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Trigonometric ratios Quiz 2

mathematics

 Find the value of $\sin b$, if ${{\cot }^{-1}}\left( \dfrac{1}{\sqrt{\sec \delta }} \right)-{{\tan }^{-1}}\left( \dfrac{1}{\sqrt{\sec \delta }} \right)=b$.

 First convert ${{\cot }^{-1}}$ to ${{\tan }^{-1}}$ and then make use of the trigonometric relation: -${{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\dfrac{a-b}{1+ab}$.

  ${{\cot }^{-1}}\left( \dfrac{1}{\sqrt{\sec \delta }} \right)-{{\tan }^{-1}}\left( \dfrac{1}{\sqrt{\sec \delta }} \right)=b \\ \Rightarrow {{\cot }^{-1}}\left( \sqrt{\cos \delta } \right)-{{\tan }^{-1}}\left( \sqrt{\cos \delta } \right)=b \\ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{\sqrt{\cos \delta }} \right)-{{\tan }^{-1}}\left( \sqrt{\cos \delta } \right)=b \\ \Rightarrow {{\tan }^{-1}}\dfrac{\dfrac{1}{\sqrt{\cos \delta }}\_\sqrt{\cos \delta }}{1+\dfrac{1}{\sqrt{\cos \delta }}.\sqrt{\cos \delta }}=b \\ \Rightarrow {{\tan }^{-1}}\dfrac{1-\cos \delta }{2\sqrt{\cos \delta }}=b \\ \Rightarrow \tan b=\dfrac{1-\cos \delta }{2\sqrt{cos\delta }} \\ \therefore \sin b=\dfrac{1-\cos \delta }{1+\cos \delta }=\dfrac{1-\left( 1-2{{\sin }^{2}}\dfrac{\delta }{2} \right)}{1+2{{\cos }^{2}}\dfrac{\delta }{2}-1} \\ \text{               }={{\tan }^{2}}\dfrac{\delta }{2} $

 Find the value of the expression  $4{{\sin }^{6}}\varphi +3{{\left( \sin \varphi -\cos \varphi  \right)}^{4}}+6{{\left( \sin \varphi +\cos \varphi  \right)}^{2}}$ , for $\varphi \left( {{45}^{\circ }},{{90}^{\circ }} \right)$. 

 First expand the expression using ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ and${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. Then use ${{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi =1$. Now convert the expression in $\cos \varphi $ and solve.

  $4{{\sin }^{6}}\varphi +3{{\left( \sin \varphi -\cos \varphi  \right)}^{4}}+6{{\left( \sin \varphi +\cos \varphi  \right)}^{2}} \\ \Rightarrow 4{{\sin }^{6}}\varphi +3{{\left( 1-2\sin \varphi \cos \varphi  \right)}^{2}}+6\left( 1+2\sin \varphi \cos \varphi  \right) \\ \Rightarrow 4{{\sin }^{6}}\varphi +3\left( 1+4{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi -4\cos \varphi \sin \varphi  \right)+6+12\sin \varphi \cos \varphi  \\ \Rightarrow 9+12{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi +4{{\sin }^{6}}\varphi  \\ \Rightarrow 9+12{{\cos }^{2}}\varphi \left( 1-{{\cos }^{2}}\varphi  \right)+4{{\left( 1-{{\cos }^{2}}\varphi  \right)}^{3}} \\ \Rightarrow 9+12{{\cos }^{2}}\varphi -12{{\cos }^{4}}\varphi +4\left( 1-{{\cos }^{6}}\varphi -3{{\cos }^{2}}\varphi +3{{\cos }^{4}}\varphi  \right) \\ \Rightarrow 9+4-4{{\cos }^{6}}\varphi  \\ \Rightarrow 13-4{{\cos }^{6}}\varphi$

 Find the value of the expression ${{\cos }^{\frac{2}{3}}}\phi +{{\sin }^{\frac{2}{3}}}\phi $ if it is given that${{\cos }^{2}}\delta =\dfrac{1}{3}\left( {{n}^{2}}-1 \right)\text{ and }{{\tan }^{2}}\dfrac{\delta }{2}={{\tan }^{\frac{2}{3}}}\phi $.

 Convert trigonometric ratios provided into other simple forms and then put given values to solve.

  ${{\cos }^{2}}\delta =\dfrac{{{n}^{2}}-1}{3},{{\tan }^{2}}\dfrac{\delta }{2}={{\tan }^{\frac{2}{3}}}\phi  \\ \text{Now,}{{\tan }^{3}}\dfrac{\delta }{2}=\tan \phi \Rightarrow \dfrac{{{\sin }^{3}}\dfrac{\delta }{2}}{{{\cos }^{3}}\dfrac{\delta }{2}}=\dfrac{\sin \phi }{\cos \phi } \\ \Rightarrow \dfrac{{{\sin }^{3}}\dfrac{\delta }{2}}{\sin \phi }=\dfrac{{{\cos }^{3}}\dfrac{\delta }{2}}{\cos \phi }=k \\ \Rightarrow {{\sin }^{3}}\dfrac{\delta }{2}=k\sin \phi .....\text{(1)} \\ \Rightarrow {{\cos }^{3}}\dfrac{\delta }{2}=k\cos \phi ....\text{(2)} \\ {{\text{k}}^{\frac{2}{3}}}{{\sin }^{\frac{2}{3}}}\phi +{{\text{k}}^{\frac{2}{3}}}{{\cos }^{\frac{2}{3}}}\phi =1 \\ \Rightarrow {{\sin }^{\frac{2}{3}}}\phi +co{{\operatorname{s}}^{\frac{2}{3}}}\phi =\dfrac{1}{{{k}^{\frac{2}{3}}}} \\ \text{Squaring and adding equation (1) and (2)} \\ \Rightarrow \left( {{\sin }^{2}}\phi +{{\cos }^{2}}\phi  \right){{k}^{2}}={{\sin }^{6}}\dfrac{\delta }{2}+{{\cos }^{6}}\dfrac{\delta }{2} \\ \text{                       =}{{\left( \text{si}{{\text{n}}^{2}}\dfrac{\delta }{2}+{{\cos }^{2}}\dfrac{\delta }{2} \right)}^{3}}-3{{\sin }^{2}}\dfrac{\delta }{2}{{\cos }^{2}}\dfrac{\delta }{2}\left( {{\sin }^{2}}\dfrac{\delta }{2}+{{\cos }^{2}}\dfrac{\delta }{2} \right) \\ \Rightarrow {{k}^{2}}=1-\dfrac{3}{4}{{\sin }^{2}}\delta =1-\dfrac{3}{4}+\dfrac{3}{4}{{\cos }^{2}}\delta  \\ \Rightarrow {{k}^{2}}=\dfrac{{{n}^{2}}}{4}\Rightarrow k=\dfrac{n}{2} \\ {{\sin }^{\frac{2}{3}}}\phi +{{\cos }^{\frac{2}{3}}}\phi = {{\left( \dfrac{2}{n}\right)}^{\frac{2}{3}}}$. 

 Find the number of points inside or on the curve ${{j}^{2}}+{{k}^{2}}={{(64)}^{\frac{2}{3}}}$ satisfying the equation given by  ${{\tan }^{4}}j+\dfrac{1}{{{\tan }^{4}}j}+1=3{{\sin }^{2}}k$.

 Make use of the trigonometric relations $-1\le \sin x\le +1$, $\tan \left( \dfrac{\pi }{4} \right)=1,\sin \left( \dfrac{\pi }{2} \right)=1$, and the algebraic relation ${{a}^{2}}+{{b}^{2}}={{\left( a-b \right)}^{2}}+2ab$. 

  ${{\tan }^{4}}j+\dfrac{1}{{{\tan }^{4}}j}+1=3{{\sin }^{2}}k \\ \Rightarrow {{\left( {{\tan }^{2}}j-{{\cot }^{2}}j \right)}^{2}}+3\ge 3\text{      }\left[ \because \text{3si}{{\text{n}}^{2}}k\le 3 \right] \\ \Rightarrow {{\tan }^{2}}j={{\cot }^{2}}k,{{\sin }^{2}}k=1 \\ \Rightarrow \tan j=\pm 1,\sin k=\pm 1 \\ \Rightarrow j=\pm \dfrac{\pi }{4},\pm \dfrac{3\pi }{4},,.. \\ \text{Again},{{j}^{2}}\le 4\Rightarrow -2\le j\le 2 \\ \Rightarrow j=\pm \dfrac{\pi }{4} \\ \text{Again,}\sin k=\pm 1\Rightarrow k=\pm \dfrac{\pi }{2},\pm \dfrac{3\pi }{2},.. \\ \text{But, }{{\text{k}}^{2}}\le 4\Rightarrow k=\pm \dfrac{\pi }{2} \\ \text{So 4 solutions are possible}\text{.}$

 Find the general solution of the equation given by $(1.732-1)\sin \gamma +(1.732+1)\cos \gamma =2$.

 Put $\sqrt{3}=1.732$ and use ${{r}^{2}}={{a}^{2}}+{{b}^{2}}$ to find radius. Use the relation $\cos \theta =\cos \alpha \Rightarrow \theta =2n\pi \pm \alpha ,\alpha \in \left[ 0,\pi  \right]$.

  $(1.732-1)\sin \gamma +(1.732+1)\cos \gamma =2 \\ \Rightarrow (\sqrt{3}-1)\sin \gamma +(\sqrt{3}+1)\cos \gamma =2\left[ \because \sqrt{3}=1.732 \right] \\ \therefore {{r}^{2}}={{(\sqrt{3}+1)}^{2}}+{{(\sqrt{3}-1)}^{2}}=8\Rightarrow r=2\sqrt{2} \\ \tan \theta =\dfrac{\sqrt{3}-1}{\sqrt{3}+1}=\dfrac{1-\dfrac{1}{\sqrt{3}}}{1+\dfrac{1}{\sqrt{3}}} \\ \Rightarrow \tan \theta =\tan ({{45}^{\circ }}-{{30}^{\circ }})=\tan {{15}^{\circ }} \\ \Rightarrow \theta ={{15}^{\circ }}=\dfrac{\pi }{12} \\ \text{Using these resuts in the given equation,} \\ r\cos (\gamma -\theta )=2 \\ \Rightarrow \cos \left( \gamma -\dfrac{\pi }{12} \right)=\dfrac{2}{r}=\dfrac{2}{2\sqrt{2}}=\cos \dfrac{\pi }{4} \\ \Rightarrow \gamma -\dfrac{\pi }{12}=2n\pi \pm \dfrac{\pi }{4} \\ \Rightarrow \gamma =2n\pi \pm \dfrac{\pi }{4}+\dfrac{\pi }{12}$