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- Question 7, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- \sin }^{4}}\theta =\dfrac{1}{\sec 2\theta }$. ====Solution==== \begin{align}L.H.S&={{\cos }^{4}}\theta -{{\s... }\dfrac{\theta }{2}=\dfrac{2}{\sin \theta }$. ====Solution==== \begin{align}L.H.S&=\tan \dfrac{\theta }{2}+c... eta }{1-\cos 2\theta }={{\cot }^{2}}\theta $. ====Solution==== \begin{align}L.H.S&=\dfrac{1+\cos 2\theta }{1... y $\cos ec2\theta -\cot 2\theta =tan\theta $. ====Solution==== \begin{align}L.H.S.&=\cos ec2\theta -\cot 2\t
- Question 7, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- eal and imaginary parts $\dfrac{2+3i}{5-2i}$. ====Solution==== \begin{align}&\dfrac{2+3i}{5-2i} \\ =&\dfrac... $\dfrac{{{\left( 1+2i \right)}^{2}}}{1-3i}$. ====Solution==== \begin{align}&\dfrac{(1+2i)^2}{1-3i}\\ =&\df... ts $\dfrac{1-i}{{{\left( 1+i \right)}^{2}}}$. ====Solution==== \begin{align}&\dfrac{1-i}{{{\left( 1+i \righ... ginary parts ${{\left( 2a-bi \right)}^{-2}}$. ====Solution==== \begin{align}&{{\left( 2a-bi \right)}^{-2}}\
- Question 1, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- }}+\cos {{37}^{\circ }}\sin {{22}^{\circ }}$ ==== Solution ==== As \begin{align} \sin (\alpha +\beta )=\si... }}+\sin {{83}^{\circ }}\sin {{53}^{\circ }}$ ====Solution==== As \begin{align}\cos (\alpha -\beta )&=\cos... }}-\sin {{19}^{\circ }}\sin {{5}^{\circ }}$ ==== Solution ==== As \begin{align}\cos (\alpha +\beta )&=\cos... }}-\cos {{40}^{\circ }}\sin {{15}^{\circ }}$ ==== Solution ==== As \begin{align}\sin (\alpha -\beta )=\sin
- Question 2, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- === Evaluate exactly: $\sin \dfrac{\pi }{12}$ ===Solution=== We rewrite $\dfrac{\pi }{12}$ as $\dfrac{\pi }... ii)=== Evaluate exactly:$\tan {{75}^{\circ }}$ ==Solution== We rewrite ${{75}^{\circ }}$ as ${{45}^{\circ }... i)=== Evaluate exactly:$\tan {{105}^{\circ }}$ ==Solution== We rewrite ${{105}^{{}^\circ }}$ as ${{60}^{{}^... v)=== Evaluate exactly:$\tan \dfrac{5\pi }{12}$ ==Solution== We rewrite $\dfrac{5\pi }{12}$ as $\dfrac{\left
- Question 6, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- s to evaluate exactly $\cos {{15}^{\circ }}$. ====Solution==== Because ${{15}^{\circ }}=\dfrac{{{30}^{\circ ... to evaluate exactly $\tan {{67.5}^{\circ }}$. ====Solution==== Because ${{67.5}^{\circ }}=\dfrac{{{135}^{\ci... to evaluate exactly $sin{{112.5}^{\circ }}$. ====Solution==== Because ${{112.5}^{\circ }}=\dfrac{{{225}^{\c... s to evaluate exactly $\cos \dfrac{\pi }{8}$. ====Solution==== Because $\dfrac{\pi }{8}=\dfrac{\dfrac{\pi }{
- Question 8, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- rline{z}=2\operatorname{Re}\left( z \right)$. ====Solution==== Assume $z=a+ib$, then $\overline{z}=a-ib$. \... line{z}=2i\operatorname{Im}\left( z \right)$. ====Solution==== Assume that $z=a+ib$, then $\overline{z}=a-i... ratorname{Im}\left( z \right) \right]}^{2}}$. ====Solution==== Suppose $z=a+ib$, then $\overline{z}=a-ib$. ... that $z=\overline{z}\Rightarrow z$ is real. ====Solution==== Suppose $z=a+bi$ ... (1) Then $\overline{z
- Question 2 & 3, Review Exercise 1 @fsc-part1-kpk:sol:unit01
- i}^{n+2}}+{{i}^{n+3}}=0$, $\forall n\in N$ \\ ====Solution==== \begin{align}{{i}^{n}}+{{i}^{n+1}}+{{i}^{n+2... left( 5+7i \right)$ in the form of $x+iy$.\\ ====Solution==== $\left( 1+3i \right)+\left( 5+7i \right)=1+5... left( 5+7i \right)$ in the form of $x+iy$.\\ ====Solution==== \begin{align}\left( 1+3i \right)-\left( 5+7i... left( 5+7i \right)$ in the form of $x+iy$.\\ ====Solution==== \begin{align}\left( 1+3i \right)\left( 5+7i
- Question 1, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- on 1(i)===== Simplify ${{i}^{9}}+{{i}^{19}}$. ====Solution==== \begin{align}{{i}^{9}}+{{i}^{19}}&=i\cdot{{i}... )===== Simplify ${{\left( -i \right)}^{23}}$. ====Solution==== \begin{align}{{\left( -i \right)}^{23}}&={{\l... lify ${{\left( -1 \right)}^{\frac{-23}{2}}}$. ====Solution==== \begin{align}{{\left( -1 \right)}^{\frac{-23}... plify ${{\left( -1 \right)}^{\frac{15}{2}}}$. ====Solution==== \begin{align}{{\left( -1 \right)}^{\frac{15}{
- Question 2 & 3, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- 107}}+{{i}^{112}}+{{i}^{122}}+{{i}^{153}}=0$. ====Solution==== \begin{align}L.H.S.&={{i}^{107}}+{{i}^{112}}+... $3\left( 1+2i \right),-2\left( 1-3i \right)$. ====Solution==== \begin{align}& 3\left( 1+2i \right)+-2\left( ... 2}-\dfrac{2}{3}i,\dfrac{1}{4}-\dfrac{1}{3}i$. ====Solution==== \begin{align}&\left( \dfrac{1}{2}-\dfrac{2}{3... sqrt{2},1 \right),\left( 1,\sqrt{2} \right)$. ====Solution==== \begin{align}&\left( \sqrt{2},1 \right)+\left
- Question 6, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- i}$ and write the answer in the form $a+ib$. ====Solution==== \begin{align}\dfrac{4+i}{3+5i}&=\dfrac{4+i}{3... i}$ and write the answer in the form $a+ib$. ====Solution==== \begin{align}\dfrac{1}{-8+i}&=\dfrac{1}{-8+i}... i}$ and write the answer in the form $a+ib$. ====Solution==== \begin{align}\dfrac{1}{7-3i}&=\dfrac{1}{7-3i}... i}$ and write the answer in the form $a+ib$. ====Solution==== \begin{align}\dfrac{6+i}{i}&=\dfrac{6+i}{i}\t
- Question 2, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- actors. $$P\left( z \right)={{z}^{3}}+6z+20$$ ====Solution==== Given: $$p\left( z \right)={{z}^{3}}+6z+20$... $P(z)$ into linear factors. $$P(z)=3z^2+7.$$ ====Solution==== \begin{align} P(z)&=3z^2+7\\ &=\left(\sqrt{3}... r factors. $$P\left( z \right)={{z}^{2}}+4$$ ====Solution==== \begin{align}P(z)&={{z}^{2}}+4\\ &={{\left( ... r factors. $$P(z)={{z}^{3}}-2{{z}^{2}}+z-2.$$ ====Solution==== Given: $$P\left( z \right)={{z}^{3}}-2{{z}^{
- Question 5, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- solutions of the equation ${{z}^{2}}+z+3=0$\\ ====Solution==== ${{z}^{2}}+z+3=0$\\ According to the quadrat... solutions of the equation ${{z}^{2}}-1=z$.\\ ====Solution==== ${{z}^{2}}-1=z$\\ ${{z}^{2}}-z-1=0$\\ Accord... lutions of the equation ${{z}^{2}}-2z+i=0$\\ ====Solution==== ${{z}^{2}}-2z+i=0$\\ According to the quadra... e solutions of the equation ${{z}^{2}}+4=0$\\ ====Solution==== ${{z}^{2}}+4=0$\\ According to the quadratic
- Question 6, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- s of the equation ${{z}^{4}}+{{z}^{2}}+1=0$\\ ====Solution==== \begin{align}{{z}^{4}}+{{z}^{2}}+1&=0\\ {{z}... he solutions of the equation ${{z}^{3}}=-8$\\ ====Solution==== \begin{align}{{z}^{3}}&=-8\\ {{z}^{3}}+{{2}^... e equation ${{\left( z-1 \right)}^{3}}=-1$.\\ ====Solution==== \begin{align}{{\left( z-1 \right)}^{3}}&=-1\... he solutions of the equation ${{z}^{3}}=1$ \\ ====Solution==== \begin{align}{{z}^{3}}&=1\\ {{z}^{3}}-{{1}^{
- Question 3, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- following exactly $\cos \left( u+v \right)$ ====Solution==== Given $\sin u=\dfrac{3}{5},$ $0\le u\le \d... e following exactly $\tan \left( u-v \right)$ ====Solution==== Given \begin{align}\sin u&=\dfrac{3}{5},\,\,... following exactly $\sin \left( u-v \right)$ ====Solution==== Given \begin{align}\sin u&=\dfrac{3}{5},\,\,\... following exactly $\cos \left( u+v \right)$ ====Solution==== Given $\sin u=\dfrac{3}{5},$ $0\le u\le \d
- Question 8, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- a +\sin \theta }{\cos \theta -\sin \theta }$ ====Solution==== \begin{align}L.H.S.&=\tan \left( \dfrac{\pi ... \right)=\dfrac{1-tan\theta }{1+tan\theta }$ ====Solution==== \begin{align}L.H.S.&=\tan \left( \dfrac{\pi ... {1-{{\tan }^{2}}\alpha {{\tan }^{2}}\beta }$ ====Solution==== \begin{align}\tan (\alpha +\beta )&=\dfrac{\... \right)}{\cos \left( \theta -\phi \right)}$ ====Solution==== \begin{align}L.H.S.&=\dfrac{1-\tan \theta \t