Question

Find the value of\[\left( {{\sin }^{2}}33{}^\circ +{{\sin }^{2}}57{}^\circ \right)\].

Answer 1

Hint: Convert any one of the terms in the given expression into terms of \[\cos \theta \] using the relation \[\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta \] . Then, we have to try to make the value of both the arguments equal such that we can use the identity \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. This will directly give us the final answer.

Complete step-by-step solution:
Since, we do not know the value of neither \[\sin {{33}^{{}^\circ }}\] nor \[\sin {{57}^{{}^\circ }}\] from the trigonometric table. So, to find the value of the \[\left( {{\sin }^{2}}33{}^\circ +{{\sin }^{2}}57{}^\circ \right)\] we need to convert any of them (i.e. \[\sin 33{}^\circ \]or\[\sin 57{}^\circ \]) to \[{{\cos }^{2}}\theta \] because we know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] and try to make the argument of both of the \[\left( {{\sin }^{2}}33{}^\circ +{{\sin }^{2}}57{}^\circ \right)\] equal.
Here, from the argument of \[\left( {{\sin }^{2}}33{}^\circ +{{\sin }^{2}}57{}^\circ \right)\] we can see that \[33{}^\circ +57{}^\circ =90{}^\circ =\dfrac{\pi }{2}\]
It can also be rewritten as:
\[33{}^\circ =\dfrac{\pi }{2}-57{}^\circ \]
Taking \[\sin \] both sides, we will get:
\[\sin 33{}^\circ =\sin \left( \dfrac{\pi }{2}-57{}^\circ \right)\]
We know that \[\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta \] where, \[\theta \in \left[ 0,\dfrac{\pi }{2} \right]\]
\[\therefore \sin 33{}^\circ =\cos 57{}^\circ \]
Now, putting \[\cos 57{}^\circ \] in place of $\sin 33{}^\circ $ in\[\left( {{\sin }^{2}}33{}^\circ +{{\sin }^{2}}57{}^\circ \right)\], we will get \[({{\cos }^{2}}57{}^\circ +{{\sin }^{2}}57{}^\circ )\].
We know the identity that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
In \[({{\cos }^{2}}57{}^\circ +{{\sin }^{2}}57{}^\circ )\] we can see that argument of both sine part and cosine part are equal.
Hence, from the identity \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]we can say that:
\[({{\cos }^{2}}57{}^\circ +{{\sin }^{2}}57{}^\circ )=1\]
\[\Rightarrow \left( {{\sin }^{2}}33{}^\circ +{{\sin }^{2}}57{}^\circ \right)=1\]
Hence, the value of \[\left( {{\sin }^{2}}33{}^\circ +{{\sin }^{2}}57{}^\circ \right)\] is 1.
This is the required solution.

Note:
Here, students need to take care that arguments of \[\sin \theta \] must lies in between \[\theta \in \left[ 0,\dfrac{\pi }{2} \right]\] otherwise, we can’t write \[\sin \left( \dfrac{\pi }{2}-\theta \right)\] as \[\cos \theta \] , there will be the chances of change of sign of the \[\cos \theta \].
For example, suppose any value of $\theta $ which is greater than $\dfrac{\pi }{2}$.
Let it be $\theta =\dfrac{2\pi }{3}$.
Then, $\sin \left( \dfrac{\pi }{2}-\theta \right)=\sin \left( \dfrac{\pi }{2}-\dfrac{2\pi }{3} \right)$
$\Rightarrow \sin \left( -\dfrac{\pi }{6} \right)$
We know that $\sin \left( -\theta \right)=-\sin \theta $:
Hence, $\sin \left( -\dfrac{\pi }{6} \right)=-\sin \left( \dfrac{\pi }{6} \right)$
 So, there is a change of sign, as can be seen by the above equation.