Question

Find the value of $\sin \left( {{15}^{\circ }} \right)$ ?

Answer 1

Hint: For answering this question we will write the $\sin \left( {{15}^{\circ }} \right)$ in forms of different angles whose values are predetermined and use them to derive its value. We can write it as $\sin \left( {{15}^{\circ }} \right)=\sin \left( {{45}^{\circ }}-{{30}^{\circ }} \right)$ and can use the formulae $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ .

Complete step-by-step answer:
Let us consider the given expression from the question $\sin \left( {{15}^{\circ }} \right)$ and simply write this as $\sin \left( {{45}^{\circ }}-{{30}^{\circ }} \right)$ .
Now let us use the formulae that we have learnt from the basic concepts it is $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ .
After applying this we will have $\sin \left( {{45}^{\circ }}-{{30}^{\circ }} \right)=\sin \left( {{45}^{\circ }} \right)\cos \left( {{30}^{\circ }} \right)-\cos \left( {{45}^{\circ }} \right)\sin \left( {{30}^{\circ }} \right)$ .
As we know the values of these angles from the trigonometric ratios table. The values are respectively $\sin \left( {{45}^{\circ }} \right)=\dfrac{1}{\sqrt{2}}$ , $\cos \left( {{45}^{\circ }} \right)=\dfrac{1}{\sqrt{2}}$ , $\sin \left( {{30}^{\circ }} \right)=\dfrac{1}{2}$ and $\cos \left( {{30}^{\circ }} \right)=\dfrac{\sqrt{3}}{2}$ .
After substituting these values we will have $\sin \left( {{45}^{\circ }}-{{30}^{\circ }} \right)=\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{\sqrt{3}}{2} \right)-\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{1}{2} \right)$ .
Now we will simplify this expression, after simplifying we will have $\sin \left( {{45}^{\circ }}-{{30}^{\circ }} \right)=\left( \dfrac{\sqrt{3}}{2\sqrt{2}} \right)-\left( \dfrac{1}{2\sqrt{2}} \right)$ .
By further simplifying this expression we will have $\sin \left( {{45}^{\circ }}-{{30}^{\circ }} \right)=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$ .
Now we have a conclusion that the value of $\sin \left( {{15}^{\circ }} \right)$ is $\dfrac{\sqrt{3}-1}{2\sqrt{2}}$ .

Note: This question can be solved in another way that is by simply write it as $\sin \left( {{15}^{\circ }} \right)=\sin \left( \dfrac{{{30}^{\circ }}}{2} \right)$ . The formulae that we can use here is $\sin \left( \dfrac{A}{2} \right)=\sqrt{\dfrac{1-\cos A}{2}}$ . By applying this formulae we will have $\sin \left( \dfrac{{{30}^{\circ }}}{2} \right)=\sqrt{\dfrac{1-\cos \left( {{30}^{\circ }} \right)}{2}}$ . By substituting the value $\cos \left( {{30}^{\circ }} \right)=\dfrac{\sqrt{3}}{2}$ we will have $\sin \left( {{15}^{\circ }} \right)=\sqrt{\dfrac{1-\dfrac{\sqrt{3}}{2}}{2}}$ . By simplifying this we will have $\sin \left( {{15}^{\circ }} \right)=\sqrt{\dfrac{2-\sqrt{3}}{4}}=\dfrac{\sqrt{2-\sqrt{3}}}{2}$ . Let us multiply and divide it with $\sqrt{2}$ after that we will have $\sin \left( {{15}^{\circ }} \right)=\dfrac{\sqrt{4-2\sqrt{3}}}{2\sqrt{2}}$ . If we observe we can say ${{\left( \sqrt{3}-1 \right)}^{2}}=\left( 4-2\sqrt{3} \right)$ so we can write $\sin \left( {{15}^{\circ }} \right)=\dfrac{\sqrt{{{\left( \sqrt{3}-1 \right)}^{2}}}}{2\sqrt{2}}=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$ . We have same answers in both the cases. So any one method can be used.