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# How do you determine $\sin \theta$ given $\cos \theta =-\dfrac{2}{3},{{90}^{0}} < \theta < {{180}^{0}}$?

Last updated date: 11th Jun 2024
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Hint: From the question given, we have been asked to find $\sin \theta$ and given $\cos \theta =-\dfrac{2}{3},{{90}^{0}} < \theta < {{180}^{0}}$. We can solve the given question by knowing the basic formulae and basic principles of trigonometry. By using them we can solve the given question very easily.

Now considering from the question we have to determine the value of $\sin \theta$ when $\cos \theta =-\dfrac{2}{3}$ in the interval ${{90}^{0}} < \theta < {{180}^{0}}$
First of all, we have to explain about the interval given in the question.
The interval given in the question is ${{90}^{0}} < \theta < {{180}^{0}}$
By the basic principles of trigonometry, we can say that the given interval indicates the second quadrant.
We know that in second quadrant sine angle is positive $\sin \theta >0$
In trigonometry, we have one basic identity that is shown below: ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
By using the above basic trigonometric identity, we have to solve the given question.
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
Shift ${{\cos }^{2}}\theta$ to the right hand side of the equation from the left hand side of the equation.
By shifting ${{\cos }^{2}}\theta$ from the left hand side of the equation to the right hand side of the equation, we get
$\Rightarrow {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta$
From the question, it has been already given that $\cos \theta =-\dfrac{2}{3},$ substitute it in the above identity.
By substituting, we get $\Rightarrow {{\sin }^{2}}\theta =1-{{\left( -\dfrac{2}{3} \right)}^{2}}$
Simplify further to get an accurate and exact answer.
By simplifying further, we get
$\Rightarrow {{\sin }^{2}}\theta =1-\dfrac{4}{9}$
$\Rightarrow {{\sin }^{2}}\theta =\dfrac{5}{9}$
$\Rightarrow \sin \theta =\pm \dfrac{\sqrt{5}}{3}$
As we have already discussed earlier, in the second quadrant sine is positive.
$\sin \theta =\dfrac{\sqrt{5}}{3}$

Hence, we determined the value of $\sin \theta$.

Note: We should be well known about the basic identities and principles of trigonometry. We should not neglect the interval given in the question. Using the interval given, we have to find the answer. Also, we should be very careful while doing the calculation part. Similarly we can determine the values of all other trigonometric ratios using one the value of one trigonometric ratio. Like here in this case we can determine the value of all trigonometric ratios as follows $\sin \theta =\dfrac{\sqrt{5}}{3}$ , $\csc \theta =\dfrac{1}{\sin \theta }=\dfrac{3}{\sqrt{5}}$ , $\sec \theta =\dfrac{1}{\cos \theta }=\dfrac{-3}{2}$ , $\tan \theta =\dfrac{\sin \theta }{\cos \theta }\Rightarrow \dfrac{\left( \dfrac{\sqrt{5}}{3} \right)}{\left( \dfrac{-2}{3} \right)}=\dfrac{-\sqrt{5}}{2}$ and $\cot \theta =\dfrac{1}{\tan \theta }=\dfrac{-2}{\sqrt{5}}$ for given $\cos \theta =-\dfrac{2}{3}$ in the interval ${{90}^{0}} < \theta < {{180}^{0}}$ .