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This is the second of our many quizzes on Double-angle identity. Click on this to access more such competitive quizzes on Trigonometry.

Double-angle identity Quiz 2

mathematics

 Find the value of $\left( {\dfrac{1}{{\cos 10^\circ }} + \dfrac{{\sqrt 3 }}{{\sin 10^\circ }}} \right)$.

 Simplify the expression to get the form $\cos A \cdot \cos B + \sin A \cdot \sin B$ and then use the identity $\cos \left( {A - B} \right) = \cos A \cdot \cos B + \sin A \cdot \sin B$. 

  $\dfrac{1}{{\cos 10^\circ }} + \dfrac{{\sqrt 3 }}{{\sin 10^\circ }} = \dfrac{{\sin 10^\circ  + \sqrt 3 \cos 10^\circ }}{{\cos 10^\circ \sin 10^\circ }} \\ $ Multiplying and dividing $2$ in the numerator, we get, $  \\   \Rightarrow \dfrac{{2\left( {\dfrac{1}{2}\sin 10^\circ  + \dfrac{{\sqrt 3 }}{2}\cos 10^\circ } \right)}}{{\cos 10^\circ \sin 10^\circ }} \\  {\text{We can write }}\dfrac{{\sqrt 3 }}{2}{\text{ }} = \cos 30^\circ  \\  {\text{And, }}\dfrac{1}{2}{\text{ }} = \sin 30^\circ  \\   \Rightarrow \dfrac{{2\left( {\cos 30^\circ \cos 10^\circ  + \sin 30^\circ \sin 10^\circ } \right)}}{{\cos 10^\circ \sin 10^\circ }} \\   \Rightarrow \dfrac{{2\left[ {\cos \left( {30 - 10} \right)} \right]}}{{\cos 10^\circ \sin 10^\circ }} \\   \Rightarrow \dfrac{{2 \times 2\cos 20^\circ }}{{2\cos 10^\circ \sin 10^\circ }} \\   \Rightarrow 4\left( {\dfrac{{\cos 20^\circ }}{{\sin 20^\circ }}} \right) \\   \Rightarrow 4\cot 20^\circ  \\  $

 Find the value of $\left( {{{\cos }^2}\theta  - {{\sin }^2}\theta } \right)$ where the value of $\theta  = 75^\circ $?

 Simplify to get the expression in the form where the double angle formula $\cos 2\theta  = \left( {\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$ can be applied. 

  $  \left( {{{\cos }^2}\theta  - {{\sin }^2}\theta } \right) \\   \Rightarrow \left( {\dfrac{{{{\cos }^2}\theta  - {{\sin }^2}\theta }}{1}} \right) \\   \Rightarrow \left( {\dfrac{{{{\cos }^2}\theta  - {{\sin }^2}\theta }}{{{{\cos }^2}\theta  + {{\sin }^2}\theta }}} \right)\left\{ {\because 1 = \left( {{{\sin }^2}\theta  + {{\cos }^2}\theta } \right)} \right\} \\  {\text{Dividing by }}{\cos ^2}\theta {\text{ in numerator and denominator}}{\text{.}} \\   \Rightarrow \dfrac{{\left( {\dfrac{{{{\cos }^2}\theta  - {{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)}}{{\left( {\dfrac{{{{\cos }^2}\theta  + {{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)}} \\   \Rightarrow \left( {\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right) \\  {\text{Where }}\theta  = 75^\circ  \\   \Rightarrow \left( {\dfrac{{1 - {{\tan }^2}75^\circ }}{{1 + {{\tan }^2}75^\circ }}} \right) \\   \Rightarrow \cos 2\left( {75^\circ } \right) \\   \Rightarrow \cos 150^\circ  \\   \Rightarrow \left( { - \dfrac{{\sqrt 3 }}{2}} \right) \\  $

 Simplify $\dfrac{{\sin 4x}}{{\sin \dfrac{x}{2}}}$.

 Use the identity $\sin 2\theta  = 2\sin \theta \cos \theta $. 

$  \dfrac{{\sin 4x}}{{\sin \dfrac{x}{2}}} \\  {\text{Solve }}\sin 4x \\   \Rightarrow \sin 4x = 2\sin 2x\cos 2x\left[ {\because {\text{By}}\sin 2x = 2\sin x\cos x} \right] \\   \Rightarrow \sin 4x = 2\left( {2\sin x\cos x} \right)\cos 2x \\   \Rightarrow \sin 4x = 4\left( {2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \right)\cos x\cos 2x \\  {\text{Putting the value of }}\sin 4x \\   \Rightarrow \dfrac{{\sin 4x}}{{\sin \dfrac{x}{2}}} = \dfrac{{8\sin \dfrac{x}{2}\cos \dfrac{x}{2}\cos x\cos 2x}}{{\sin \dfrac{x}{2}}} \\   \Rightarrow \dfrac{{\sin 4x}}{{\sin \dfrac{x}{2}}} = \left( {8\cos \dfrac{x}{2}\cos x\cos 2x} \right) \\  $

 Find the value of $\cos 75^\circ \cos 37{{\dfrac{1}{2}}^\circ }\sin 37{{\dfrac{1}{2}}^\circ }$.

 Use the identity $\sin 2\theta  = 2\sin \theta  \cdot \cos \theta $. 

 $  \left( {\cos 75^\circ \cos 37{{\dfrac{1}{2}}^\circ }\sin 37{{\dfrac{1}{2}}^\circ }} \right) \\  {\text{Dividing and multiplying by }}2 \\   \Rightarrow \left( {\dfrac{1}{2}\cos 75^\circ } \right)\left( {2\cos 37{{\dfrac{1}{2}}^\circ }\sin 37{{\dfrac{1}{2}}^\circ }} \right) \\   \Rightarrow \left( {\dfrac{1}{2}\cos 75^\circ } \right)\left( {\sin 75^\circ } \right) \\  {\text{Again multiplying and dividing by }}2 \\    \dfrac{1}{4}\left( {2\cos 75^\circ \sin 75^\circ } \right) \\   \Rightarrow \dfrac{1}{4}\left( {\sin 150^\circ } \right) \\   \Rightarrow \dfrac{1}{4}\left( {\dfrac{{1}}{{2}}} \right) = \dfrac{1}{{8}} $

 Find the value of $\left( {{{\sin }^2}2x - {{\sin }^2}x} \right)$.

 Use $\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$. 

 $  {\sin ^2}2x - {\sin ^2}x = \left( {\sin 2x + \sin x} \right)\left( {\sin 2x - \sin x} \right) \\   \Rightarrow \left( {2\sin \dfrac{{3x}}{2}\cos \dfrac{x}{2}} \right)\left( {2\cos \dfrac{{3x}}{2}\sin \dfrac{x}{2}} \right) \\   \Rightarrow \left( {2\sin \dfrac{{3x}}{2}\cos \dfrac{{3x}}{2}} \right)\left( {2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \right) \\   \Rightarrow \sin 3x\sin x \\  $ 