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

Question

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

Elise Winters 2022-04-29 Answered

I calculated ${\sin 75}^{\circ}$ 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 ${\sin 75}^{\circ}$ as follows:
${\sin 75}^{\circ} = \sin (30^{\circ} + 45^{\circ})$
$= \sin 30 ° \cos 45 ° + \cos 30 ° \sin 45 °$
$= \frac{1}{2} \cdot \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}$
$= \frac{1}{2 \sqrt{2}} + \frac{\sqrt{3}}{2 \sqrt{2}}$
The actual answer is
$\frac{\sqrt{2} + \sqrt{6}}{4}$

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Answers (1)

Killian Curry

Answered 2022-04-30 Author has 18 answers

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

\sin 45 = \cos 45 = \frac{\sqrt{2}}{2}

(rationalizing

\frac{1}{\sqrt{2}}

) to get the answer more easily.

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