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This is the second of our many quizzes on Trigonometric values on the unit circle. Click on this to access more such competitive quizzes on Trigonometry.

Trigonometric values on the unit circle Quiz 2

mathematics

 Simplify $\sqrt[3]{81}\tan \left( 210{}^\circ \right)\cot (240{}^\circ )$.

 Find the coordinate values at the given degree and find the trigonometric values.

  $\sqrt[3]{81}\tan \left( 210{}^\circ \right)\cot (240{}^\circ ) \\ \Rightarrow \sqrt[3]{81}\left( \dfrac{\sin \left( 210{}^\circ \right)}{\cos \left( 210{}^\circ \right)} \right)\left( \dfrac{\cos \left( 240{}^\circ \right)}{\sin \left( 240{}^\circ \right)} \right) \\ \Rightarrow 3\sqrt[3]{3}\left( \dfrac{-\dfrac{1}{2}}{-\dfrac{\sqrt{3}}{2}} \right)\left( \dfrac{-\dfrac{1}{2}}{-\dfrac{\sqrt{3}}{2}} \right) \\ \Rightarrow \sqrt[3]{3} $

 Find the value of $\sqrt{3}\sin \left( {{210}^{{}^\circ }} \right)$ on a unit circle.

 Convert the given sine angle into known values, then apply the corresponding formula to simplify the expression. Then use the unit circle to substitute the values and simplify.

  $ \sqrt{3}\sin \left( {{210}^{{}^\circ }} \right) \\ \Rightarrow \sqrt{3}\sin \left( 270{}^\circ -60{}^\circ \right) \\ \Rightarrow \sqrt{3}\left[ \sin 270{}^\circ \cos 60{}^\circ -\cos 270{}^\circ \sin 60{}^\circ \right] \\ \text{Substituting values referring to unit circle, we get} \\ \Rightarrow \sqrt{3}\left[ -1\left( \dfrac{1}{2} \right)-0 \right] \\ \Rightarrow \dfrac{-\sqrt{3}}{2} $

 Find the value of $\tan \left( \dfrac{\pi }{6} \right)+\cos \left( \dfrac{13\pi }{6} \right)$ .

 $ \tan \left( \dfrac{\pi }{6} \right)+\cos \left( \dfrac{13\pi }{6} \right) \\ =\tan \left(\dfrac{\pi }{6}\right)+\cos \left(\dfrac{\pi }{6}\right)     [\because cos\dfrac{13\pi }{6} = cos(390) = cos(360+30)=cos 30^\circ] \\ =\dfrac{1}{\sqrt{3}}+\dfrac{\sqrt{3}}{2} \\ =\dfrac{5\sqrt{3}}{6}$

 What is the ratio of $\cot \left( \dfrac{5\pi }{6} \right)$ and $\tan \left( \dfrac{3\pi }{4} \right)$.

 Use $\tan \theta =\dfrac{\sin \theta }{\cos \theta },\cot \theta =\dfrac{\cos \theta }{\sin \theta }$. Then substitute values from the unit circle separately. Then find the ratio.

 $ \cot \left( \dfrac{5\pi }{6} \right) \\ =\dfrac{\cot \left(\dfrac{5\pi }{6}\right)}{\tan \left(\dfrac{3\pi }{4}\right)}\\ \dfrac{\cot \left(-\dfrac{\pi }{6}+\pi \right)}{-tan\left(\dfrac{\pi}{4}\right)} \\ =\dfrac{-\sqrt{3}}{-1} = \sqrt{3}$.

 What is twice the value of $\dfrac{\sin \left( \dfrac{7\pi }{6} \right)\cos \pi }{\tan \left( \dfrac{\pi }{3} \right)}\times \dfrac{\text{cosec}\left( \dfrac{5\pi }{4} \right)\sec (2\pi )}{\cot \left( \dfrac{7\pi }{4} \right)}$. 

 Find the coordinate values on the unit circle for the given radians and determine the respective trigonometric values.

$\dfrac{\sin \left(\dfrac{7\pi }{6}\right)\cos \left(\pi \right)}{\tan \left(\dfrac{\pi }{3}\right)}\times \dfrac{\csc \left(\dfrac{5\pi }{4}\right)\sec \left(2\pi \right)}{\cot \left(\dfrac{7\pi }{4}\right)} \\ =\dfrac{\sin \left(\dfrac{7\pi }{6}\right)\cos \left(\pi \right)}{\tan \left(\dfrac{\pi }{3}\right)}\times \dfrac{\dfrac{1}{\sin \left(\dfrac{5\pi }{4}\right)}\times \dfrac{1}{\cos \left(2\pi \right)}}{\cot \left(\dfrac{7\pi }{4}\right)} \\ =\dfrac{\sin \left(\dfrac{7\pi }{6}\right)\cos \left(\pi \right)}{\tan \left(\dfrac{\pi }{3}\right)}\times \dfrac{\dfrac{1}{\sin \left(\dfrac{5\pi }{4}\right)}\times \dfrac{1}{\left(-1+2\cos ^2\left(\pi \right)\right)}}{\cot \left(\dfrac{7\pi }{4}\right)} \\ = \dfrac{\cos \left(\pi \right)\sin \left(\dfrac{7\pi }{6}\right)}{\tan \left(\dfrac{\pi }{3}\right)\sin \left(\dfrac{5\pi }{4}\right)\cot \left(\dfrac{3\pi }{4}\right)\left(-1+2\cos ^2\left(\pi \right)\right)} \\  = \dfrac{\left(-1\right)\left(-\dfrac{1}{2}\right)}{\sqrt{3}\left(-\dfrac{\sqrt{2}}{2}\right)\left(-1\right)\left(-1+2\left(-1\right)^2\right)} = \dfrac{1}{\sqrt{6}}$