# I calculated \sin 75^{\circ} as \frac{1}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}}, but the answer is

I calculated ${\mathrm{sin}75}^{\circ }$ as $\frac{1}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}}$, but the answer is $\frac{\sqrt{2}+\sqrt{6}}{4}$. What went wrong?
I calculated the exact value of ${\mathrm{sin}75}^{\circ }$ as follows:
${\mathrm{sin}75}^{\circ }=\mathrm{sin}\left({30}^{\circ }+{45}^{\circ }\right)$
$=\mathrm{sin}30°\mathrm{cos}45°+\mathrm{cos}30°\mathrm{sin}45°$
$=\frac{1}{2}·\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}·\frac{1}{\sqrt{2}}$
$=\frac{1}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}}$
$\frac{\sqrt{2}+\sqrt{6}}{4}$

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Killian Curry
They’re the same value. Multiply the numerator and denominator of your answer by $\sqrt{2}$ to see why.
$\frac{1+\sqrt{3}}{2\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{2}}\cdot \frac{1+\sqrt{3}}{2\sqrt{2}}=\frac{\sqrt{2}+\sqrt{6}}{4}$
You can also use $\mathrm{sin}45=\mathrm{cos}45=\frac{\sqrt{2}}{2}$ (rationalizing $\frac{1}{\sqrt{2}}$) to get the answer more easily.