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

Find the value of(sin⁡233∘+sin⁡257∘)\left( {{\sin }^{2}}33{}^\circ +{{\sin }^{2}}57{}^\circ \right).

Explanation

Solution

Convert any one of the terms in the given expression into terms of cos⁡θ\cos \theta using the relation sin⁡(π2−θ)=cos⁡θ\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⁡2θ+cos⁡2θ=1{{\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∘\sin {{33}^{{}^\circ }} nor sin⁡57∘\sin {{57}^{{}^\circ }} from the trigonometric table. So, to find the value of the (sin⁡233∘+sin⁡257∘)\left( {{\sin }^{2}}33{}^\circ +{{\sin }^{2}}57{}^\circ \right) we need to convert any of them (i.e. sin⁡33∘\sin 33{}^\circ orsin⁡57∘\sin 57{}^\circ ) to cos⁡2θ{{\cos }^{2}}\theta because we know that sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 and try to make the argument of both of the (sin⁡233∘+sin⁡257∘)\left( {{\sin }^{2}}33{}^\circ +{{\sin }^{2}}57{}^\circ \right) equal.
Here, from the argument of (sin⁡233∘+sin⁡257∘)\left( {{\sin }^{2}}33{}^\circ +{{\sin }^{2}}57{}^\circ \right) we can see that 33∘+57∘=90∘=π233{}^\circ +57{}^\circ =90{}^\circ =\dfrac{\pi }{2}
It can also be rewritten as:
33∘=π2−57∘33{}^\circ =\dfrac{\pi }{2}-57{}^\circ
Taking sin⁡\sin both sides, we will get:
sin⁡33∘=sin⁡(π2−57∘)\sin 33{}^\circ =\sin \left( \dfrac{\pi }{2}-57{}^\circ \right)
We know that sin⁡(π2−θ)=cos⁡θ\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta where, θ∈[0,π2]\theta \in \left[ 0,\dfrac{\pi }{2} \right]
∴sin⁡33∘=cos⁡57∘\therefore \sin 33{}^\circ =\cos 57{}^\circ
Now, putting cos⁡57∘\cos 57{}^\circ in place of sin⁡33∘\sin 33{}^\circ in(sin⁡233∘+sin⁡257∘)\left( {{\sin }^{2}}33{}^\circ +{{\sin }^{2}}57{}^\circ \right), we will get (cos⁡257∘+sin⁡257∘)({{\cos }^{2}}57{}^\circ +{{\sin }^{2}}57{}^\circ ).
We know the identity that sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1
In (cos⁡257∘+sin⁡257∘)({{\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⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1we can say that:
(cos⁡257∘+sin⁡257∘)=1({{\cos }^{2}}57{}^\circ +{{\sin }^{2}}57{}^\circ )=1
⇒(sin⁡233∘+sin⁡257∘)=1\Rightarrow \left( {{\sin }^{2}}33{}^\circ +{{\sin }^{2}}57{}^\circ \right)=1
Hence, the value of (sin⁡233∘+sin⁡257∘)\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⁡θ\sin \theta must lies in between θ∈[0,π2]\theta \in \left[ 0,\dfrac{\pi }{2} \right] otherwise, we can’t write sin⁡(π2−θ)\sin \left( \dfrac{\pi }{2}-\theta \right) as cos⁡θ\cos \theta , there will be the chances of change of sign of the cos⁡θ\cos \theta .
For example, suppose any value of θ\theta which is greater than π2\dfrac{\pi }{2}.
Let it be θ=2π3\theta =\dfrac{2\pi }{3}.
Then, sin⁡(π2−θ)=sin⁡(π2−2π3)\sin \left( \dfrac{\pi }{2}-\theta \right)=\sin \left( \dfrac{\pi }{2}-\dfrac{2\pi }{3} \right)
⇒sin⁡(−π6)\Rightarrow \sin \left( -\dfrac{\pi }{6} \right)
We know that sin⁡(−θ)=−sin⁡θ\sin \left( -\theta \right)=-\sin \theta :
Hence, sin⁡(−π6)=−sin⁡(π6)\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.