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- Question 8 & 9, Review Exercise 10 @fsc-part1-kpk:sol:unit10
- n. =====Question 8===== Prove the identity $\sin \left( \dfrac{\pi }{4}-\theta \right)\sin \left( \dfrac{\pi }{4}+\theta \right)=\dfrac{1}{2}\cos 2\theta $.... ==== We know that $2\sin \alpha \sin \beta =\cos \left( \alpha -\beta \right)-\cos \left( \alpha +\beta \right)$ \begin{align}L.H.S.&=\sin \left( \dfrac{\p
- Question 2, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- end{align} \begin{align} \Rightarrow \quad \sin \left( \frac{\pi }{3}-\frac{\pi }{4} \right) & =\sin \f... }{4}-\cos \frac{\pi }{3}\sin \frac{\pi }{4} \\ &=\left( \frac{\sqrt{3}}{2} \right)\left( \frac{1}{\sqrt{2}} \right)-\left( \frac{1}{2} \right)\left( \frac{1}{\sqrt{2}} \right) \\ & =\frac{\sq
- Question 5, Exercise 10.3 @fsc-part1-kpk:sol:unit10
- == We know that\\ $2\cos \alpha \cos \beta =\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)$\\ \begin{align}L.H.S.&=\cos {{20}^{\circ }}... c }}\\ &=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2} \right)\cos {{80}^{\circ }}\\ &=\df... 0}^{\circ }}\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( 2\,\cos {{80}^{\circ }}\cos {{40}^{\circ }} \rig
- Question 5, Exercise 10.3 @fsc-part1-kpk:sol:unit10
- == We know that\\ $2\cos \alpha \cos \beta =\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)$\\ \begin{align}L.H.S.&=\cos {{20}^{\circ }}... c }}\\ &=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2} \right)\cos {{80}^{\circ }}\\ &=\df... 0}^{\circ }}\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( 2\,\cos {{80}^{\circ }}\cos {{40}^{\circ }} \rig
- Question 6, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- egin{align}{{z}^{4}}+{{z}^{2}}+1&=0\\ {{z}^{4}}+2\left( \dfrac{1}{2} \right){{z}^{2}}+\dfrac{1}{4}-\dfrac{1}{4}+1&=0\\ {{\left( {{z}^{2}}+\dfrac{1}{2} \right)}^{2}}+\dfrac{4-1}{4}&=0\\ {{\left( {{z}^{2}}+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}=0\\ {{\left( {{z}^{2}}+\dfrac{1}{2} \right)}^{2}}&=-\dfrac{3}
- Question 3 & 4, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- dition and multiplicative. \begin{align}{{z}_{1}}\left( {{z}_{2}}+{{z}_{3}} \right)&={{z}_{1}}{{z}_{2}}+... {{z}_{2}}+{{z}_{3}}&=\sqrt{2}-\sqrt{3}i+2+3i\\ &=\left( \sqrt{2}+2 \right)+\left( 3-\sqrt{3} \right)i\\ L.H.S.&={{z}_{1}}\left( {{z}_{2}}+{{z}_{3}} \right)\\ &=\left( \sqrt{3}+\sqrt{2
- Question 5, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- $\beta$ in the first Quadrant, then find: $\sin \left( \alpha +\beta \right)$. ====Solution==== Given:... a }\\ \Rightarrow \quad \sec \alpha &=-\sqrt{1+{{\left(\frac{3}{4} \right)}^{2}}}=-\sqrt{1+\frac{9}{16}}... \cos \alpha \\ \Rightarrow \quad \sin \alpha &=\left(\frac{3}{4}\right)\left(-\frac{4}{5} \right)\\ \Rightarrow \quad \sin\alpha &= \frac{3}{5}\end{align}
- Question 7, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- &={{\cos }^{4}}\theta -{{\sin }^{4}}\theta \\ &=\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)\\ &=\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)\left( 1 \right)\\ &=\cos 2\theta \quad \text{(By usin
- Question 2 & 3, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- t i^{106}+i^{112}+i^{122}+i\cdot i^{152}\\ &=i.{{\left( {{i}^{2}} \right)}^{53}}+{{\left( {{i}^{2}} \right)}^{56}}+{{\left( {{i}^{2}} \right)}^{61}}+i.{{\left( {{i}^{2}} \right)}^{76}}\\ &=i.{{\left( -1 \right)}^{53}}+{{\left(
- Question 2 & 3, Review Exercise 1 @fsc-part1-kpk:sol:unit01
- {{i}^{2}}+{{i}^{n}}\cdot {{i}^{3}}\\ &={{i}^{n}}\left( 1+i+{{i}^{2}}+{{i}^{3}} \right)\\ &={{i}^{n}}\left( 1+i+{{i}^{2}}+{{i}^{3}} \right)\\ &=i\left( 1+i+{{\left( i \right)}^{2}}+i{{\left( i \right)}^{2}} \right)\\ &=i\left( 1+i+\left( -1 \right)+i\lef
- Question 8, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- tan. =====Question 8(i)===== Prove that: $\tan \left( \dfrac{\pi }{4}+\theta \right)=\dfrac{\cos \the... }$ ====Solution==== \begin{align}L.H.S.&=\tan \left( \dfrac{\pi }{4}+\theta \right)\\ &=\dfrac{\sin \left( \dfrac{\pi }{4}+\theta \right)}{\cos \left( \dfrac{\pi }{4}+\theta \right)}\\ &=\dfrac{\sin\dfrac{\
- Question 9 and 10, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- \cos 4\theta +\cos \theta \sin 4\theta \\ &=\sin \left( \theta +4\theta \right)\\ &=\sin 5\theta =R.H.S... =====Question 10===== Show that: $\dfrac{\sin \left( {{180}^{\circ }}-\alpha \right)\cos \left( {{270}^{\circ }}-\alpha \right)}{\sin \left( {{180}^{\circ }}+\alpha \right)\cos \left( {{270}^{\cir
- Question 9 & 10, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- =Question 9===== Find the conjugate of $\dfrac{\left( 3-2i \right)\left( 2+3i \right)}{\left( 1+2i \right)\left( 2-i \right)}$. ====Solution==== Let \begin{align}z&=\dfrac{\left( 3-2i \right)\left
- Question 7, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- Separate into real and imaginary parts $\dfrac{{{\left( 1+2i \right)}^{2}}}{1-3i}$. ====Solution==== \b... ate into real and imaginary parts $\dfrac{1-i}{{{\left( 1+i \right)}^{2}}}$. ====Solution==== \begin{align}&\dfrac{1-i}{{{\left( 1+i \right)}^{2}}}\\ =&\dfrac{1-i}{1-1+2i}\\ =&\... )===== Separate into real and imaginary parts ${{\left( 2a-bi \right)}^{-2}}$. ====Solution==== \begin{
- Question 1, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- }}&=i\cdot{{i}^{8}}+i\cdot{{i}^{18}}\\ &=i\cdot{{\left( {{i}^{2}} \right)}^{4}}+i\cdot{{\left( {{i}^{2}} \right)}^{9}}\\ &=i\cdot{{\left( -1 \right)}^{4}}+i\cdot{{\left( -1 \right)}^{9}}\\ &=i\cdot1+i\cdot\left( -1 \right)\\ &=i-i\\ &=0\end