Question

Find the value of $\sin {75}^{\circ }$

• A. $\frac{\sqrt{3}+1}{2\sqrt{2}}$
• B. $\frac{\sqrt{3}-1}{2\sqrt{2}}$
• C. $\frac{1-\sqrt{3}}{2\sqrt{2}}$
• D. $\frac{\sqrt{3}+1}{2+\sqrt{2}}$

Answer 1

Answer: (A), (D)

$\frac{\sqrt{3}+1}{2\sqrt{2}}$
$\frac{\sqrt{3}+1}{2+\sqrt{2}}$

Consider the given value,$\sin {75}^0$

Now,

$\sin {75}^0=\sin ({45}^0+{30}^0)$

$=\sin {45}^0\cos {30}^0+\cos {45}^0\sin {30}^0$

$=\frac{1}{\sqrt{2}}.\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}.\frac{1}{2}$

$\sin {75}^0=\frac{\sqrt{3}+1}{2\sqrt{2}}$

Hence, this is the answer