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# Given $sin\theta$ = $\dfrac{2}{3}$ and $\dfrac{\pi }{2} < \theta < \pi$ how do you find the value of the other 5 trigonometric functions?

Last updated date: 28th Feb 2024
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Hint: To solve these types of problems, we will use the relationship between the trigonometric ratios and some trigonometric identities. We should know the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Also, the relationship such as, $\tan \theta =\dfrac{\sin \theta }{\cos \theta },\csc \theta =\dfrac{1}{\sin \theta },\sec \theta =\dfrac{1}{\cos \theta }\And \cot \theta =\dfrac{1}{\tan \theta }$. We should also know that in the second quadrant only sine and cosecant ratios are positive and others are negative.

Complete step by step solution:
We are given that the $\sin \theta =\dfrac{2}{3}$. Here the angle is $\dfrac{\pi }{2} < \theta < \pi$. As the angle lies in this range it means that the angle lies in the second quadrant. We know that in the second quadrant only sine and cosecant ratios are positive and others are negative.
We know the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, substituting $\sin \theta =\dfrac{2}{3}$ in this identity, we get
$\Rightarrow {{\left( \dfrac{2}{3} \right)}^{2}}+{{\cos }^{2}}\theta =1$
$\Rightarrow \dfrac{4}{9}+{{\cos }^{2}}\theta =1$
Subtracting $\dfrac{4}{9}$ from both sides of the above equation, we get
$\Rightarrow {{\cos }^{2}}\theta =1-\dfrac{4}{9}=\dfrac{5}{9}$
taking the square root of both sides of the above equation, we get
\begin{align} & \Rightarrow \cos \theta =\pm \sqrt{\dfrac{5}{9}} \\ & \Rightarrow \cos \theta =\pm \dfrac{\sqrt{5}}{3} \\ \end{align}
As we already said that the angle is in the third quadrant so cosine ratios will be negative.
$\Rightarrow \cos \theta =-\dfrac{\sqrt{5}}{3}$
Now, we can find the other trigonometric ratios using their relationships as,
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$, substituting the value of the ratios, we get
$\Rightarrow \tan \theta =\dfrac{\dfrac{2}{3}}{-\dfrac{\sqrt{5}}{3}}=-\dfrac{2}{\sqrt{5}}$
$\Rightarrow \csc \theta =\dfrac{1}{\sin \theta }=\dfrac{1}{\dfrac{2}{3}}=\dfrac{3}{2}$
$\Rightarrow \sec \theta =\dfrac{1}{\cos \theta }=\dfrac{1}{-\dfrac{\sqrt{5}}{3}}=-\dfrac{3}{\sqrt{5}}$
$\Rightarrow \cot \theta =\dfrac{1}{\tan \theta }=\dfrac{1}{-\dfrac{2}{\sqrt{5}}}=-\dfrac{\sqrt{5}}{2}$
Thus, we have found all the trigonometric ratios.

Note:
To solve these types of problems, one should know the trigonometric identities and the relationship between the ratios. Here we used $\tan \theta =\dfrac{\sin \theta }{\cos \theta },\csc \theta =\dfrac{1}{\sin \theta },\sec \theta =\dfrac{1}{\cos \theta }\And \cot \theta =\dfrac{1}{\tan \theta }$ and ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Also, we should know which trigonometric ratios are positive in which quadrants. In the first quadrant, all ratios are positive, in the third quadrant tangent and cotangent are positive, and in the fourth quadrant only cosine and secant are positive.