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- Question 1, Exercise 8.1 @math-11-nbf:sol:unit08
- os \alpha \cos \beta - \sin \alpha \sin \beta \\ \implies \cos (180+60) & = \cos 180 \cos 60 - \sin 180 \sin 60 \\ \implies \cos (180+60) & = (-1)\left(\frac{1}{2}\right) - ... os \alpha \cos \beta + \sin \alpha \sin \beta \\ \implies \cos (180-60) & = \cos 180 \cos 60 + \sin 180 \sin 60 \\ \implies \cos (180-60) & = (-1)\left(\frac{1}{2}\right) +
- Question 4 Exercise 8.2 @math-11-nbf:sol:unit08
- frac{3}{5} \right)\\ \end{align*} \begin{align*} \implies \boxed{\sin 2\theta = \frac{24}{25}}. \end{align... ac{16}{25} \right)\\ \end{align*} \begin{align*} \implies \boxed{\cos 2\theta = -\frac{7}{25}}. \end{align... frac{24/25}{-7/25}\\ \end{align*} \begin{align*} \implies \boxed{\tan 2\theta = -\frac{24}{7}}. \end{align... 1-\cos\theta}{2}}$$ As $0<\theta<\dfrac{\pi}{2}$ implies $0<\dfrac{\theta}{2}<\dfrac{\pi}{4}$, that is, $\
- Question 7, Exercise 1.4 @math-11-nbf:sol:unit01
- as \begin{align*} &\arg (z-1)=-\dfrac{\pi}{4} \\ \implies & \arg(x+iy-1) = -\dfrac{\pi}{4} \\ \implies & \arg(x-1+iy) = -\dfrac{\pi}{4} \\ \implies & \tan^{-1}\left(\dfrac{y}{x-1}\right) = -\dfrac{\pi}{4} \\ \implies & \dfrac{y}{x-1} = \tan\left(-\dfrac{\pi}{4}\righ
- Question 8, Exercise 1.2 @math-11-nbf:sol:unit01
- x+i y$, we have \begin{align} & |2(x+iy)-i|=4 \\ \implies & |2x+i(2y-1)|=4 \\ \implies & \sqrt{(2x)^2+(2y-1)^2}=4 \end{align} Squaring on both sides \begin{align} & (2x)^2+(2y-1)^2 = 16\\ \implies & 4x^2+4y^2-4y+1-16=0 \\ \implies & 4x^2+4y^2-4y-15=0, \end{align} as required. GOOD ====Question 8(ii)===
- Question 4, Exercise 1.3 @math-11-nbf:sol:unit01
- ign} &2(1-i)z-(1-i) (2+5 i)\omega=(1-i) (2+3i)\\ \implies & (2-2i)z-(2+5+5i-2i)\omega=2+3+3i-2i \\ \implies & (2-2i)z-(7+3i)\omega=5+i \quad \cdots (4) \end{align} $(3)-(4)$ implies \begin{align} (9+5i) \omega=1-i \end{align} \begin{align} \implies \omega & =\dfrac{1-i}{9+5i}\\ &=\dfrac{1-i}{9+5i}
- Question 12, Exercise 8.1 @math-11-nbf:sol:unit08
- gin{align*} & \alpha+\beta=180^{\circ}-\gamma \\ \implies & \tan(\alpha+\beta) = \tan(180^{\circ}-\gamma) \\ \implies & \frac{\tan\alpha + \tan\beta}{1-\tan\alpha \tan\beta} = \tan(2(90)-\gamma) \\ \implies & \tan\alpha + \tan\beta = -\tan\gamma[1-\tan\alpha \tan\beta] \\ \implies & \tan\alpha + \tan\beta = -\tan\gamma+\tan\alpha
- Question 5 and 6, Review Exercise @math-11-nbf:sol:unit08
- tan \theta \cdot \tan 45^{\circ}} =\frac{1}{3}\\ \implies & \frac{\tan \theta - 1}{1 + \tan \theta}= \frac{1}{3}\\ \implies & 3 \tan \theta - 3 = 1 + \tan \theta \\ \implies & 2 \tan \theta = 4 \\ \implies & \tan \theta = 2 \end{align*} GOOD =====Question 6(i)===== If $\sin (\a
- Question 20, 21 and 22, Exercise 4.3 @math-11-nbf:sol:unit04
- .\\ As \begin{align} &S_n=\frac{n}{2}[a_1+a_n]\\ \implies & 876=\frac{n}{2}[7+139]\\ \implies & 1752=146n\\ \implies & n=\frac{1752}{146}=12. \end{align} Also we have \begin{align} &a_n=a_1+(n-1)d\\ \implies & 139=7+(12-1)d\\ \implies & 139-7=11d\\ \implies
- Question 17, 18 and 19, Exercise 4.3 @math-11-nbf:sol:unit04
- $.\\ We have \begin{align} & a_n=a_1+(n-1)d \\ \implies & 96=6+(n-1)(6) \\ \implies & 96=6+6n-6 \\ \implies & 6n=96 \\ \implies & n = 24. \end{align} Now \begin{align} S_n&=\frac{n}{2}[a_1+a_n] \\ \implies S_{24}&
- Question 2, Review Exercise @math-11-nbf:sol:unit08
- ft(\frac{3}{5}\right)^2\\ & =1-\frac{9}{25} \\ \implies \cos^2 \theta&=\frac{16}{25} \\ \cos \theta&=\pm... \theta$ is obtuse, so $\theta$ lies in II Q. This implies $\cos \theta <0$, thus $$\cos \theta=-\frac{4}{5}... \frac{5}{13}\right)^2\\ & =1-\frac{25}{169} \\ \implies \cos^2 \theta&=\frac{144}{169} \\ \cos \phi&=\pm... } As $\phi$ is acute, so $\phi$ lies in I Q. This implies $\cos\pi >0$, thus $$\cos \phi=\frac{12}{13}$$ As
- Question 13, Exercise 8.1 @math-11-nbf:sol:unit08
- 2+(-5)^2=r^2 \cos^2\varphi+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{align*} Also \begin{align*} & \frac{-5}{12}=\frac{r\sin \varphi }{r\cos \varphi } \\ \implies & \frac{-5}{12}=\tan \varphi \\ \implies & \varph
- Question 5 Exercise 8.2 @math-11-nbf:sol:unit08
- \sqrt{\frac{16}{25}} \end{align*} \begin{align*} \implies \boxed{\sin\theta = \frac{4}{5}} \end{align*} Als... \sqrt{\frac{9}{25}} \end{align*} \begin{align*} \implies \boxed{\cos\theta=\frac{3}{5}} \end{align*} Now \... & = \frac{4/5}{3/5} \end{align*} \begin{align*} \implies \boxed{\tan\theta=\frac{4}{3}} \end{align*} GOOD ... a}{2}}. \] As \(\pi < 2\theta < \frac{3\pi}{2}\) implies \(\frac{\pi}{2} < \theta < \pi\), i.e., \(\theta\
- Question 20 and 21, Exercise 4.4 @math-11-nbf:sol:unit04
- 1}. $$ This gives \begin{align*} &a_5=a_1 r^4 \\ \implies & 48=3r^4 \\ \implies & r^4 = 16 \\ \implies & r^4 = 2^4 \\ \implies & r = 2. \end{align*} Thus \begin{align*} & a_2=a_1 r= (3)(2) = 6 \\ & a_3=a_1 r^2
- Question 7, Review Exercise @math-11-nbf:sol:unit08
- in (\alpha - \theta) = \cos (\alpha + \theta) \\ \implies & \sin \alpha \cos \theta - \cos \alpha \sin \the... \alpha \sin \theta}{\cos \alpha \cos \theta} \\ \implies & \tan \alpha -\tan \theta= 1-\tan \alpha \tan \theta\\ \implies & \tan \alpha +\tan \alpha \tan \theta =1+\tan \theta\\ \implies & \tan \alpha(1 + \tan \theta) =1+\tan \theta\\ \
- Question 4, Exercise 1.1 @math-11-nbf:sol:unit01
- olution.** \begin{align}&(2+3i)x+(1+3i)y+2=0\\ \implies &(2x+y+2)+(3x+3y)i=0.\end{align} Comparing real a... egin{align}&\dfrac{x}{(1+i)}+\dfrac{y}{1-2i}=1\\ \implies &\dfrac{x(1-2i)+y(1+i)}{(1+i)(1-2i)}=1\\ \implies &\dfrac{x-i2x+y+iy}{1-2i^2+i-2i}=1\\ \implies &\dfrac{(x+y)+(y-2x)i}{3-i}=1\\ \implies & (x+y)+(y-2x)i=3-i