Use an appropriate Half-Angle Formula to find the exact value of $\sin 105^\circ$
Is this correct? $$\sin 105^\circ = \frac{\sqrt{2+\sqrt{3}}}{4}$$
Because $105^\circ$ is in the second quadrant the square root after the equal sign is $+$. However, because $\cos 210^\circ$ is negative, it value is negative $-\frac{\sqrt{3}}{2}$.
Is my thought process correct?
Also, can I factor out $\frac{1}{4}$?
$\endgroup$ 13 Answers
$\begingroup$You are right about the cosine of $210^\circ$. The formula as currently written is not correct, there should be a $2$ in the denominator instead of a $4$. But the following would be correct:
$$\sqrt{\frac{2+\sqrt{3}}{4}},$$ and the current version is likely due to a typo. It is more common to take the square root of the $4$, and write the answer as $\dfrac{\sqrt{2+\sqrt{3}}}{2}$.
We now give a derivation, that undoubtedly goes along similar lines to yours.
By using the double-angle formula $\cos 2x=1-2\sin^2 x$, and letting $2x=210^{\circ}$, we then get
$$2\sin^2 105^{\circ}=1-\left(-\frac{\sqrt{3}}{2}\right)=\frac{2+\sqrt{3}}{2}.$$
This yields
$$\sin^2 105^{\circ}=\frac{2+\sqrt{3}}{4}.$$
Find the positive square root. We get $\sin 105^{\circ}=\dfrac{\sqrt{2+\sqrt{3}}}{2}$.
Remark: In a comment, OP refers to a half angle formula. The formula we used can equally well be written as $\cos y=1-2\sin^2(y/2)$, or various equivalent forms. One of these forms is $\sin(y/2)=\pm\sqrt{\frac{1-\cos y}{2}}$. I prefer working from the double-angle formula each time. A difficulty with the plethora of equivalent formulas is that they get memorized, often imperfectly, and are not readily recalled when needed later.
$\endgroup$ 9 $\begingroup$This is an equivalent way to get the same answer as above:
$\displaystyle\sin 105^{\circ}=\sqrt{\frac{1-\cos210^{\circ}}{2}} =\sqrt{\frac{1-(-\frac{\sqrt{3}}{2})}{2}}=\sqrt{\frac{2+\sqrt{3}}{4}}=\frac{\sqrt{2+\sqrt{3}}}{2}$.
Notice that using the addition formula for the sine gives a different-looking answer:
$\sin 105^{\circ}=\sin(60^{\circ}+45^{\circ})=\sin60^{\circ}\cos45^{\circ}+\cos60^{\circ}\sin45^{\circ}=\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}+\frac{1}{2}\cdot\frac{\sqrt{2}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}$.
$\endgroup$ $\begingroup$$$ \sin(105^\circ) = \sin(90^\circ + 15^\circ) = \cos(15^\circ) = \sqrt{ \frac{1 + \cos(30^\circ)}{2}} = \sqrt{ \frac{1 + \frac{1}{2} \sqrt{3}}{2}} = \sqrt{ \frac{2 + \sqrt{3}}{4}}. $$
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