} GOOD ===== Question 3(b)=====... }-60^{\circ}\right) \\ &= \sin 60 ^{\circ}\\ &= \dfrac{\sqrt{3}}{2}. \end{align*} Also, we have \begin{... 60^{\circ}\right) \\ &= - \cos 60 ^{\circ}\\ &= -\dfrac{1}{2}. \end{align*} Now \begin{align*} \tan 120^
- Question 8, Exercise 8.1
- kistan. ===== Question 8===== If $\sin \alpha=\dfrac{3}{5}$, where $0<\alpha<\dfrac{\pi}{2}$ and $\cos \beta=\dfrac{12}{13}$, where $\dfrac{3 \pi}{2}<\beta<2 \pi$ find: \\ (i) $\csc (\alpha+\beta)$ (ii) $\sec (\alpha+\be
- Question 2, Review Exercise
- =====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 $\sin (... eta-\phi)$. ** Solution. ** Given: $\sin \theta=\dfrac{3}{5}$ and $\sin \phi=\dfrac{5}{13}$, where $\theta$ is obtuse and $\phi$ is acute. As\begin{align*} \co
- Question 6 Exercise 8.2
- $$1-2\sin^2 \alpha=\cos 2\alpha.$$ Put $\alpha= \dfrac{\pi}{8}$, we have \begin{align*} 1-2\sin^2 \left(... $$2\cos^2 \alpha -1=\cos 2\alpha.$$ Put $\alpha= \dfrac{\pi}{12}$, we have \begin{align*} 2\cos^2 \left(\... t values for the expression: $\frac{2 \tan \left(\dfrac{\pi}{12}\right)}{1-\tan ^{2}\left(\dfrac{\pi}{12}\right)}$ ** Solution. ** We have a double-angle iden
- Question 8(x, xi & xii) Exercise 8.2
- stion 8(x)===== Verify the identities: $\sec 2 x=\dfrac{\cos x}{\cos x+\sin x}+\dfrac{\sin x}{\cos x-\sin x}$ ** Solution. ** \begin{align*} RHS &= \dfrac{\cos x}{\cos x+\sin x}+\dfrac{\sin x}{\cos x-\sin x}\\ &=\dfrac{\cos x(\cos x-\sin x)+\sin x(\cos x+\sin