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- Question 8 & 9, Review Exercise 10 @fsc-part1-kpk:sol:unit10
- he identity $\sin \left( \dfrac{\pi }{4}-\theta \right)\sin \left( \dfrac{\pi }{4}+\theta \right)=\dfrac{1}{2}\cos 2\theta $. ====Solution==== We know that $2... in \alpha \sin \beta =\cos \left( \alpha -\beta \right)-\cos \left( \alpha +\beta \right)$ \begin{align}L.H.S.&=\sin \left( \dfrac{\pi }{4}-\theta \right)\s
- Question 5, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- uadrant, then find: $\sin \left( \alpha +\beta \right)$. ====Solution==== Given: $\tan\alpha =\dfrac{3... lign}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alpha\\ \Rightarrow \quad \sec \alpha &=\pm \sqrt{1+{{\tan}^{2}}... gn}\sec \alpha &=-\sqrt{1+{{\tan}^{2}}\alpha }\\ \Rightarrow \quad \sec \alpha &=-\sqrt{1+{{\left(\frac{3}{4} \right)}^{2}}}=-\sqrt{1+\frac{9}{16}}=-\sqrt{\frac{25}{1
- Question 2, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- eta -\cos \alpha \sin.\end{align} \begin{align} \Rightarrow \quad \sin \left( \frac{\pi }{3}-\frac{\pi }{4} \right) & =\sin \frac{\pi }{3}\cos \frac{\pi }{4}-\cos \... in \frac{\pi }{4} \\ &=\left( \frac{\sqrt{3}}{2} \right)\left( \frac{1}{\sqrt{2}} \right)-\left( \frac{1}{2} \right)\left( \frac{1}{\sqrt{2}} \right) \\ & =\fra
- Question 5, Exercise 10.3 @fsc-part1-kpk:sol:unit10
- os \alpha \cos \beta =\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)$\\ \begin{align}L.H.S.&=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos... {\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2} \right)\cos {{80}^{\circ }}\\ &=\dfrac{1}{2}\cos {{80}^{... eft( 2\,\cos {{80}^{\circ }}\cos {{40}^{\circ }} \right)\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( \cos
- Question 5, Exercise 10.3 @fsc-part1-kpk:sol:unit10
- os \alpha \cos \beta =\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)$\\ \begin{align}L.H.S.&=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos... {\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2} \right)\cos {{80}^{\circ }}\\ &=\dfrac{1}{2}\cos {{80}^{... eft( 2\,\cos {{80}^{\circ }}\cos {{40}^{\circ }} \right)\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( \cos
- Question 6, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- +{{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}{4}\end{align}\\ Take square ro
- Question 3 & 4, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- \begin{align}{{z}_{1}}\left( {{z}_{2}}+{{z}_{3}} \right)&={{z}_{1}}{{z}_{2}}+{{z}_{1}}{{z}_{3}}\\ {{z}_{... }&=\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}i \right)\left( \lef
- Question 3, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- e each of the following exactly $\cos \left( u+v \right)$ ====Solution==== Given $\sin u=\dfrac{3}{5},$ ... \cos $ is $+ve$ in first quadrant . \begin{align}\Rightarrow \,\,\,\,\,\cos u&=\sqrt{1-{{\sin }^{2}}u}\\ &=\sqrt{1-{{\left( \frac{3}{5} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{9}{25}}\,\,=\sqrt{\dfrac{16}{25}}\\ \Rightarrow \,\,\,\,\,\cos u&=\dfrac{4}{5}\end{align} Al
- Question 7, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- \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 using double angle identity)}\\ &=\dfrac{1}{\sec 2\th
- Question 2 & 3, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- +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( -1 \right)}^{56}}+
- Question 2 & 3, Review Exercise 1 @fsc-part1-kpk:sol:unit01
- {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\left( -1 \right) \right)\\
- Question, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- d the exact value of $\sin \left( \alpha -\beta \right)$. ====Solution==== Given: $\sin \alpha=-\dfrac{4... \sin^2\alpha}\\ &=-\sqrt{1-{{\left(-\frac{4}{5} \right)}^{2}}}\\ &=-\sqrt{1-\dfrac{16}{25}}=\sqrt{\dfrac{9}{25}}\\ \Rightarrow \quad \cos \alpha&=-\dfrac{3}{5}.\end{align}... , \begin{align}&=\sqrt{1-{{\left(-\dfrac{12}{13} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{144}{169}}=\sqrt{\dfra
- Question 8, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- ove that: $\tan \left( \dfrac{\pi }{4}+\theta \right)=\dfrac{\cos \theta +\sin \theta }{\cos \theta -\... lign}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{\pi }{4}\cos \theta +\cos\dfrac{\p
- Question 9 and 10, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- a \sin 4\theta \\ &=\sin \left( \theta +4\theta \right)\\ &=\sin 5\theta =R.H.S.\end{align} =====Quest... : $\dfrac{\sin \left( {{180}^{\circ }}-\alpha \right)\cos \left( {{270}^{\circ }}-\alpha \right)}{\sin \left( {{180}^{\circ }}+\alpha \right)\cos \left( {{270}^{\circ }}+\alpha \right)}=1$ ====Solu
- Question 9 & 10, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- ==== 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( 2+3i \right)