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This is one of our many quizzes on Half-angle identity. Click on this to access more such competitive quizzes on Trigonometry.

Half-angle identity Quiz 2

mathematics

 Find the value of $\text{cosec}x$ if $\sin \left( \dfrac{x}{2} \right)=-\dfrac{7}{8}$ where $\dfrac{x}{2}$ lies between $\pi \text{ and }\dfrac{3\pi }{2}$.

 Use the identity $\sin \left( \dfrac{x}{2} \right)=\pm \sqrt{\dfrac{1-\cos x}{2}}$then by Pythagoras theorem, find the required value.

  $\sin \left( \dfrac{x}{2} \right)=\pm \sqrt{\dfrac{1-\cos x}{2}}=-\dfrac{7}{8} \\   \Rightarrow \dfrac{1-\cos x}{2}=\dfrac{49}{64}\Rightarrow \cos x=-\dfrac{17}{32} \\   \text{And, }\pi < \dfrac{x}{2} < \dfrac{3\pi }{2}\Rightarrow 0{}^\circ < x < 180{}^\circ  \\   \text{Hence, }\text{cosec}x=\dfrac{32}{7\sqrt{15}} \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/8cdf7373-8a93-4704-a430-6cd37eb013fb2280302095902825193.png"/>

 If $\tan \left( \dfrac{a}{2} \right)=4$ find the value of $\sec a$. 

 Use the identity ${{\tan }^{2}}\left( \dfrac{a}{2} \right)=\dfrac{1-\cos a}{1+\cos a}$.

  $\text{We have, }\tan \left( \dfrac{a}{2} \right)=4 \\  \sqrt{\dfrac{1-\cos a}{1+\cos a}}=4\left[ \because {{\tan }^{2}}\left( \dfrac{a}{2} \right)=\dfrac{1-\cos a}{1+\cos a} \right] \\   \Rightarrow \dfrac{1-\cos a}{1+\cos a}=16 \\   \Rightarrow 17\cos a=-15 \\   \Rightarrow \cos a=\dfrac{-15}{17} \\   \therefore \sec a=-\dfrac{17}{15}$ 

 Given that, $\sin a=\dfrac{24}{25}$, find the minimum value that $\cot \dfrac{a}{2}$ can have.

 First find the value of cosine function by using Pythagoras theorem then use the identity ${{\cot }^{2}}\left( \dfrac{a}{2} \right)=\dfrac{1+\cos a}{1-\cos a}$.

  $\text{Here, }\sin a=\dfrac{24}{25}\Rightarrow \cos a=\dfrac{7}{25} \\  {{\cot }^{2}}\left( \dfrac{a}{2} \right)=\dfrac{1+\cos a}{1-\cos a} \\   \Rightarrow \cot \left( \dfrac{a}{2} \right)=\pm \sqrt{\dfrac{1+\dfrac{7}{25}}{1-\dfrac{7}{25}}} \\ \Rightarrow \pm \sqrt{\dfrac{32}{18}}=\pm \dfrac{4}{3}$

 If $\sin 2x=\cos 4x$ where $''x''$ lies between $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$, find the largest possible value of $''x''$.

 Use the identity $\cos 4x=1-2{{\sin }^{2}}2x$then form a quadratic equation, solve that equation for the required value.

  $\text{Given, }\sin 2x=\cos 4x \\  \Rightarrow \sin 2x=1-2{{\sin }^{2}}2x \\   \Rightarrow 2{{\sin }^{2}}2x+\sin 2x-1=0 \\   \Rightarrow \sin 2x=-1\text{ or }\dfrac{1}{2} \\   \Rightarrow 2x=-\dfrac{\pi }{2}\text{ or }\dfrac{\pi }{6}\text{ or }\dfrac{\pi }{6} \\   \therefore x=-\dfrac{\pi }{4}\text{or }\dfrac{\pi }{12}\text{ or }\dfrac{5\pi }{12}$

 Given that $16\left( 1-\cos 2a \right)=9\left( 1+\cos 2a \right)$ find the difference between $\dfrac{\left( 1-{{\tan }^{2}}A \right)}{\left( 1+{{\tan }^{2}}A \right)}$ and ${{\cos }^{2}}A-{{\sin }^{2}}A$.

 Use the identity ${{\cot }^{2}}\left( \dfrac{a}{2} \right)=\dfrac{1+\cos a}{1-\cos a}$ for $\cot A$ then by using Pythagoras theorem find other trigonometric ratios.

  $\text{Here, } 16\left( 1-\cos 2a \right)=9\left( 1+\cos 2a \right)\Rightarrow \dfrac{16}{9}={{\left( \dfrac{4}{3} \right)}^{2}}=\dfrac{1+\cos 2a}{1-\cos 2a} \\   \cot A=\dfrac{4}{3}\left[ \because {{\cot }^{2}}a=\dfrac{1+\cos 2a}{1-\cos 2a} \right] \\ \text{By using Pythagoras theorem, we will get;} \\ \sin A=\dfrac{3}{5}\text{ and }\cos A=\dfrac{4}{5} \\  \dfrac{\left( 1-{{\tan }^{2}}A \right)}{\left( 1+{{\tan }^{2}}A \right)}=\dfrac{\left[ 1-{{\left( \dfrac{3}{4} \right)}^{2}} \right]}{\left[ 1+{{\left( \dfrac{3}{4} \right)}^{2}} \right]}\Rightarrow \dfrac{\left[ 1-\left( \dfrac{9}{16} \right) \right]}{\left[ 1+\left( \dfrac{9}{16} \right) \right]}\text{=}\dfrac{7}{25}  \\ {{\cos }^{2}}A - {{\sin }^{2}}A={{\left( \dfrac{4}{5} \right)}^{2}}-{{\left( \dfrac{3}{5} \right)}^{2}}=\dfrac{7}{25}  \\   \text{ Difference}=0 $ 