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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 13, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- &r^2\cos^2 \varphi+r^2\sin^2\varphi = 4^2+3^2\\ \implies &r^2\left(\cos^2\varphi+\sin^2\varphi\right) = 16+9\\ \implies &r^2 = 25\\ \implies &r=5 \end{align} Also \begin{align}r\cos\varphi = 4 \implies 5\cos\varphi = 4 \implies \cos\varphi =\dfrac{4}{
- Question 13, Exercise 10.1 @math-11-kpk:sol:unit10
- &r^2\cos^2 \varphi+r^2\sin^2\varphi = 4^2+3^2\\ \implies &r^2\left(\cos^2\varphi+\sin^2\varphi\right) = 16+9\\ \implies &r^2 = 25\\ \implies &r=5 \end{align} Also \begin{align}r\cos\varphi = 4 \implies 5\cos\varphi = 4 \implies \cos\varphi =\dfrac{4}{
- Question 6 Exercise 4.1 @math-11-kpk:sol:unit04
- 0$ \begin{align}&P_{0+1}=\dfrac{5-0}{0+1} P_0\\ \implies &P_1=5.\end{align} For $r=1$ \begin{align}&P_{1+1}=\dfrac{5-1}{1+1} P_1\\ \implies &P_2=2\cdot 5=10.\end{align} For $r=2$ \begin{align}&P_{2+1}=\dfrac{5-2}{2+1} P_2\\ \implies &P_3=1\cdot 10=10\end{align} For $r=3$ \begin{align}&P_{3+1}=\dfrac{5-3}{3+1} P_3\\ \implies &P_4=\dfrac{10}{2}=5.\end{align} For $r=4$ \begin
- 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 2 Exercise 4.3 @math-11-kpk:sol:unit04
- begin{align}&S_{21}=\dfrac{21}{2}(a_1+a_{21}) \\ \implies &210=\dfrac{21}{2}(-40+a_{21}) \\ \implies &-40+a_{21}=\dfrac{2 \times 210}{21}=20 \\ \implies &a_{21}=20+40=60.\end{align} Also $a_{21}=a_1+20 d$, then\\ \begin{align}&20d=60-(-40)=100 \\ \implies &d=\dfrac{100}{20}=5 \end{align} Hence $a_{21}=60
- Question11 and 12, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- \begin{align}&\alpha+\beta +\gamma =180^\circ\\ \implies &\alpha +\beta =180^\circ-\gamma \\ \implies &\dfrac{\alpha +\beta }{2}=\dfrac{180^\circ-\gamma }{2}\\ \implies &\dfrac{\alpha}{2}+\dfrac{\beta}{2}=90^\circ-\dfr... ght)=\tan \left( 90-\dfrac{\gamma }{2} \right)\\ \implies &\dfrac{\tan\dfrac{\alpha}{2}+\tan\dfrac{\beta}{2
- Question11 and 12, Exercise 10.1 @math-11-kpk:sol:unit10
- \begin{align}&\alpha+\beta +\gamma =180^\circ\\ \implies &\alpha +\beta =180^\circ-\gamma \\ \implies &\dfrac{\alpha +\beta }{2}=\dfrac{180^\circ-\gamma }{2}\\ \implies &\dfrac{\alpha}{2}+\dfrac{\beta}{2}=90^\circ-\dfr... ght)=\tan \left( 90-\dfrac{\gamma }{2} \right)\\ \implies &\dfrac{\tan\dfrac{\alpha}{2}+\tan\dfrac{\beta}{2
- 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}&