# Prove the given result: $\sin {{15}^{\circ }}=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$.

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Hint: We start solving the problem by considering the L.H.S (Left hand side) of the given result. We then write ${{15}^{\circ }}$ as $\left( {{45}^{\circ }}-{{30}^{\circ }} \right)$ and make use of the result $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ to proceed through the problem. We then make use of the results $\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$ , $\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$ , $\sin {{30}^{\circ }}=\dfrac{1}{2}$ and $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$ to proceed further through the problem. We then make the necessary calculations to complete the proof of given result.

According to the problem, we are asked to prove the given result $\sin {{15}^{\circ }}=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$.
Let us consider L.H.S (Left hand side) of the given result.
So, we have $\sin {{15}^{\circ }}$.
$\Rightarrow \sin {{15}^{\circ }}=\sin \left( {{45}^{\circ }}-{{30}^{\circ }} \right)$ ---(1).
We know that $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ . Let us use this result in equation (1).
$\Rightarrow \sin {{15}^{\circ }}=\sin {{45}^{\circ }}\cos {{30}^{\circ }}-\cos {{45}^{\circ }}\sin {{30}^{\circ }}$ ---(2).
We know that $\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$ , $\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$ , $\sin {{30}^{\circ }}=\dfrac{1}{2}$ and $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$ . Let us use this results in equation (2).
$\Rightarrow \sin {{15}^{\circ }}=\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{\sqrt{3}}{2} \right)-\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{1}{2} \right)$ .
$\Rightarrow \sin {{15}^{\circ }}=\left( \dfrac{\sqrt{3}}{2\sqrt{2}} \right)-\left( \dfrac{1}{2\sqrt{2}} \right)$ .
$\Rightarrow \sin {{15}^{\circ }}=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$ .
We can see that the L.H.S (Left Hand Side) of the given result is equal to the R.H.S (Right Hand Side) of it, which means that we have proved the given result.
$\therefore$ We have proved that the value of $\sin {{15}^{\circ }}$ as $\dfrac{\sqrt{3}-1}{2\sqrt{2}}$ .

Note:
We can also prove the given result of the problem as shown below:
We know that $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$ .
$\Rightarrow \cos \left( 2\left( {{15}^{\circ }} \right) \right)=\dfrac{\sqrt{3}}{2}$ ---(3).
We know that $\cos 2A=1-2{{\sin }^{2}}A$ . Let us use this result in equation (3).
$\Rightarrow 1-2{{\sin }^{2}}\left( {{15}^{\circ }} \right)=\dfrac{\sqrt{3}}{2}$ .
$\Rightarrow 1-\dfrac{\sqrt{3}}{2}=2{{\sin }^{2}}\left( {{15}^{\circ }} \right)$ .
$\Rightarrow 2{{\sin }^{2}}\left( {{15}^{\circ }} \right)=\dfrac{3}{4}+\dfrac{1}{4}-2\left( \dfrac{\sqrt{3}}{2} \right)\left( \dfrac{1}{2} \right)$ .
$\Rightarrow 2{{\sin }^{2}}\left( {{15}^{\circ }} \right)={{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}+{{\left( \dfrac{1}{2} \right)}^{2}}-2\left( \dfrac{\sqrt{3}}{2} \right)\left( \dfrac{1}{2} \right)$ ---(6).
We can see that the R.H.S (Right hand side) of the equation (6) resembles ${{a}^{2}}+{{b}^{2}}-2ab$ which is equal to ${{\left( a-b \right)}^{2}}$ .
$\Rightarrow 2{{\sin }^{2}}\left( {{15}^{\circ }} \right)={{\left( \dfrac{\sqrt{3}}{2}-\dfrac{1}{2} \right)}^{2}}$.
$\Rightarrow 2{{\sin }^{2}}\left( {{15}^{\circ }} \right)={{\left( \dfrac{\sqrt{3}-1}{2} \right)}^{2}}$.
$\Rightarrow {{\sin }^{2}}\left( {{15}^{\circ }} \right)=\dfrac{1}{2}\times {{\left( \dfrac{\sqrt{3}-1}{2} \right)}^{2}}$.
$\Rightarrow \sin \left( {{15}^{\circ }} \right)=\sqrt{\dfrac{1}{2}\times {{\left( \dfrac{\sqrt{3}-1}{2} \right)}^{2}}}$.
$\Rightarrow \sin \left( {{15}^{\circ }} \right)=\dfrac{1}{\sqrt{2}}\times \left( \dfrac{\sqrt{3}-1}{2} \right)$.
$\Rightarrow \sin \left( {{15}^{\circ }} \right)=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$, which is the given result.