Question

Evaluate the following expression, ${{\sin }^{2}}30{}^\circ +{{\sin }^{2}}45{}^\circ +{{\sin }^{2}}60{}^\circ +{{\sin }^{2}}90{}^\circ $.

Answer 1

Hint:When we look at the question, we know that the angles are standard angles So, we should know that $\sin 30{}^\circ =\dfrac{1}{2},\sin 45{}^\circ =\dfrac{1}{\sqrt{2}},\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ and $\sin 90{}^\circ =1$. By substituting these values in the given expression, we can find the answer.

Complete step-by-step answer:
In this question, we are asked to evaluate an expression, that is, ${{\sin }^{2}}30{}^\circ +{{\sin }^{2}}45{}^\circ +{{\sin }^{2}}60{}^\circ +{{\sin }^{2}}90{}^\circ $.
We must have the basic knowledge of trigonometric ratios, which are the ratios of two of the three sides of a right angled triangle. We can say that $\sin \theta =\dfrac{Perpendicular}{Hypotenuse}$.
To solve the given expression, we will put the values of the standard sine angles, which are, $\sin 30{}^\circ ,\sin 45{}^\circ ,\sin 60{}^\circ $ and $\sin 90{}^\circ $ respectively. After putting the values in the expression, we will simplify it further to get the desired answer.
Now, we know that $\sin 30{}^\circ $ is expressed as $\dfrac{1}{2}$, $\sin 45{}^\circ $ is expressed as $\dfrac{1}{\sqrt{2}}$, $\sin 60{}^\circ $ is expressed as $\dfrac{\sqrt{3}}{2}$, and $\sin 90{}^\circ $ is expressed as 1. So, we can substitute these values in the given expression, so we get,
${{\sin }^{2}}30{}^\circ +{{\sin }^{2}}45{}^\circ +{{\sin }^{2}}60{}^\circ +{{\sin }^{2}}90{}^\circ ={{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}+{{\left( 1 \right)}^{2}}$
And we know that, ${{\left( \dfrac{1}{2} \right)}^{2}}=\dfrac{1}{4},{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}=\dfrac{1}{2},{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}=\dfrac{3}{4},{{\left( 1 \right)}^{2}}=1$. So, we will substitute these values in the expression and get as,
$\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{3}{4}+1$
Now, we will take the LCM of the above values. So, we will get,
$\dfrac{1+2+3+4}{4}$
And on further simplification, we will get the value of the expression as,
$\dfrac{10}{4}=\dfrac{5}{2}$
Hence, we get the value of the given expression, ${{\sin }^{2}}30{}^\circ +{{\sin }^{2}}45{}^\circ +{{\sin }^{2}}60{}^\circ +{{\sin }^{2}}90{}^\circ $ as $\dfrac{5}{2}$.

Note: While solving this question, one can think of applying the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, which gives ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $. This is also a correct way of solving the question, but it will become lengthy and so the chances of calculation mistakes will also increase. So, it is better to remember that, $\sin 30{}^\circ =\dfrac{1}{2},\sin 45{}^\circ =\dfrac{1}{\sqrt{2}},\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ and $\sin 90{}^\circ =1$.