AHL 3.10—Compound angle identities

The cosine compound angle identity is $\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta$. Note that the right-hand side uses a minus sign when the left-hand side has a plus.

Substituting directly: $\cos \left(\frac{\pi }{4}+\frac{\pi }{6}\right)=\cos \frac{\pi }{4}\cos \frac{\pi }{6}-\sin \frac{\pi }{4}\sin \frac{\pi }{6}$.

The expansion $\cos \frac{\pi }{4}\cos \frac{\pi }{6}-\sin \frac{\pi }{4}\sin \frac{\pi }{6}$ matches the first option exactly.

For $\cos (\alpha +\beta )$, the right-hand side must have a minus sign between the two product terms. Any option with a plus between cosine-cosine and sine-sine products is wrong for a sum of angles.

The option $\cos \frac{\pi }{4}\cos \frac{\pi }{6}+\sin \frac{\pi }{4}\sin \frac{\pi }{6}$ corresponds to $\cos (\alpha -\beta )$, not $\cos (\alpha +\beta )$. Eliminated.

The options $\sin \frac{\pi }{4}\cos \frac{\pi }{6}+\cos \frac{\pi }{4}\sin \frac{\pi }{6}$ and $\sin \frac{\pi }{4}\cos \frac{\pi }{6}-\cos \frac{\pi }{4}\sin \frac{\pi }{6}$ are expansions of $\sin (\alpha +\beta )$ and $\sin (\alpha -\beta )$ respectively — not cosine. Both eliminated.

The only remaining option, $\cos \frac{\pi }{4}\cos \frac{\pi }{6}-\sin \frac{\pi }{4}\sin \frac{\pi }{6}$, correctly expands $\cos (\alpha +\beta )$ with the required minus sign.

We write $105°=60°+45°$, since both are standard angles with known exact trigonometric values.

$\sin (105°)=\sin (60°+45°)=\sin 60°\cos 45°+\cos 60°\sin 45°$

$=\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}+\frac{1}{2}\cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}=\frac{\sqrt{6}+\sqrt{2}}{4}$

The exact value is $\frac{\sqrt{6}+\sqrt{2}}{4}$. Numerically, $\frac{\sqrt{6}+\sqrt{2}}{4}\approx 0.9659=\sin (105°)$ ✓

$105°$ is in the second quadrant, where sine is positive. We expect $\sin (105°)\approx 0.966$, which is close to 1.

$\frac{\sqrt{2}+1}{4}\approx \frac{1.414+1}{4}\approx 0.604$. This is too small for $\sin (105°)\approx 0.966$. Eliminated.

$\frac{\sqrt{6}-\sqrt{2}}{4}\approx \frac{2.449-1.414}{4}\approx 0.259$. This equals $\sin (15°)$, not $\sin (105°)$. Eliminated.

$\frac{\sqrt{3}+1}{2\sqrt{2}}\approx \frac{2.732}{2.828}\approx 0.966$. While numerically close, this is not the simplified standard form — it equals $\frac{\sqrt{6}+\sqrt{2}}{4}$ but written differently; checking: $\frac{\sqrt{3}+1}{2\sqrt{2}}=\frac{(\sqrt{3}+1)\sqrt{2}}{4}=\frac{\sqrt{6}+\sqrt{2}}{4}$, so both are equal. However, the standard simplified form is $\frac{\sqrt{6}+\sqrt{2}}{4}$.

The conventional simplified exact form is $\frac{\sqrt{6}+\sqrt{2}}{4}$, which is the accepted IB answer format for this type of expression.