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This is one of our many quizzes on Sine and cosine of complementary angles. Click on this to access more such competitive quizzes on Trigonometric Ratios.

Sine and cosine of complementary angles Quiz 2

mathematics

 Simplify: $\dfrac{{{\sin }^{6}}{{\left( 89.5 \right)}^{\circ }}}{2{{\cos }^{6}}{{\left( 0.5 \right)}^{\circ }}}+\dfrac{{{\cos }^{6}}{{\left( 89.5 \right)}^{\circ }}}{2{{\sin }^{6}}{{\left( 0.5 \right)}^{\circ }}}$

 Use $\sin \left( {{90}^{\circ }}-\theta  \right)=\cos \theta $  and  $\cos \left( {{90}^{\circ }}-\theta  \right)=\sin \theta $ to solve the expression.

$\begin{align} &nbsp;&nbsp;& Consider\text{ }the\text{ }given\text{ }expression. \\ &nbsp;& \dfrac{{{\sin }^{6}}{{\left( 89.5 \right)}^{\circ }}}{2{{\cos }^{6}}{{\left( 0.5 \right)}^{\circ }}}+\dfrac{{{\cos }^{6}}{{\left( 89.5 \right)}^{\circ }}}{2{{\sin }^{6}}{{\left( 0.5 \right)}^{\circ }}} \\ &nbsp;& =\dfrac{{{\sin }^{6}}\left( {{90}^{\circ }}-{{0.5}^{\circ }} \right)}{2{{\cos }^{6}}{{\left( 0.5 \right)}^{\circ }}}+\dfrac{{{\cos }^{6}}\left( {{90}^{\circ }}-{{0.5}^{\circ }} \right)}{2{{\sin }^{6}}{{\left( 0.5 \right)}^{\circ }}} \\ &nbsp;& =\dfrac{{{\cos }^{6}}{{\left( 0.5 \right)}^{\circ }}}{2{{\cos }^{6}}{{\left( 0.5 \right)}^{\circ }}}+\dfrac{{{\sin }^{6}}{{\left( 0.5 \right)}^{\circ }}}{2{{\sin }^{6}}{{\left( 0.5 \right)}^{\circ }}} \\ &nbsp;& =\dfrac{1}{2}+\dfrac{1}{2}=1 \\ \end{align}$

 Which of the following is equal to $\dfrac{7}{2}\sin {{71}^{\circ }}\cos {{19}^{\circ }}+\dfrac{7}{2}{{\sin }^{2}}{{19}^{\circ }}+\dfrac{9}{2}\tan {{45}^{\circ }}$?

 First apply the relation between sine and cosine of complementary angles and then apply suitable trigonometric identity to simplify.

$\begin{align} &nbsp;&nbsp;& Write\text{ }the\text{ }given\text{ }value \\ &nbsp;& \dfrac{7}{2}\sin {{71}^{\circ }}\cos {{19}^{\circ }}+\dfrac{7}{2}{{\sin }^{2}}{{19}^{\circ }}+\dfrac{9}{2}\tan {{45}^{\circ }} \\ &nbsp;& =\dfrac{7}{2}\sin \left( {{90}^{\circ }}-{{19}^{\circ }} \right)\cos {{19}^{\circ }}+\dfrac{7}{2}{{\sin }^{2}}{{19}^{\circ }}+\dfrac{9}{2}\tan {{45}^{\circ }} \\ &nbsp;& =\dfrac{7}{2}{{\cos }^{2}}{{19}^{\circ }}+\dfrac{7}{2}{{\sin }^{2}}{{19}^{\circ }}+\dfrac{9}{2}\times 1\text{ }\left( \because \tan {{45}^{\circ }}=1 \right) \\ &nbsp;& =\dfrac{7}{2}\left( {{\cos }^{2}}{{19}^{\circ }}+{{\sin }^{2}}{{19}^{\circ }} \right)+\dfrac{9}{2} \\ &nbsp;& =\dfrac{7}{2}+\dfrac{9}{2}=8\text{ }\left( \because {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \right) \\ \end{align}$

 Evaluate: ${{e}^{{{\left[ {{\sin }^{2}}\left( {{90}^{\circ }}-\dfrac{u}{2} \right)+{{\cos }^{2}}\left( {{90}^{\circ }}-\dfrac{u}{2} \right) \right]}^{3}}}}$

 Use relation between sine and cosine of complementary angles in power of the given number. 

$\begin{align} &nbsp;&nbsp;& Consider\text{ }the\text{ }given\text{ }number,\text{ }{{e}^{{{\left[ {{\sin }^{2}}\left( {{90}^{\circ }}-\dfrac{u}{2} \right)+{{\cos }^{2}}\left( {{90}^{\circ }}-\dfrac{u}{2} \right) \right]}^{3}}}} \\ &nbsp;& ={{e}^{{{\left( {{\cos }^{2}}\dfrac{u}{2}+{{\sin }^{2}}\dfrac{u}{2} \right)}^{3}}}} \\ &nbsp;& ={{e}^{{{\left( 1 \right)}^{3}}}}\text{ }\left( \because {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \right) \\ &nbsp;& =e \\ \end{align}$

 Let $T=\sin \left( \dfrac{A}{2}+5 \right).\cos \left[ {{90}^{\circ }}-\left( \dfrac{A+10}{2} \right) \right]$ and $Z=\cos \left( \dfrac{A}{2}+5 \right).sin\left[ {{90}^{\circ }}-\left( \dfrac{A+10}{2} \right) \right]$,  then write the value of ${{\left( T+Z \right)}^{\dfrac{5}{2}}}$.

 Use the relations  $\cos \left( {{90}^{\circ }}-\theta  \right)=\sin \theta $ and  $\sin \left( {{90}^{\circ }}-\theta  \right)=\cos \theta $ to solve further.

$\begin{align} &nbsp;&nbsp;& Write\text{ }the\text{ }value\text{ }of\text{ T}\text{.} \\ &nbsp;& T=\sin \left( \dfrac{A}{2}+5 \right).\cos \left[ {{90}^{\circ }}-\left( \dfrac{A+10}{2} \right) \right] \\ &nbsp;& \Rightarrow T=\sin \left( \dfrac{A+10}{2} \right).\sin \left( \dfrac{A+10}{2} \right)={{\sin }^{2}}\left( \dfrac{A+10}{2} \right).........\left( i \right) \\ &nbsp;& \left[ \because \cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta \right] \\ &nbsp;& Write\text{ }the\text{ }value\text{ }of\text{ z} \\ &nbsp;& z=\cos \left( \dfrac{A}{2}+5 \right).\sin \left[ {{90}^{\circ }}-\left( \dfrac{A+10}{2} \right) \right] \\ &nbsp;& \Rightarrow z=\cos \left( \dfrac{A+10}{2} \right).\cos \left( \dfrac{A+10}{2} \right)={{\cos }^{2}}\left( \dfrac{A+10}{2} \right)......\left( ii \right) \\ &nbsp;& \left( \because \sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta \right) \\ &nbsp;& \text{Add values of T and Z using equations }\left( i \right)\text{ and }\left( ii \right) \\ &nbsp;& T+Z={{\sin }^{2}}\left( \dfrac{A+10}{2} \right)+{{\cos }^{2}}\left( \dfrac{A+10}{2} \right) \\ &nbsp;& =1 \\ &nbsp;& \left( \because {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \right) \\ &nbsp;& \Rightarrow {{\left( T+Z \right)}^{^{\dfrac{5}{2}}}}={{\left( 1 \right)}^{^{\dfrac{5}{2}}}}=1 \\ \end{align}$

 Find the value of $log\left[ \sin \left( {{90}^{\circ }}-\dfrac{17m}{2} \right)\cos \dfrac{17m}{2}+\cos \left( {{90}^{\circ }}-\dfrac{17m}{2} \right)sin\dfrac{17m}{2} \right]$.

 First write  $\sin \left( {{90}^{\circ }}-\theta  \right)=\cos \theta $  and  $\cos \left( {{90}^{\circ }}-\theta  \right)=\sin \theta $. Then apply suitable trigonometric identity to simplify.

$\begin{align} &nbsp;&nbsp;& Consider\text{ }the\text{ }given\text{ }expression. \\ &nbsp;& \log \left[ \sin \left( {{90}^{\circ }}-\dfrac{17m}{2} \right)\cos \dfrac{17m}{2}+\cos \left( {{90}^{\circ }}-\dfrac{17m}{2} \right)\sin \left( \dfrac{17m}{2} \right) \right] \\ &nbsp;& =\log \left[ \cos \dfrac{17m}{2}.\cos \dfrac{17m}{2}+\sin \dfrac{17m}{2}.\sin \dfrac{17m}{2} \right] \\ &nbsp;& =\log \left( {{\cos }^{2}}\dfrac{17m}{2}+{{\sin }^{2}}\dfrac{17m}{2} \right) \\ &nbsp;& =\log 1 \\ &nbsp;& \left( \because {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \right) \\ &nbsp;& =0 \\ \end{align}$