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- Question 4 Exercise 8.2
- theta<\dfrac{\pi}{2}$, i.e. $\theta$ lies in QI. We have $$\sin\theta = \pm \sqrt{1-\cos^2}.$$ Since ... {7}}. \end{align*} (d) $\sin \dfrac{\theta}{2}$ We have $$\sin\left(\frac{\theta}{2} \right) = \pm \... {5}}} \end{align*} (e) $\cos \dfrac{\theta}{2}$ We have $$\cos\left(\frac{\theta}{2} \right) = \pm \... \frac{3\pi}{2}\), i.e., \(\theta\) lies in QIII. We have: \begin{align*} \sec \theta &= \pm \sqrt{1+
- Question 5 Exercise 8.2
- $\sin 2\theta=\dfrac{24}{25}$, $2\theta$ in QII. We have $$\cos 2\theta = \pm \sqrt{1-\sin^2 2\theta}... frac{49}{625}} = -\frac{7}{25} \end{align*} Also we have $$\sin\theta = \pm \sqrt{\frac{1-\cos 2\thet... -\dfrac{7}{25}\) and \(2\theta\) lies in QIII. We have: \[\sin 2\theta = \pm \sqrt{1 - \cos^2 2\theta}\] Since \(2\theta\) lies in QIII, we know that \(\sin 2\theta < 0\). Therefore: \begin
- Question 6 Exercise 8.2
- sin 15^{\circ} \cos 15^{\circ}$ ** Solution. ** We have double-angle identity: $$\sin 2 \theta = 2\s... 5^{\circ}-\sin ^{2} 15^{\circ}$ ** Solution. ** We have double-angle identity: $$\cos^2\theta -\sin^... ^{2}\left(\frac{\pi}{8}\right)$ ** Solution. ** We have a double-angle identity: $$\cos 2\alpha = 1-... pha=\cos 2\alpha.$$ Put $\alpha= \dfrac{\pi}{8}$, we have \begin{align*} 1-2\sin^2 \left(\frac{\pi}{8}
- Question 2, Exercise 8.1
- 360^{\circ}-15^{\circ}\right)$. ** Solution. ** We are given that $\cos 15^\circ = \dfrac{\sqrt{3}+1... \circ$ is not given)** To find $\cos 345^\circ$, we use the identity $\cos(360^\circ - \theta) = \cos... t(90^{\circ}-A\right) ... (1)$$ Put $A=15^\circ$, we get \begin{align*} & \sin \left(90-15\right) =
- Question 9, Exercise 8.1
- $, $\beta$ is obtuse angle, i.e. it is in QII.\\ We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^... $, $\beta$ is obtuse angle, i.e. it is in QII.\\ We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^... $, $\beta$ is obtuse angle, i.e. it is in QII.\\ We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^
- Question 2, Review Exercise
- \cos\pi >0$, thus $$\cos \phi=\frac{12}{13}$$ As, we have \begin{align*} \sin(\theta -\phi)&=\sin \the... \tan \phi & = \frac{5}{12} \end{align*} Finally, we have \begin{align*} \tan(\theta - \phi) &= \frac{... \tan \phi & = \frac{5}{12} \end{align*} Finally, we have \begin{align*} \tan(\theta + \phi) &= \frac{
- Question 5 and 6, Exercise 8.1
- $\tan \beta=-\dfrac{5}{12}$, $\beta$ is in QII. We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^... d $\cot\beta=\dfrac{15}{8}$, $\beta$ is in QIII. We have an identity: $$\sin \alpha=\pm \sqrt{1-\cos^
- Question 14, Exercise 8.1
- ngle $\alpha$ with the ground as shown in figure, we have \begin{align*} &\tan\alpha = \frac{\overline... {\sqrt{58}} \end{align*} **c.** From the figure, we see $\theta-\alpha$ is and angle between the wire
- Question 3, Exercise 8.1
- rc}\\ &= \dfrac{\sqrt{3}}{2}. \end{align*} Also, we have \begin{align*} \cos 120^{\circ} & = \cos \le
- Question 7, Exercise 8.1
- }$, where $\beta$ is acute angle, i.e. is in QI. We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^
- Question 10, Exercise 8.1
- 2) \end{align*} Combining L.H.S with (1) and (2), we have $$\tan \left(\gamma+\frac{\pi}{4}\right)=\fr
- Question 1, 2 and 3 Exercise 8.2
- Given: $\sin \alpha=y$ and $\alpha$ lies in QII. We have an identity: $$\cos \alpha = \pm \sqrt{1-\si