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- Question 1, Exercise 8.1
- in 180 \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{a... in 180 \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{al
- Question 4 Exercise 8.2
- in QI, so \begin{align*} \sin\theta & = \sqrt{1-\left(\frac{3}{5} \right)^2} \\ &= \sqrt{1-\frac{9}{25}... \sin 2\theta & = 2 \sin\theta \cos\theta \\ &= 2\left(\frac{4}{5} \right) \left(\frac{3}{5} \right)\\ \end{align*} \begin{align*} \implies \boxed{\sin 2\thet... ign*} \cos 2\theta & = 1-2\sin^2\theta \\ &= 1-2\left(\frac{4}{5} \right)^2\\ &= 1-2\left(\frac{16}{25}
- Question 9, Exercise 8.1
- s\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=-\sqrt{1-{{\left(\dfrac{1}{\sqrt{2}}\right)}^{2}}}\\ &=-\sqrt{1-\d... cos \beta & =\sqrt{1-\cos^2\beta} \\ &=\sqrt{1-{{\left(-\dfrac{3}{5} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{9... \alpha \cos \beta + \cos \alpha \sin \beta \\ &= \left( \frac{1}{\sqrt{2}} \right) \left( -\frac{3}{5} \right) + \left( -\frac{1}{\sqrt{2}} \right) \left( \fr
- Question 11, Exercise 8.1
- ==== Question 11(i)===== 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
- Question 5 and 6, Exercise 8.1
- s\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=-\sqrt{1-{{\left(-\frac{4}{5} \right)}^{2}}}\\ &=-\sqrt{1-\dfrac{1... c \beta & =-\sqrt{1+\tan^2\beta} \\ &=-\sqrt{1+{{\left(-\dfrac{5}{12} \right)}^{2}}}\\ &=-\sqrt{1+\dfrac... mplies \sin\beta & = \tan\beta \cos\beta \\ & = \left(-\frac{5}{12} \right) \left(-\frac{12}{13} \right) \\ \implies \sin\beta & = \frac{5}{13}. \end{align*
- Question 7 Exercise 8.2
- gin{align*} \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(\frac{2-1-\cos 4\alpha}{2} \right) \\ &=\frac{1-\
- Question 3, Exercise 8.1
- exact value of $\cos 120^{\circ}$ 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 ^{\... } Also \begin{align*} \cos 120^{\circ} & = \cos \left(90^{\circ}+30^{\circ}\right) \\ &= - \sin 30^{\ci
- Question 2(i, ii, iii, iv and v) Exercise 8.3
- \sin 70^{\circ} + \sin 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*
- Question 2, Exercise 8.1
- exact value of $\cos 15^{\circ}$ 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... t{2}}$ to find $\cos 165^{\circ}$ 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
- Question 6 Exercise 8.2
- d exact values for the expression: $1-2 \sin ^{2}\left(\frac{\pi}{8}\right)$ ** Solution. ** We have a ... dfrac{\pi}{8}$, we 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{
- Question 4 Exercise 8.3
- \cos 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 (80^\circ - 40
- Question 13, Exercise 8.1
- r^2 \sin^2 \varphi \\ \implies & 144+25={{r}^{2}}\left( {{\cos }^{2}}\varphi +{{\sin }^{2}}\varphi \right) \\ \implies & 169={{r}^{2}}\left( 1 \right) \\ \implies & r=\sqrt{169}=13 \end{al... 12}=\tan \varphi \\ \implies & \varphi =\tan^{-1}\left(-\frac{5}{12}\right) \end{align*} Now \begin{al... hi \sin \theta +r\sin \varphi \cos \theta \\ =& r\left( \cos \varphi \sin \theta +\sin \varphi \cos \the
- Question 9, Review Exercise
- . =====Question 9===== Simplify: $$\sqrt{\frac{\left(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\
- Question 4, Exercise 8.1
- (iii)===== 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 \le
- Question 10, Exercise 8.1
- tan. ===== 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}{... gn*} ===== Question 10(iii)===== Verify: $\cos \left(\alpha+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}(\c... ** Solution. ** \begin{align*} L.H.S &= \cos \left(\alpha+\frac{\pi}{4}\right) \\ & = \cos\alpha \co