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- Question 8 & 9, Review Exercise 10
- 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
- uadrant, then find: $\sin \left( \alpha +\beta \right)$. ====Solution==== Given: $\tan\alpha =\dfrac{3... lign}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alpha\\ \Rightarrow \quad \sec \alpha &=\pm \sqrt{1+{{\tan}^{2}}... gn}\sec \alpha &=-\sqrt{1+{{\tan}^{2}}\alpha }\\ \Rightarrow \quad \sec \alpha &=-\sqrt{1+{{\left(\frac{3}{4} \right)}^{2}}}=-\sqrt{1+\frac{9}{16}}=-\sqrt{\frac{25}{1
- Question 2, Exercise 10.1
- eta -\cos \alpha \sin.\end{align} \begin{align} \Rightarrow \quad \sin \left( \frac{\pi }{3}-\frac{\pi }{4} \right) & =\sin \frac{\pi }{3}\cos \frac{\pi }{4}-\cos \... in \frac{\pi }{4} \\ &=\left( \frac{\sqrt{3}}{2} \right)\left( \frac{1}{\sqrt{2}} \right)-\left( \frac{1}{2} \right)\left( \frac{1}{\sqrt{2}} \right) \\ & =\fra
- Question 5, Exercise 10.3
- os \alpha \cos \beta =\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)$\\ \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
- 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 3, Exercise 10.1
- 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
- \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, Exercise 10.1
- 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
- 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
- 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 2, Exercise 10.3
- in \beta =2\sin \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right).$$ Put $\alpha ={{37}^{\circ }}$, $\beta ={{43}^{\circ }}$... left( \dfrac{{{37}^{\circ }}+{{43}^{\circ }}}{2} \right)\cos \left( \dfrac{{{37}^{\circ }}-{{43}^{\circ }}}{2} \right)\\ &=2\sin \left( \dfrac{{{80}^{\circ }}}{2} \
- Question 13, Exercise 10.1
- wing in the form $r\,\,\sin \left( \theta +\phi \right)$ where terminal ray of $\theta$ and $\phi$ are i... \\ \implies &r^2\left(\cos^2\varphi+\sin^2\varphi\right) = 16+9\\ \implies &r^2 = 25\\ \implies &r=5 \en... ft(\dfrac{4}{5}\sin\theta +\dfrac{3}{5}\cos\theta\right)\\ &=r\left(\cos\varphi\sin\theta +\sin\varphi\cos\theta\right)\\ &=r\sin(\theta+\varphi),\\ &\text{ where } \si
- Question 6 & 7, Review Exercise 10
- n}L.H.S&=\cos 4\theta \\ &=\cos 2\left( 2\theta \right)\\ &=1-2si{{n}^{2}}2\theta \\ &=1-2{{\left( 2sin\theta \cos \theta \right)}^{2}}\\ &=1-8si{{n}^{2}}\theta {{\cos }^{2}}\the... dfrac{1}{2}\left( 2\sin 6x\sin x+2\cos 4x\cos 3x \right)\\ &=\dfrac{1}{2}\left[ \cos \left( 6x-x \right)-\cos \left( 6x+x \right)+\left( \cos \left( 4x+3x \righ
- Question 8 and 9, Exercise 10.2
- \cos}^{4}}\theta &={{\left( {{\cos }^{2}}\theta \right)}^{2}}\\ &={{\left( \dfrac{1+\cos 2\theta }{2} \right)}^{2}}\\ &=\dfrac{1+2\cos 2\theta +{{\cos }^{2}}2... {4}\left[ 1+2\cos 2\theta +{{\cos }^{2}}2\theta \right]\\ &=\dfrac{1}{4}\left[ 1+2\cos 2\theta +\dfrac{1+\cos 4\theta }{2} \right]\\ &=\dfrac{1}{8}\left[ 2+4\cos 2\theta +1+\cos 4
- Question 6, Exercise 10.1
- ha }{2}-\left( 1-{{\cos }^{2}}\dfrac{\alpha }{2} \right)\\ \cos \alpha &=2{{\cos }^{2}}\dfrac{\alpha }{2... -1\\ &=2\left( 1-{{\sin }^{2}}\dfrac{\alpha }{2} \right)-1\\ &=2-2{{\sin }^{2}}\dfrac{\alpha }{2}-1\\ \co... ii)===== Show that: $\sin \left( \alpha +\beta \right)\sin \left( \alpha -\beta \right)={{\cos }^{2}}\beta -{{\cos }^{2}}\alpha$ ====Solution==== \begin{alig