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- Question 1, Exercise 8.1
- \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{alig... \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
- Question 4 Exercise 8.2
- lign*} \sin\theta & = \sqrt{1-\left(\frac{3}{5} \right)^2} \\ &= \sqrt{1-\frac{9}{25}}\\ &= \sqrt{\frac{... 2 \sin\theta \cos\theta \\ &= 2\left(\frac{4}{5} \right) \left(\frac{3}{5} \right)\\ \end{align*} \begin{align*} \implies \boxed{\sin 2\theta = \frac{24}{25}}.... & = 1-2\sin^2\theta \\ &= 1-2\left(\frac{4}{5} \right)^2\\ &= 1-2\left(\frac{16}{25} \right)\\ \end{ali
- Question 9, Exercise 8.1
- \alpha}\\ &=-\sqrt{1-{{\left(\dfrac{1}{\sqrt{2}}\right)}^{2}}}\\ &=-\sqrt{1-\dfrac{1}{2}}=\sqrt{\dfrac{1... -\cos^2\beta} \\ &=\sqrt{1-{{\left(-\dfrac{3}{5} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{9}{25}} \\ & =\sqrt{\d... 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) \\ &= -\fra
- Question 5 and 6, Exercise 8.1
- \sin^2\alpha}\\ &=-\sqrt{1-{{\left(-\frac{4}{5} \right)}^{2}}}\\ &=-\sqrt{1-\dfrac{16}{25}}=\sqrt{\dfrac{9}{25}}\\ \Rightarrow \quad \cos \alpha&=-\dfrac{3}{5}.\end{align*... tan^2\beta} \\ &=-\sqrt{1+{{\left(-\dfrac{5}{12} \right)}^{2}}}\\ &=-\sqrt{1+\dfrac{25}{144}} \\ & =-\sqr... } =-\dfrac{13}{12} \end{align*} \begin{align*} \Rightarrow \quad \cos \beta=\frac{1}{\sec \beta}=-\frac
- Question 11, Exercise 8.1
- Show that: $\dfrac{\sin \left(180^{\circ}+\lambda\right) \cos \left(270^{\circ}+\lambda\right)}{\sin \left(180^{\circ}-\lambda\right) \cos \left(270^{\circ}-\lambda\right)}=1$ ** Solution. ** \begin{align*} L.H.S & = \dfrac{\sin \left(
- Question 7 Exercise 8.2
- os ^{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(\frac{2-1-\cos 4\alpha}{2} \right) \\ &=\frac{1-\cos 4\alpha}{8} \end{align*} GOOD
- Question 3, Exercise 8.1
- irc}$ by using $\cos \left(180^{\circ}-60^{\circ}\right)$ and $\cos \left(90^{\circ}+30^{\circ}\right)$. ** Solution. ** \begin{align*} \cos 120^{\circ} & = \cos \left(180^{\circ}-60^{\circ}\right) \\ &= - \cos 60 ^{\circ}\\ &= -\dfrac{1}{2}. \en... 120^{\circ} & = \cos \left(90^{\circ}+30^{\circ}\right) \\ &= - \sin 30^{\circ}\\ &= -\dfrac{1}{2}. \end
- Question 2(i, ii, iii, iv and v) Exercise 8.3
- n 30^{\circ} \\ & = 2 \sin \left(\frac{70+30}{2} \right) \cos \left(\frac{70-30}{2} \right) \\ & = 2 \sin \left(\frac{100}{2} \right) \cos \left(\frac{40}{2} \right) \\ & = 2 \sin 50^\circ \cos 20^\circ \end{align*} GOOD =====Question
- Question 2, Exercise 8.1
- circ}$ by using $\cos \left(45^{\circ}-30^{\circ}\right)$. ** Solution. ** \begin{align*} \cos 15^{\circ} & = \cos \left(45^{\circ}-30^{\circ}\right)\\ &= \cos 45 \cos 30 + \sin 45 \sin 30 \\ &= \df... irc}$ by using $\cos \left(180^{\circ}-15^{\circ}\right)$. ** Solution. ** \begin{align*} \cos 165^{\circ} & = \cos \left(180^{\circ}-15^{\circ}\right)\\ &= -\cos 15 \quad (\because \cos(180-\theta)=-
- Question 6 Exercise 8.2
- the expression: $1-2 \sin ^{2}\left(\frac{\pi}{8}\right)$ ** Solution. ** We have a double-angle identit... have \begin{align*} 1-2\sin^2 \left(\frac{\pi}{8}\right)&=\cos 2\left(\frac{\pi}{8}\right)\\ &=\cos \left(\frac{\pi}{4}\right)\\ \end{align*} \begin{align*} \implies \boxed{1-2\sin^2 \left(\frac
- Question 4 Exercise 8.3
- os 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 (80^\circ - 40^\circ) \right) \cos 20^\circ \\ &=\frac{1}{4}\left( \cos {{120}
- Question 13, Exercise 8.1
- eft( {{\cos }^{2}}\varphi +{{\sin }^{2}}\varphi \right) \\ \implies & 169={{r}^{2}}\left( 1 \right) \\ \implies & r=\sqrt{169}=13 \end{align*} Also \begin{al... \implies & \varphi =\tan^{-1}\left(-\frac{5}{12}\right) \end{align*} Now \begin{align*} & 12\sin \thet... s \varphi \sin \theta +\sin \varphi \cos \theta \right) \\ =& r\sin \left( \theta +\varphi \right), \en
- Question 9, Review Exercise
- (1-\tan ^{2} x \cos (-x) \cos \left(360^{\circ}-x\right)\right) \tan 45^{\circ}}{\left\{\sin 90^{\circ}-\sin \left(180^{\circ}+x\right)\right\}\left\{\sin 90^{\circ}-\cos \left(90^{\circ}-x\right)\right\}}}$$ ** Solution. ** \begin{align
- Question 4, Exercise 8.1
- 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(\frac{\theta}{3}\righ
- Question 10, Exercise 8.1
- )===== 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... i)===== Verify: $\cos \left(\alpha+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}(\cos \alpha-\sin \alpha)$ ... {align*} L.H.S &= \cos \left(\alpha+\frac{\pi}{4}\right) \\ & = \cos\alpha \cos\frac{\pi}{4} - \sin\alph