Identities from compound angles

When angles are tied together by a con­di­tion, such as being the angles of a tri­an­gle ($A + B + C = 180°$) or adding up to $45°$, the com­pound angle for­mu­lae pro­duce strik­ing iden­ti­ties. This les­son proves the stan­dard ones and uses them to eval­u­ate expres­sions like $(1 + \tan 22°)(1 + \tan 23°)$ that would be hard to find any other way. The method is always the same: write the con­di­tion as "one angle equals some­thing", take the tan­gent (or cotan­gent) of both sides, and expand.

Results for the angles of a tri­an­gle#

Results. If $A + B + C = 180°$ (for exam­ple, the angles of a tri­an­gle $ABC$), then

1. $\Sigma\tan A = \Pi\tan A$, that is, $\tan A + \tan B + \tan C = \tan A\tan B\tan C$
2. $\Sigma\cot A\cot B = 1$, that is, $\cot A\cot B + \cot B\cot C + \cot C\cot A = 1$
3. $\displaystyle \Sigma\tan A\tan B = \frac{1 + \cos A\cos B\cos C}{\cos A\cos B\cos C}$

Proof of 1. From $A + B = 180° - C$,

$$\displaystyle \begin{aligned} \tan(A + B) &= \tan(180° - C) \\ \frac{\tan A + \tan B}{1 - \tan A\tan B} &= -\tan C \\ \tan A + \tan B &= -\tan C + \tan A\tan B\tan C \\ \tan A + \tan B + \tan C &= \tan A\tan B\tan C \end{aligned}$$

Proof of 2. Take cotan­gents instead:

$$\displaystyle \begin{aligned} \cot(A + B) &= \cot(180° - C) \\ \frac{\cot A\cot B - 1}{\cot B + \cot A} &= -\cot C \\ \cot A\cot B - 1 &= -\cot B\cot C - \cot A\cot C \\ \cot A\cot B + \cot B\cot C + \cot C\cot A &= 1 \end{aligned}$$

Proof of 3. Here $\cos(A + B + C) = \cos 180° = -1$. Divide the Group VII for­mula $\cos(A + B + C) = \Pi\cos A - \Sigma\cos A\sin B\sin C$ by $\cos A\cos B\cos C$:

$$\displaystyle \frac{-1}{\cos A\cos B\cos C} = 1 - \Sigma\tan B\tan C$$

$$\displaystyle \Sigma\tan A\tan B = 1 + \frac{1}{\cos A\cos B\cos C} = \frac{\cos A\cos B\cos C + 1}{\cos A\cos B\cos C}$$

Results when the angles add to $90°$#

Results. If $A + B + C = 90°$, then

1. $\Sigma\cot A = \Pi\cot A$
2. $\Sigma\tan A\tan B = 1$

Hint for the proof. Start from $A + B = 90° - C$ and take $\cot(A + B) = \cot(90° - C) = \tan C$ for the first, or $\tan(A + B) = \cot C$ for the sec­ond, then expand as before.

Results when two angles add to $45°$#

Results. If $A + B = 45°$ (or $225°$), then

1. $(1 + \tan A)(1 + \tan B) = 2$
2. $(1 - \cot A)(1 - \cot B) = 2$
3. $(1 + \cot A)(1 + \cot B) = 2\cot A\cot B$

Proof of 1. From $A + B = 45°$,

$$\displaystyle \begin{aligned} \tan(A + B) &= \tan 45° \\ \frac{\tan A + \tan B}{1 - \tan A\tan B} &= 1 \\ \tan A + \tan B &= 1 - \tan A\tan B \\ \tan A + \tan B + \tan A\tan B &= 1 \end{aligned}$$

Add $1$ to both sides and fac­torise:

$$1 + \tan A + \tan B + \tan A\tan B = 2 \implies (1 + \tan A) + \tan B(1 + \tan A) = 2 \implies (1 + \tan A)(1 + \tan B) = 2$$

Results 2 and 3 come the same way from $\cot(A + B) = 1$, that is, $\cot A\cot B - 1 = \cot A + \cot B$. Since $\tan 225° = \tan 45° = 1$, every­thing also holds for $A + B = 225°$.

Exam­ple 1#

Find $(1 + \tan 22°)(1 + \tan 23°)$.

$22° + 23° = 45°$, so by result 1 the value is $2$.

Exam­ple 2#

Find $\displaystyle \sin^2 52\tfrac{1}{2}° - \sin^2 22\tfrac{1}{2}°$.

Use $\sin^2 A - \sin^2 B = \sin(A + B)\sin(A - B)$:

$$\displaystyle \sin 75°\sin 30° = \cos 15°\sin 30° = \frac{\sqrt{3} + 1}{2\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{4\sqrt{2}}$$

Exam­ple 3#

Prove that $\displaystyle \frac{\cos 9° + \sin 9°}{\cos 9° - \sin 9°} = \cot 36°$.

By Group VI, the left side is

$$\displaystyle \frac{1 + \tan 9°}{1 - \tan 9°} = \tan(45° + 9°) = \tan 54° = \cot 36°$$

since $54° + 36° = 90°$.

Exam­ple 4#

Find $\tan 100° + \tan 125° + \tan 100°\tan 125°$.

Here $100° + 125° = 225°$, and $\tan 225° = 1$:

$$\displaystyle \begin{aligned} \frac{\tan 100° + \tan 125°}{1 - \tan 100°\tan 125°} &= 1 \\ \tan 100° + \tan 125° &= 1 - \tan 100°\tan 125° \end{aligned}$$

So $\tan 100° + \tan 125° + \tan 100°\tan 125° = 1$.

Exam­ple 5#

Prove that $\tan 70° - \tan 20° = 2\tan 50°$.

Since $70° - 20° = 50°$,

$$\displaystyle \tan(70° - 20°) = \tan 50° \implies \frac{\tan 70° - \tan 20°}{1 + \tan 70°\tan 20°} = \tan 50°$$

Now $70° + 20° = 90°$, so $\tan 70°\tan 20° = 1$ and the denom­i­na­tor is $2$:

$$\displaystyle \frac{\tan 70° - \tan 20°}{2} = \tan 50° \implies \tan 70° - \tan 20° = 2\tan 50°$$

Prac­tice#

1. Find $(1 + \tan 10°)(1 + \tan 35°)$.
2. Find $(1 - \cot 23°)(1 - \cot 22°)$.
3. Find $\tan 20° + \tan 25° + \tan 20°\tan 25°$.
4. Prove that $\tan 50° - \tan 40° = 2\tan 10°$.
5. Prove that $\tan 3A - \tan 2A - \tan A = \tan 3A\tan 2A\tan A$.
6. In a tri­an­gle $ABC$, $\tan A = 1$ and $\tan B = 2$. Find $\tan C$.
7. If $A + B + C = 90°$, $\cot A = 2$ and $\cot B = 3$, find $\cot C$.
8. Find $\displaystyle \cos^2 52\tfrac{1}{2}° - \sin^2 22\tfrac{1}{2}°$.
9. Write $\displaystyle \frac{\cos 11° + \sin 11°}{\cos 11° - \sin 11°}$ as a sin­gle tan­gent.
10. Prove that $\tan 65° - \tan 25° = 2\tan 40°$.

Answers#