Question

sin 15=root3-1/2root2​

Answer 1

So we have to Prove that

$Sin15^{\circ} = \dfrac{\sqrt{3} -1}{2\sqrt{2}}$

♦We will use this Identity in Proof :-

• Sin(a - b) = Sina • Cosb - Sinb •Cosa

→ As 15° = 45 - 30

♦Values To be used :-

$\bullet Sin 45^{\circ} = \dfrac{1}{\sqrt{2}} = Cos 45^{\circ}$

$\bullet Sin 30^{\circ} = \dfrac{1}{2}$

$\bullet Cos 30^{\circ} = \dfrac{\sqrt{3}}{2}$

♦ Then

$Sin(15^{\circ}) = Sin(45 -30)$

$\boxed{\tiny{\implies Sin(15^{\circ}) = Sin(45^{\circ}) \times Cos(30^{\circ}) - Sin(30^{\circ}) \times Cos(45^{\circ})}}$

$\boxed{\small{\implies \dfrac{\sqrt{3} -1}{2\sqrt{2}}= \left(\dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{3}}{2}\right) - \left(\dfrac{1}{2} \times \dfrac{1}{\sqrt{2}}\right)}}$

$\boxed{ \implies \dfrac{\sqrt{3} -1}{2\sqrt{2}} =\left( \dfrac{\sqrt{3}}{2\sqrt{2}} - \dfrac{1}{2\sqrt{2}}\right)}$

$\boxed{\dfrac{\sqrt{3} -1}{2\sqrt{2}} = \dfrac{\sqrt{3} -1}{2\sqrt{2}}}$

So as LHS = RHS

Hence Proved