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- Question 2, Exercise 10.3 @fsc-part1-kpk:sol:unit10
- c }}+\sin {{43}^{\circ }}.$$ ====Solution==== We have an identity: $$\sin \alpha +\sin \beta =2\sin \le... ht)\end{align} Since $cos(-\theta)=cos\theta$, we have $$\sin {{37}^{\circ }}+\sin {{43}^{\circ }}=2\sin... rc }}-\cos {{82}^{\circ }}$. ====Solution==== We have an identity: $$\cos \alpha -\cos \beta =-2\sin \l... &=-2\sin(59^\circ)\sin(-23^\circ) \end{align} We have $\sin(-\theta)=-\sin\theta$, therefore $$\cos {{3
- Question 5, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- }+z+3=0$\\ According to the quadratic formula, we have\\ $a=1,\,\,\,b=1$ and $c=3$\\ Quadratic formula ... }-z-1=0$\\ According to the quadratic formula, we have\\ $a=1,\quad b=-1$ and $c=-1$\\ Quadratic formul... -2z+i=0$\\ According to the quadratic formula, we have\\ $a=1,\,\,\,b=-2$ and $c=i$\\ Quadratic formula... 2}}+4=0$\\ According to the quadratic formula, we have\\ $a=1,\,\,\,b=0$ and $c=4$\\ Quadratic formula i
- Question 13, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- rphi =\dfrac{3}{5}.\end{align} Thus, from (1), we have \begin{align}&4\sin\theta +3\cos\theta \\ &=5 \le... phi =\dfrac{8}{17}.\end{align} Thus, from (1), we have \begin{align}&15\sin\theta +8\cos\theta \\ &=17 \... rac{-5}{\sqrt{29}}.\end{align} Thus, from (1), we have \begin{align}&2\sin\theta -5\cos\theta \\ &=\sqrt... dfrac{1}{\sqrt{2}}.\end{align} Thus, from (1), we have \begin{align}&\sin\theta+\cos\theta \\ &=\sqrt{2}
- Question 2, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- $\implies \cos\theta = -\dfrac{12}{13}$$ Thus, we have the following by using double angle identities: \... $\implies \cos\theta = -\dfrac{12}{13}$$ Thus, we have the following by using double angle identities: \... $\implies \cos\theta = -\dfrac{12}{13}$$ Thus, we have the following by using double angle identities: Thus, we have the following by using double angle identities. \
- Question 1, Exercise 10.3 @fsc-part1-kpk:sol:unit10
- difference $2\sin 6x\sin x$. ====Solution==== We have an identity: $$-2\sin \alpha \sin \beta =\cos (\a... rc }}\cos {{123}^{\circ }}$. ====Solution==== We have an identity: $$2\sin \alpha \cos \beta =\sin (\a... +B}{2}\cos \dfrac{A-B}{2}.$$ ====Solution==== We have an identity: $$2\sin \alpha \cos \beta =\sin (\... +Q}{2}\cos \dfrac{P-Q}{2}.$$ ====Solution==== We have an identity: $$2\cos \alpha \cos \beta =\cos \le
- Question 5, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- &=3-i \ldots (2)\end{align} From (1) and (2), we have $$\overline{z_1+z_2}=\overline{z_1}+\overline{z_... \ &=13 \ldots (2)\end{align} From (1) and (2), we have $$\overline{z_1 z_2}=\overline{z_1}\overline{z_2}... {2}}} \ldots (2) \end{align} From (1) and (2), we have $$\overline{\left(\dfrac{z_1}{z_2}\right)}=\dfrac
- Question 6, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- ight)=0$\\ According to the quadratic formula, we have\\ $a=1,\,\,\,b=-2$ and $c=4$ \\ Quadratic formula... -3z+3=0$\\ According to the quadratic formula, we have\\ $a=1,\,\,\,b=-3$ and $c=3$ \\ Quadratic formula... end{align} According to the quadratic formula, we have\\ $a=1,\quad\quad b=1$ and $c=1$ \\ Quadratic for
- Question, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- =-\dfrac{12}{13}$, $\beta$ is in 2nd quadrant. We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^2\alp... -\dfrac{12}{13}$, $\beta$ is in 2nd quadrant. We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^2\alp... -\dfrac{12}{13}$, $\beta$ is in 2nd quadrant. We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^2\alp
- Question 11, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- dfrac{-2}{5}+\dfrac{11i}{5} \end{align} Hence, we have $${\rm Re}\left( \dfrac{{{z}_{1}}{{z}_{2}}}{\over... c{1}{z_1\overline{z_1}}=\dfrac{1}{5}.$$ Hence, we have $${\rm Im}\left(\dfrac{1}{z_1\overline{z_1}}\righ
- Question 2, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- z}^{3}}+6z+20$.\\ By using synthetic division, we have $$\begin{array}{c|cccc} -2 & 1 & 0 & 6 & 20 \\ ... {{z}^{2}}+z-2$.\\ By using synthetic division, we have $$\begin{array}{c|cccc} 2 & 1 & -2 & 1 & -2 \\
- Question 3, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- We find: $\cos \theta =-\dfrac{3}{5}$. Thus, we have the following by using double angle identity: \be... We find: $\cos \theta =-\dfrac{3}{5}$. Thus, we have the following by using half angle identities: \be
- Question 4 and 5, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- \pi }{3}$.\\ By using double angle identities, we have \begin{align}\sin 2\theta &=2\sin \theta \cos \th... \pi }{3}$ \\ By using double angle identities, we have \begin{align}\cos 2\theta &=2{{\cos }^{2}}\theta
- Question 5, Exercise 10.3 @fsc-part1-kpk:sol:unit10
- ac{4\pi }{9}=\dfrac{3}{16}.$$ ====Solution==== We have an identities: $$2\sin \alpha \sin \beta =\cos \l... 70}^{\circ }}=\dfrac{1}{16}$. ====Solution==== We have an identities: $$2\sin \alpha \sin \beta =\cos \l
- Unit 1: Complex Numbers (Solutions)
- xtbook Board, Peshawar, Pakistan. On this page we have provided the solutions of the questions. After r
- Unit 10: Trigonometric Identities of Sum and Difference of Angles (Solutions)
- xtbook Board, Peshawar, Pakistan. On this page we have provided the solutions of the questions. After r