Question

If \[\sin \theta =\dfrac{1}{2}\] and angle \[\theta \] is acute, how do you find the other five ratios?

Answer 1

Hint: We are given \[\sin \theta =\dfrac{1}{2}\] and also we have \[\theta \] is acute. Using them we will find the other ratio. We will learn how each identity is connected to the other. Using this identity, we will solve our problem. We will also need to have knowledge of which ratio is positive and negative and also lie in which quadrant. These play key roles in the solution of ours.

Complete step by step answer:
We are given that \[\sin \theta =\dfrac{1}{2}\] and we are given that \[\theta \] is acute angle means the angles that lie between 0 and 90 degrees. So, \[\theta \] must lie between 0 and 90 degrees. Now, we can see that \[\sin \theta =\dfrac{1}{2}\] and we also know that \[\sin {{30}^{\circ }}=\dfrac{1}{2}\] as 30 degrees lie between 0 to 90 degrees. So, it means we have \[\theta ={{30}^{\circ }}.\] Now we can use this value of \[\theta ={{30}^{\circ }}\] and find our other ratio. Now, we get our other ratio as
\[\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}\]
\[\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}\]
\[\sec {{30}^{\circ }}=\dfrac{2}{\sqrt{3}}\]
\[\operatorname{cosec}{{30}^{\circ }}=\dfrac{2}{1}\]
\[\cot {{30}^{\circ }}=\dfrac{\sqrt{3}}{1}\]

Note:
This one was easy as we get the particular \[\theta ={{30}^{\circ }}.\] So, we will learn another way which will work for all types of problems. To use this we will learn that each ratio is connected to one another. For our problem, we need,
1. Reciprocal Identity
\[\sin \theta =\dfrac{1}{\operatorname{cosec}\theta },\cos \theta =\dfrac{1}{\sec \theta },\tan \theta =\dfrac{1}{\cot \theta }\]
2. Three Identity
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
\[{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \]
\[{{\operatorname{cosec}}^{2}}\theta =1+{{\cot }^{2}}\theta \]
3. Knowledge of sign of ratio in a different quadrant, we know

Now, we have \[\sin \theta =\dfrac{1}{2},\] \[\theta \] is acute, so \[\theta \] is less than 90 degrees, so \[\theta \] will lie in the first quadrant. In the first quadrant, all ratios are positive. Now, we know \[\sin \theta =\dfrac{1}{\operatorname{cosec}\theta }\] or \[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }\]
So, \[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }=\dfrac{1}{\dfrac{1}{2}}=\dfrac{2}{1}.....\left( i \right)\]
Now, using \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1,\] we put \[\sin \theta =\dfrac{1}{2},\] so we get,
\[\Rightarrow {{\left( \dfrac{1}{2} \right)}^{2}}+{{\cos }^{2}}\theta =1\]
On simplifying, we get,
\[\cos \theta =\pm \sqrt{1-\dfrac{1}{4}}\]
So,
\[\Rightarrow \cos \theta =\pm \dfrac{\sqrt{3}}{2}\]
As \[\cos \theta \] is positive in the first quadrant, so, we get,
\[\cos \theta =\dfrac{\sqrt{3}}{2}.....\left( ii \right)\]
Now as \[\cos \theta =\dfrac{1}{\sec \theta }\] or \[\sec \theta =\dfrac{1}{\cos \theta }\]
So we get using (ii) that
\[\Rightarrow \sec \theta =\dfrac{1}{\dfrac{\sqrt{3}}{2}}=\dfrac{2}{\sqrt{3}}.....\left( iii \right)\]
Now as we know that \[\tan \theta =\dfrac{\sin \theta }{\cos \theta },\] using \[\sin \theta =\dfrac{1}{2}\] and \[\cos \theta =\dfrac{\sqrt{3}}{2},\] we get
\[\Rightarrow \tan \theta =\dfrac{1}{\sqrt{3}}\]
Now as \[\cot \theta =\dfrac{1}{\tan \theta },\] so
\[\cot \theta =\dfrac{1}{\dfrac{1}{\sqrt{3}}}=\sqrt{3}\]
So, we get all our ratio as \[\cos \theta =\dfrac{\sqrt{3}}{2},\operatorname{cosec}\theta =2,\sec \theta =\dfrac{2}{\sqrt{3}},\tan \theta =\dfrac{1}{\sqrt{3}},\cot \theta =\sqrt{3}.\]