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# Find the value of the expression $2{{\left( \sin 15+\sin 75 \right)}^{2}}$?

Last updated date: 25th Jul 2024
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Hint: We first try to convert all the trigonometric ratios into forms of equal angles to apply the formulas and identities like $2\sin \theta \cos \theta =\sin 2\theta$ and ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. We convert $\sin 75$ into $\cos 15$. We break the square part using the formula of ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. We place the values in the formula and find the final solution.

We first convert all the given trigonometric ratios into forms of equal angles. We choose angles of 15.
We know that $\sin \alpha =\cos \left( \dfrac{\pi }{2}-\alpha \right)$. Putting the value of $\alpha =75$, we get
$\sin 75=\cos \left( \dfrac{\pi }{2}-75 \right)=\cos 15$.
Therefore, we have $2{{\left( \sin 15+\sin 75 \right)}^{2}}=2{{\left( \sin 15+\cos 15 \right)}^{2}}$.
We now apply the formula of ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
We get ${{\left( \sin 15+\cos 15 \right)}^{2}}={{\sin }^{2}}15+{{\cos }^{2}}15+2\sin 15\cos 15$.
We have the formula of multiple angles where we get $2\sin \theta \cos \theta =\sin 2\theta$.
So, we get $2\sin 15\cos 15=\sin \left( 15\times 2 \right)=\sin 30=\dfrac{1}{2}$.
We also have the identity formula of ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.
Applying the formula, we get ${{\sin }^{2}}15+{{\cos }^{2}}15=1$.
Putting all the values we get
\begin{align} & {{\left( \sin 15+\cos 15 \right)}^{2}} \\ & ={{\sin }^{2}}15+{{\cos }^{2}}15+2\sin 15\cos 15 \\ & =1+\dfrac{1}{2} \\ & =\dfrac{3}{2} \\ \end{align}
At the end we multiply with 2 to get $2{{\left( \sin 15+\sin 75 \right)}^{2}}=2\times \dfrac{3}{2}=3$.
The value of the expression $2{{\left( \sin 15+\sin 75 \right)}^{2}}$ is 3.

Note: we can also convert $\sin 15$ into $\cos 75$. But in that case the multiple angle formula gives us the $\sin \left( 75\times 2 \right)=\sin 150$ instead of $\sin 30=\dfrac{1}{2}$. We have to convert the associative angle using other formulas to simplify it. The problem becomes unnecessarily longer and that’s why we used an angle of 15.