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- Question 4 Exercise 8.2
- \cos 2 \theta$ %%(c)%% $\tan 2 \theta$ (d) $\sin \frac{\theta}{2}$ (e) $\cos \frac{\theta}{2}$ (f) $\tan \frac{\theta}{2}$ when: $\cos \theta=\frac{3}{5}$ where $0<\theta<\frac{\pi}{2}$ ** Solution. ** Given: $\c
- Question 1, Exercise 8.1
- \sin 60 \\ \implies \cos (180+60) & = (-1)\left(\frac{1}{2}\right) - (0)\left(\frac{\sqrt{3}}{2}\right) \\ & = -\frac{1}{2} - 0 = -\frac{1}{2} \end{align*} \begin{align*} \cos (\alpha - \beta) & = \cos \alpha \cos \be
- Question 5 Exercise 8.2
- eta$ using the information given: $\sin 2 \theta=\frac{24}{25}, 2 \theta$ in QII ** Solution. ** Given: $\sin 2\theta=\dfrac{24}{25}$, $2\theta$ in QII. We have $$\cos 2\the... = - \sqrt{1-\sin^2 2\theta}\\ &=- \sqrt{1-\left(\frac{24}{25}\right)^2} \\ &=- \sqrt{\frac{49}{625}} = -\frac{7}{25} \end{align*} Also we have $$\sin\theta
- Question 12, Exercise 8.1
- +\beta) = \tan(180^{\circ}-\gamma) \\ \implies & \frac{\tan\alpha + \tan\beta}{1-\tan\alpha \tan\beta} ... pha+\beta+\gamma=180^{\circ}$, prove that: $\cot \frac{\alpha}{2}+\cot \frac{\beta}{2}+\cot \frac{\gamma}{2}=\cot \frac{\alpha}{2} \cot \frac{\beta}{2} \cot \frac{\gamma}{2}$ ** S
- 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 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 $\sin ... ta-\phi)$. ** Solution. ** Given: $\sin \theta=\dfrac{3}{5}$ and $\sin \phi=\dfrac{5}{13}$, where $\theta$ is obtuse and $\phi$ is acute. As\begin{align*} \c
- Question 5 and 6, Exercise 8.1
- stan. ===== Question 5===== For $\sin \alpha=\dfrac{4}{5}, \tan \beta=-\dfrac{5}{12}$ with terminal side of an angles in QII, find $\cos (\alpha+\beta)$ an... -\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 4 Exercise 8.3
- \cos 60^{\circ} \cos 40^{\circ} \cos 20^{\circ}=\dfrac{1}{16}$ ** Solution. ** \begin{align*} LHS &= \c... 40^\circ \cos 20^\circ \\ &= \cos 80^\circ \left(\frac{1}{2}\right) \cos 40^\circ \cos 20^\circ \\ &= \frac{1}{2} \left( \cos 80^\circ \cos 40^\circ \right) \cos 20^\circ \\ &= \frac{1}{4} \left(2 \cos 80^\circ \cos 40^\circ \right
- Question 11, Exercise 8.1
- istan. ===== Question 11(i)===== Show that: $\dfrac{\sin \left(180^{\circ}+\lambda\right) \cos \left(... }=1$ ** Solution. ** \begin{align*} L.H.S & = \dfrac{\sin \left(180^{\circ}+\lambda\right) \cos \left(... t) \cos \left(270^{\circ}-\lambda\right)} \\ &= \dfrac{\sin \left(2(90)+\lambda\right) \cos \left(3(90)+... da\right) \cos \left(3(90)-\lambda\right)}\\ &= \dfrac{(-\sin \lambda) (\sin\lambda)}{(\sin\lambda)(-\si
- Question 10, Exercise 8.1
- ===== Question 10(i)===== Verify: $\sin \left(\dfrac{\pi}{2}-\alpha\right)=\cos \alpha$ ** Solution. ** \begin{align*} L.H.S & = \sin \left(\frac{\pi}{2}-\alpha\right) \\ & =\sin\frac{\pi}{2} \cos \alpha - \cos \frac{\pi}{2} \sin\alpha \\ & = 1\times \cos \alpha - 0 \times \sin\alpha \\
- Question 2, Exercise 8.1
- t)\\ &= \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{\s
- Question 3(vi, vii, viii, ix & x) Exercise 8.3
- n{align*} LHS & = 2\tan y \cos 3y \\ & = 2 \cdot \frac{\sin y}{\cos y} \cos 3y \\ & = \sec y (2 \cos 3y ... OD =====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) \
- 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} \
- Question 8, Exercise 8.1
- istan. ===== 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+\b
- Question 6 Exercise 8.2
- cos\theta$$ This gives $$\sin\theta \cos\theta = \frac{1}{2}\sin 2\theta$$ Put $\theta = 15^{\circ}$ \begin{align*} \sin 15^{\circ} \cos 15^{\circ} & = \frac{1}{2}\sin 2(15^{\circ}) \\ & \frac{1}{2}\sin 30^{\circ} = \frac{1}{2} \times \frac{1}{2} \end{align*} \begin{align*} \implies \boxed{\sin