Finding sine of five pi over twelve using compound angles

Key concept: The exact value of $\sin\left(\frac{5\pi}{12}\right)$ comes from a compound-angle identity, and the cleanest route is to split $\frac{5\pi}{12}$ into angles with known sine and cosine values.

Step by step

Break the angle into known parts

The angle $\frac{5\pi}{12}$ can be written as

$\frac{5\pi}{12}=\frac{\pi}{4}+\frac{\pi}{6}$.

That is the key step, because both $\frac{\pi}{4}$ and $\frac{\pi}{6}$ have exact trig values.

Now apply the compound-angle formula:

$\sin(A+B)=\sin A\cos B+\cos A\sin B$.

So

$\sin\left(\frac{5\pi}{12}\right)=\sin\left(\frac{\pi}{4}+\frac{\pi}{6}\right)$

$=\sin\frac{\pi}{4}\cos\frac{\pi}{6}+\cos\frac{\pi}{4}\sin\frac{\pi}{6}$.

Substitute the exact values

Use the standard exact values:

$\sin\frac{\pi}{4}=\frac{\sqrt2}{2}$, $\cos\frac{\pi}{6}=\frac{\sqrt3}{2}$,

$\cos\frac{\pi}{4}=\frac{\sqrt2}{2}$, $\sin\frac{\pi}{6}=\frac12$.

Then

$\sin\left(\frac{5\pi}{12}\right)=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\cdot\frac12$.

Factor out $\frac{\sqrt2}{2}$:

$=\frac{\sqrt2}{2}\left(\frac{\sqrt3}{2}+\frac12\right)$

$=\frac{\sqrt2}{2}\cdot\frac{\sqrt3+1}{2}$

$=\frac{\sqrt2(\sqrt3+1)}{4}$.

You can also write this as

$\boxed{\frac{\sqrt6+\sqrt2}{4}}$.

Check the sign and the quadrant

The angle $\frac{5\pi}{12}$ is between $\frac{\pi}{4}$ and $\frac{\pi}{2}$, so it lies in Quadrant I. That means sine must be positive. This is why the final answer is the positive root, not a plus-or-minus expression.

The square-root method shown in the prompt can work too, but it is easy to lose the sign if you do not check the quadrant. The compound-angle formula is usually cleaner here because it gives the exact value directly.

A useful identity to remember

For angles like $75^\circ$ or $\frac{5\pi}{12}$, splitting into $45^\circ+30^\circ$ is often the fastest route. The pattern appears often in exams because it tests both exact trig values and angle-sum identities. If you remember $\sin(a+b)$ and the exact values for $\pi/6$ and $\pi/4$, this whole type of question becomes very manageable.

Pitfall alert

The mistake most students make with $\sin\left(\frac{5\pi}{12}\right)$ is trying to force the half-angle formula first and then forgetting which sign to choose. Since $\frac{5\pi}{12}$ is in Quadrant I, the sine must be positive, so any answer with a negative sign is immediately wrong. Another trap is simplifying $\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\cdot\frac12$ and stopping too early, leaving it as a partially factored expression when the exam expects a fully simplified exact value. It is also easy to confuse $\frac{5\pi}{12}$ with $\frac{\pi}{5}$ or $\frac{5\pi}{6}$ under time pressure, so always rewrite the angle as $\frac{\pi}{4}+\frac{\pi}{6}$ before evaluating.

Try different conditions

If the question changed to find $\sin\left(\frac{7\pi}{12}\right)$, the same compound-angle idea still works, but you would write $\frac{7\pi}{12}=\frac{\pi}{3}+\frac{\pi}{4}$. Then $\sin\left(\frac{7\pi}{12}\right)=\sin\frac{\pi}{3}\cos\frac{\pi}{4}+\cos\frac{\pi}{3}\sin\frac{\pi}{4}$. That gives $\frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2}+\frac12\cdot\frac{\sqrt2}{2}=\frac{\sqrt6+\sqrt2}{4}$ again, but now the quadrant reasoning changes because the angle is still in Quadrant I. If instead the angle were $\frac{13\pi}{12}$, the same algebra would produce a reference-angle value, but the final sign would be negative because the angle lies in Quadrant III.

Further reading

compound angle formula, exact trigonometric values, reference angle quadrant