How to Find the Exact Value of sin 15° Without a Calculator

Tables give sine and cosine for $30°$, $45°$ and $60°$, but what about $15°$, $75°$ or $22.5°$? Compound angle formulae answer this by building new angles from known ones, and their special cases produce double, triple and half angle results.

This part covers the addition and subtraction formulae, double and triple angle formulae, half and one-third angle formulae, and mixed problems.

**The compound angle formulae are $\sin (A\pm B)=\sin A\cos B\pm \cos A\sin B$, $\cos (A\pm B)=\cos A\cos B\mp \sin A\sin B$ and $\tan (A\pm B)=\frac{\tan A\pm \tan B}{1\mp \tan A\tan B}$, and applying the tangent formula twice gives $\tan (A+B+C)$.

Derivation idea.** $\cos (A-B)$ comes from equating two expressions for the distance between the points at angles A and B on the unit circle; replacing B with $-B$, or using $\sin x=\cos (90°-x)$, gives the rest.

Three angles:

$$\tan (A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}$$

Worked example. Find $\sin 15°$ exactly:

$$\sin (45°-30°)=\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\cdot \frac{1}{2}=\frac{\sqrt{3}-1}{2\sqrt{2}}\approx 0.2588$$

Worked example 2. $\tan 75°=\tan (45°+30°)=\frac{\sqrt{3}+1}{\sqrt{3}-1}=2+\sqrt{3}\approx 3.732$.

An everyday example. A surveyor measuring a hill from two positions combines angles of elevation, and these formulae turn the combined angle into usable ratios.

The substance. **$\sin (A+B)\ne \sin A+\sin B$** — with $A=B=30°$, the left side is $\sin 60°\approx 0.866$ but the right side is $1$.

**Putting $B=A$ in the compound angle formulae gives $\sin 2A=2\sin A\cos A$, $\cos 2A={\cos }^{2}A-{\sin }^{2}A$ and $\tan 2A=\frac{2\tan A}{1-{\tan }^{2}A}$, and writing $3A=2A+A$ gives the triple angle formulae.

Double angle formulae:**

- $\sin 2A=2\sin A\cos A=\frac{2\tan A}{1+{\tan }^{2}A}$
- $\cos 2A={\cos }^{2}A-{\sin }^{2}A=2{\cos }^{2}A-1=1-2{\sin }^{2}A$
- $\tan 2A=\frac{2\tan A}{1-{\tan }^{2}A}$

Triple angle formulae:

- $\sin 3A=3\sin A-4{\sin }^{3}A$
- $\cos 3A=4{\cos }^{3}A-3\cos A$
- $\tan 3A=\frac{3\tan A-{\tan }^{3}A}{1-3{\tan }^{2}A}$

Worked example. If $\sin A=\frac{3}{5}$ with A acute, then $\cos A=\frac{4}{5}$, so

$$\sin 2A=2\times \frac{3}{5}\times \frac{4}{5}=\frac{24}{25},\cos 2A=1-2\times \frac{9}{25}=\frac{7}{25}$$

and $\sin 3A=3\times \frac{3}{5}-4\times \frac{27}{125}=\frac{225-108}{125}=\frac{117}{125}$.

An everyday example. The distance a ball thrown on a level ground travels depends on $\sin 2\theta$ when air resistance is ignored, which is why a launch angle of $45°$ gives the longest throw.

The substance. **$\sin 2A$ is not $2\sin A$** — doubling the angle does not double the sine.

**Replacing A with $\frac{A}{2}$ in the double angle formulae gives results such as $\cos A=2{\cos }^2\frac{A}{2}-1$, and replacing A with $\frac{A}{3}$ in the triple angle formulae gives results such as $\cos A=4{\cos }^3\frac{A}{3}-3\cos \frac{A}{3}$.

Half angle results:**

- $\sin A=2\sin \frac{A}{2}\cos \frac{A}{2}$
- $\sin \frac{A}{2}=\pm \sqrt{\frac{1-\cos A}{2}}$ and $\cos \frac{A}{2}=\pm \sqrt{\frac{1+\cos A}{2}}$
- $\tan \frac{A}{2}=\frac{1-\cos A}{\sin A}=\frac{\sin A}{1+\cos A}$
- The sign depends on the quadrant of $\frac{A}{2}$

One-third angle results:

- $\sin A=3\sin \frac{A}{3}-4{\sin }^3\frac{A}{3}$
- $\cos A=4{\cos }^3\frac{A}{3}-3\cos \frac{A}{3}$

Worked example. Find $\cos 22.5°$ and $\tan 22.5°$:

$$\cos 22.5°=\sqrt{\frac{1+\cos 45°}{2}}=\sqrt{\frac{1+0.7071}{2}}\approx 0.9239$$

The positive root is taken because $22.5°$ is in the first quadrant. Also $\tan 22.5°=\frac{1-\cos 45°}{\sin 45°}=\sqrt{2}-1\approx 0.414$.

An everyday example. Folding a square paper napkin corner to corner, then again halves the angle each time — from $90°$ to $45°$ to $22.5°$ — exactly the steps these formulae handle.

The substance. One-third angle formulae lead to cubic equations — given $\cos A$, finding $\cos \frac{A}{3}$ means solving $4c^3-3c=\cos A$.

**Mixed problems are solved by rewriting every angle in terms of a single angle, choosing the form of $\cos 2A$ that removes awkward constants, and recognising standard formulae in reverse.

Strategy:**

- Express all angles in terms of one angle
- Use $1-\cos 2A=2{\sin }^{2}A$ and $1+\cos 2A=2{\cos }^{2}A$ to remove the 1s

Worked example. Prove $\frac{\sin 2A}{1+\cos 2A}=\tan A$.

$$\frac{\sin 2A}{1+\cos 2A}=\frac{2\sin A\cos A}{2{\cos }^{2}A}=\frac{\sin A}{\cos A}=\tan A$$

Worked example 2. Evaluate ${\cos }^{2}15°-{\sin }^{2}15°$. This is $\cos 30°=\frac{\sqrt{3}}{2}\approx 0.866$, with no need to find $\sin 15°$ at all.

Worked example 3. If $\tan A=\frac{1}{2}$ and $\tan B=\frac{1}{3}$, then $\tan (A+B)=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}=1$, so $A+B=45°$ for acute A and B.

An everyday example. An engineer designing a staircase with two flights at different slopes combines those angles, and these identities simplify the resulting expressions.

The substance. Seeing a formula in reverse is the key skill — $2\sin A\cos A$ should be recognised instantly as $\sin 2A$.