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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... = - \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 = \pm \sqrt{\frac{1-\cos 2\theta}{2}}
- 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 2, Review Exercise
- n*} \cos^2 \theta &= 1-\sin^2\theta\\ &= 1-\left(\frac{3}{5}\right)^2\\ & =1-\frac{9}{25} \\ \implies \cos^2 \theta&=\frac{16}{25} \\ \cos \theta&=\pm\frac{4}{5}\\ \end{align*} As $\theta$ is obtuse, so $\theta$ lies in II
- Question 9, Exercise 8.1
- \cos \beta + \cos \alpha \sin \beta \\ &= \left( \frac{1}{\sqrt{2}} \right) \left( -\frac{3}{5} \right) + \left( -\frac{1}{\sqrt{2}} \right) \left( \frac{4}{5} \right) \\ &= -\frac{3}{5\sqrt{2}} - \frac{4}{5\sqrt{2}} \\ &=
- Question 4 Exercise 8.3
- 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) \cos 20^\circ \\ &= \frac{1}{4} \left( \cos (80^\circ + 40^\circ) + \cos (8
- Question 11, Exercise 8.1
- }=-1$ ** Solution. ** \begin{align*} L.H.S & = \frac{\sin \left(90^{\circ}+\alpha\right)-\cos \left(36... ght)+\cos \left(90^{\circ}+\alpha\right)} \\ & = \frac{\cos \alpha + \sin \alpha - \cos(360^{\circ} - \a... c} - \alpha) + \cos(90^{\circ} + \alpha)} \\ & = \frac{\cos \alpha + \sin \alpha - \cos \alpha}{\sin(180... c} - \alpha) + \cos(90^{\circ} + \alpha)} \\ & = \frac{\sin \alpha}{\sin(180^{\circ} - \alpha) + \sin(27
- Question 5 and 6, Exercise 8.1
- &=-\sqrt{1-\sin^2\alpha}\\ &=-\sqrt{1-{{\left(-\frac{4}{5} \right)}^{2}}}\\ &=-\sqrt{1-\dfrac{16}{25}}... n*} \begin{align*} \Rightarrow \quad \cos \beta=\frac{1}{\sec \beta}=-\frac{12}{13}.\end{align*} This gives \begin{align*} \frac{\sin\beta}{\cos\beta} & = \tan\beta \\ \implies \
- Question 10, Exercise 8.1
- olution. ** \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 \\... uestion 10(iii)===== Verify: $\cos \left(\alpha+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}(\cos \alpha-\si
- Question 7 Exercise 8.2
- ign*} \sin ^{2} \alpha \cos ^{2} \alpha &= \left(\frac{1-\cos 2\alpha}{2} \right)\left(\frac{1+\cos 2\alpha}{2} \right)\\ &= \frac{1}{4}(1-\cos^2 2\alpha) \\ &=\frac{1}{4}\left(1-\frac{1+\cos 4\alpha}{2} \right) \\ &=\frac{1}{4}\left(
- Question 1, Review Exercise
- 5^{\circ}-30^{\circ}\right)=\ldots$\\ * (a) $\frac{\sqrt{6}-\sqrt{2}}{4}$\\ * (b) $\frac{\sqrt{6}+\sqrt{2}}{4}$\\ * %%(c)%% $\frac{\sqrt{6}-\sqrt{2}}{2}$\\ * (d) $\frac{\sqrt{3}-\sqrt{2}}{2}$ \\ <btn type="link" collapse="a1">See
- Question 8, Exercise 8.1
- \sqrt{1 - \sin^2 \alpha} \\ &= \sqrt{1 - \left( \frac{3}{5} \right)^2} \\ &= \sqrt{\frac{16}{25}}\\ \cos \alpha & = \frac{4}{5}. \end{align*} $\cos \beta=\dfrac{12}{13}$, where $\dfrac{3 \pi}{... -\sqrt{1 - \cos^2 \beta} \\ &= -\sqrt{1 - \left( \frac{12}{13} \right)^2} \\ &= -\sqrt{\frac{25}{169}}\\
- 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
- Question 4, Exercise 8.1
- ==== Rewrite as a single expression. $\sin \left(\frac{\theta}{3}\right) \cos \left(\frac{\theta}{6}\right)+\cos \left(\frac{\theta}{3}\right) \sin \left(\frac{\theta}{6}\right)$ ** Solution. ** \begin{align*} & \sin \left(\fr