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Question 5 Exercise 8.2
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s \theta$ and $\tan \theta$ 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.... s \theta$ and $\tan \theta$ using the information given: $\cos 2 \theta=-\frac{7}{25}, 2 \theta$ in QIII ** Solution. ** Given: \(\cos 2\theta = -\dfrac{7}{25}\) and \(2\theta\
Question 1, Exercise 8.1
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=180^{\circ}, \beta=60^{\circ}$ ** Solution. ** Given: $\alpha=180^{\circ}$, $\beta=60^{\circ}$. \begi... a=60^{\circ}, \beta=90^{\circ}$ ** Solution. ** Given: \(\alpha = 60^\circ\), \(\beta = 90^\circ\) \beg... =180^{\circ}, \beta=30^{\circ}$ ** Solution. ** Given: $\alpha = 180^\circ$, $\beta = 30^\circ$. \begin... pha=\pi, \beta=\frac{2 \pi}{3}$ ** Solution. ** Given: $\alpha = \pi$, $\beta = \frac{2\pi}{3}$. \begin
Question 9, Exercise 8.1
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, Islamabad, Pakistan. ===== Question 9(i)===== Given $\alpha$ and $\beta$ are obtuse angles with $\sin... ind: $\sin (\alpha \pm \beta)$ ** Solution. ** Given: $\sin \alpha=\dfrac{1}{\sqrt{2}}$, $\alpha$ is o... sqrt{2}}. \end{align*} ===== Question 9(ii)===== Given $\alpha$ and $\beta$ are obtuse angles with $\sin... find: $\cos (\alpha \pm \beta)$ ** Solution. ** Given: $\sin \alpha=\dfrac{1}{\sqrt{2}}$, $\alpha$ is o
Question 4 Exercise 8.2
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where $0<\theta<\frac{\pi}{2}$ ** Solution. ** Given: $\cos\theta=\dfrac{3}{5}$ where $0<\theta<\dfrac... re $\pi<\theta<\frac{3 \pi}{2}$ ** Solution. ** Given: \(\tan \theta = \frac{12}{5}\) where \(\pi < \th... $\frac{3 \pi}{2}<\theta<2 \pi$ ** Solution. ** Given: \(\sin \theta = -\frac{7}{25}\) where \(\frac{3\... $\frac{3 \pi}{2}<\theta<2 \pi$ ** Solution. ** Given: \(\sec \theta = \sqrt{5}\) where \(\frac{3\pi}{2
Question 2, Review Exercise
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d, Islamabad, Pakistan. =====Question 2(i)===== Given that $\sin \theta=\dfrac{3}{5}, \sin \phi=\dfrac{... values of $\sin (\theta-\phi)$. ** Solution. ** Given: $\sin \theta=\dfrac{3}{5}$ and $\sin \phi=\dfrac... ac{56}{65} \end{align*} =====Question 2(ii)===== Given that $\sin \theta=\dfrac{3}{5}, \sin \phi=\dfrac{... values of $\tan (\theta-\phi)$. ** Solution. ** Given $\sin \theta=\dfrac{3}{5}$ and $\sin \phi=\dfrac{
Question 2, Exercise 8.1
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**Alternative Method (if $\cos 15^{\circ}$ is not given)** \begin{align*} \cos 165^{\circ} & = \cos \left... irc}-15^{\circ}\right)$. ** Solution. ** We are given that $\cos 15^\circ = \dfrac{\sqrt{3}+1}{2\sqrt{2... **Alternative Method (if $\cos 15^\circ$ is not given)** To find $\cos 345^\circ$, we use the identity... nd then find $\tan 75^{\circ}$. ** Solution. ** Given $$\cos A=\sin \left(90^{\circ}-A\right) ... (1)$$
Question 12, Exercise 8.1
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alpha \tan \beta \tan \gamma$. ** Solution. ** Given: $$\alpha+\beta+\gamma=180^{\circ}$$ This gives \... beta}{2} \cot \frac{\gamma}{2}$ ** Solution. ** Given: $$\alpha+\beta+\gamma=180^{\circ}$$ This gives \... }{2} \tan \frac{\alpha}{2}+1=0$ ** Solution. ** Given: $$\alpha+\beta+\gamma=180^{\circ}$$ This gives \
Question 5 and 6, Exercise 8.1
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ta)$ and $\cos (\alpha-\beta)$. ** Solution. ** Given: $\sin \alpha=\dfrac{4}{5}$, $\alpha$ is in QII a... a)$ (iii) $\tan(\alpha-\beta)$. ** Solution. ** Given: $\cos\alpha=-\dfrac{7}{25}$, $\alpha$ is in QII
Question 7, Exercise 8.1
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rd, Islamabad, Pakistan. ===== Question 7===== Given $\alpha$ and $\beta$ are acute angles with $\sin ... a)$ (iii) $\tan(\alpha+\beta)$. ** Solution. ** Given: $\sin \alpha=\dfrac{12}{13}$, where $\alpha$ is
Question 1, 2 and 3 Exercise 8.2
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h-11-nbf:sol:unit08:math-11-nbf-ex8-2-q1.svg |}} Given: $x=-3$ and $y=4$. \begin{align*} r&= \sqrt{(-3)... \tan 2 \alpha$ in terms of $y$. ** Solution. ** Given: $\sin \alpha=y$ and $\alpha$ lies in QII. We ha
Question 8, Exercise 8.1
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)$ (iii) $\cot (\alpha+\beta)$ ** Solution. ** Given: $\sin \alpha=\dfrac{3}{5}$, where $0<\alpha<\dfr
Question 5 and 6, Review Exercise
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a)$ prove that $\tan \alpha=1$. ** Solution. ** Given: \begin{align*} & \sin (\alpha - \theta) = \cos
Question 7, Review Exercise
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a)$ prove that $\tan \alpha=1$. ** Solution. ** Given: \begin{align*} & \sin (\alpha - \theta) = \cos