We know that, $\sin(75^\circ)=\sin(30^\circ+45^\circ)=\sin45^\circ.\cos30^\circ+\sin30^\circ.\cos45^\circ=\frac{1}{\sqrt{2}}.\frac{\sqrt{3}}{2}+\frac{1}{2}.\frac{1}{\sqrt{2}}=\frac{\sqrt{3}+1}{2\sqrt{2}}$
And, $\cos(75^\circ)=\cos(30^\circ+45^\circ)=\cos45^\circ.\cos30^\circ+\sin30^\circ.\sin45^\circ=\frac{1}{\sqrt{2}}.\frac{\sqrt{3}}{2}-\frac{1}{2}.\frac{1}{\sqrt{2}}=\frac{\sqrt{3}-1}{2\sqrt{2}}$
Now, $\cos\theta=\sqrt{1-\sin^2\theta}$
So, $\cos(75^\circ)=\sqrt{1-\sin^2(75^\circ)}=\sqrt{1-(\frac{\sqrt{3}+1}{2\sqrt{2}})^2}=\sqrt{1-\frac{(\sqrt{3}+1)^2}{8}}=\sqrt{\frac{8-(\sqrt{3}+1)^2}{8}}=\sqrt{\frac{8-(4+2\sqrt{3})}{8}}=\sqrt{\frac{4-2\sqrt{3}}{8}}=\sqrt{\frac{2-\sqrt{3}}{4}}$
Thus, $\frac{\sqrt{3}-1}{2\sqrt{2}}=\sqrt{\frac{2-\sqrt{3}}{4}}$
Now, how do we simplify the RHS to represent the LHS? I have checked that LHS and RHS both have a value of $0.2588190451$. But, I am unable to show that by simplification.
Best Answer
To verify $$\frac{\sqrt3-1}{2\sqrt2}=\sqrt{\frac{2-\sqrt3}{4}}$$
Note that the LHS is positive, it is equvalent to verifying that
$$\left( \frac{\sqrt3-1}{2\sqrt2}\right)^2=\frac{2-\sqrt3}{4}$$
$$ \frac{3+1-2\sqrt3}{8}=\frac{2-\sqrt3}{4}$$
which is clearly true.