Results with sixty degrees

Three angles of the form $\theta$, $60° + \theta$ and $60° - \theta$ (or $120° \pm \theta$) fit together remark­ably well: their sines, cosines and tan­gents mul­ti­ply and add to give sim­ple answers in terms of $3\theta$. These results let you find val­ues such as $\cos 20°\cos 40°\cos 60°\cos 80°$ or $\tan 20°\tan 40°\tan 80°$ in two lines. This les­son proves the main results and shows how to spot when they apply.

Prod­ucts with $60° \pm \theta$#

Result. If $x = 60°$ (or $120°$, $240°$, $300°$), then

1. $\displaystyle \sin\theta\sin(x + \theta)\sin(x - \theta) = \frac{1}{4}\sin 3\theta$
2. $\displaystyle \cos\theta\cos(x + \theta)\cos(x - \theta) = \frac{1}{4}\cos 3\theta$
3. $\tan\theta\tan(x + \theta)\tan(x - \theta) = \tan 3\theta$
4. $\cot\theta\cot(x + \theta)\cot(x - \theta) = \cot 3\theta$

Proof of 1 for $x = 60°$. Use $\sin(A + B)\sin(A - B) = \sin^2 A - \sin^2 B$:

$$\displaystyle \begin{aligned} \sin\theta\sin(60° + \theta)\sin(60° - \theta) &= \sin\theta\left[\sin^2 60° - \sin^2\theta\right] \\ &= \sin\theta\left[\frac{3}{4} - \sin^2\theta\right] \\ &= \sin\theta \cdot \frac{3 - 4\sin^2\theta}{4} \\ &= \frac{1}{4}(3\sin\theta - 4\sin^3\theta) \\ &= \frac{1}{4}\sin 3\theta \end{aligned}$$

Result 2 is proved the same way with $\cos(A + B)\cos(A - B) = \cos^2 A - \sin^2 B$, and result 3 is result 1 divided by result 2. For $x = 120°$, $240°$, $300°$ the angles $x \pm \theta$ dif­fer from $60° \mp \theta$ only by sign changes, which can­cel in pairs.

How to spot it. Take the small­est angle as $\theta$ and check whether the other two are $60° - \theta$ and $60° + \theta$. For $20°$, $40°$, $80°$: $60° - 20° = 40°$ and $60° + 20° = 80°$, so the result applies.

Exam­ple 1#

Find $\cos 20°\cos 40°\cos 60°\cos 80°$.

With $\theta = 20°$, the angles $40° = 60° - 20°$ and $80° = 60° + 20°$ appear. And $\displaystyle \cos 60° = \frac{1}{2}$. So

$$\displaystyle \frac{1}{2}\left[\cos 20°\cos(60° + 20°)\cos(60° - 20°)\right] = \frac{1}{2} \cdot \frac{1}{4}\cos 60° = \frac{1}{8} \cdot \frac{1}{2} = \frac{1}{16}$$

Exam­ple 2#

If $A = \tan 6°\tan 42°$ and $B = \cot 66°\cot 78°$, find $\displaystyle \frac{A}{B}$.

$$\displaystyle \frac{A}{B} = \frac{\tan 6°\tan 42°}{\cot 66°\cot 78°} = \tan 6°\tan 42°\tan 66°\tan 78°$$

Group the angles by result 3. With $\theta = 6°$: $\tan 6°\tan 54°\tan 66° = \tan 18°$, so $\displaystyle \tan 6°\tan 66° = \frac{\tan 18°}{\tan 54°}$. With $\theta = 18°$: $\tan 18°\tan 42°\tan 78° = \tan 54°$, so $\displaystyle \tan 42°\tan 78° = \frac{\tan 54°}{\tan 18°}$. Mul­ti­ply:

$$\displaystyle \frac{A}{B} = \frac{\tan 18°}{\tan 54°} \cdot \frac{\tan 54°}{\tan 18°} = 1, \quad \text{so } A = B$$

Sums of tan­gents with $60° \pm \theta$#

Result. If $x = 60°$ (or $120°$, $240°$, $300°$), then

1. $\tan\theta + \tan(x + \theta) - \tan(x - \theta) = 3\tan 3\theta = 3\tan\theta\tan(x + \theta)\tan(x - \theta)$
2. $\cot\theta + \cot(x + \theta) - \cot(x - \theta) = 3\cot 3\theta = 3\cot\theta\cot(x + \theta)\cot(x - \theta)$
3. $\cot(x + \theta)\cot(x - \theta) + \cot(x - \theta)\cot\theta - \cot\theta\cot(x + \theta) = 3$
4. $\tan(x + \theta)\tan(x - \theta) + \tan(x - \theta)\tan\theta - \tan\theta\tan(x + \theta) = 3$

Proof of 1 for $x = 60°$. Write $t = \tan\theta$ and use $\tan(A \pm B)$ with $\tan 60° = \sqrt{3}$:

$$\displaystyle t + \frac{\sqrt{3} + t}{1 - \sqrt{3}t} - \frac{\sqrt{3} - t}{1 + \sqrt{3}t}$$

Putting the last two frac­tions over the com­mon denom­i­na­tor $1 - 3t^2$ gives $\displaystyle \frac{8t}{1 - 3t^2}$, and adding $t$:

$$\displaystyle t + \frac{8t}{1 - 3t^2} = \frac{t - 3t^3 + 8t}{1 - 3t^2} = \frac{3(3t - t^3)}{1 - 3t^2} = 3\tan 3\theta$$

The sec­ond form fol­lows from the prod­uct result $\tan\theta\tan(60° + \theta)\tan(60° - \theta) = \tan 3\theta$. Divid­ing result 1 by the prod­uct $\tan\theta\tan(60° + \theta)\tan(60° - \theta)$ gives result 3, and result 2 divided by the cotan­gent prod­uct gives result 4.

Exam­ple 3#

Show that $\cot 76°\cot 44° + \cot 44°\cot 16° - \cot 16°\cot 76° = 3$.

Here $\theta = 16°$, $60° + \theta = 76°$, $60° - \theta = 44°$. This is exactly result 3.

Sums with $120° \pm \theta$#

Result.

1. $\cos\theta + \cos(120° + \theta) + \cos(120° - \theta) = 0$
2. $\displaystyle \cos^3\theta + \cos^3(120° + \theta) + \cos^3(120° - \theta) = \frac{3}{4}\cos 3\theta$

Proof. For 1, use $\cos(A + B) + \cos(A - B) = 2\cos A\cos B$:

$$\displaystyle \cos\theta + 2\cos 120°\cos\theta = \cos\theta + 2\left(-\frac{1}{2}\right)\cos\theta = 0$$

For 2, use the alge­bra fact: if $a + b + c = 0$ then $a^3 + b^3 + c^3 = 3abc$. By result 1 the three cosines add to $0$, so

$$\displaystyle \cos^3\theta + \cos^3(120° + \theta) + \cos^3(120° - \theta) = 3\cos\theta\cos(120° + \theta)\cos(120° - \theta) = \frac{3}{4}\cos 3\theta$$

Sums of squares with $60° \pm \theta$#

Result.

1. $\displaystyle \sin^2\theta + \sin^2(60° + \theta) + \sin^2(60° - \theta) = \frac{3}{2}$
2. $\displaystyle \cos^2\theta + \cos^2(60° + \theta) + \cos^2(60° - \theta) = \frac{3}{2}$

Proof of 1. Use $\displaystyle \sin^2 A = \frac{1 - \cos 2A}{2}$:

$$\displaystyle \frac{1 - \cos 2\theta}{2} + \frac{1 - \cos(120° + 2\theta)}{2} + \frac{1 - \cos(120° - 2\theta)}{2} = \frac{3}{2} - \frac{1}{2}\left[\cos 2\theta + \cos(120° + 2\theta) + \cos(120° - 2\theta)\right]$$

The bracket is $0$ by the pre­vi­ous result, so the sum is $\displaystyle \frac{3}{2}$. For exam­ple, $\displaystyle \sin^2 10° + \sin^2 70° + \sin^2 50° = \frac{3}{2}$.

Two angles adding to $60°$#

Result. If $A + B = 60°$, then

1. $\displaystyle \cos^2 A + \cos^2 B - \cos A\cos B = \frac{3}{4}$
2. $\displaystyle \sin^2 A + \sin^2 B + \sin A\sin B = \frac{3}{4}$

Proof of 1.

$$\displaystyle \begin{aligned} &\cos^2 A + 1 - \sin^2 B - \frac{1}{2}(2\cos A\cos B) \\ &= 1 + \cos(A + B)\cos(A - B) - \frac{1}{2}\left[\cos(A + B) + \cos(A - B)\right] \\ &= 1 + \frac{1}{2}\cos(A - B) - \frac{1}{4} - \frac{1}{2}\cos(A - B) = \frac{3}{4} \end{aligned}$$

The same steps work when $A - B = 60°$, since $\cos(A - B)$ and $\cos(A + B)$ sim­ply swap roles. For exam­ple, $\displaystyle \cos^2 76° + \cos^2 16° - \cos 76°\cos 16° = \frac{3}{4}$, because $76° - 16° = 60°$.

Prac­tice#

1. Find $\sin 20°\sin 40°\sin 80°$.
2. Find $\tan 20°\tan 40°\tan 80°$.
3. Find $\cos 10°\cos 50°\cos 70°$.
4. Find $\sin 10°\sin 50°\sin 70°$.
5. Find $\tan 10°\tan 50°\tan 70°$.
6. Find $\sin^2 20° + \sin^2 80° + \sin^2 40°$.
7. Find $\cos^2 10° + \cos^2 50° + \cos^2 70°$.
8. Find $\cos^2 40° + \cos^2 20° - \cos 40°\cos 20°$.
9. Find $\sin^2 50° + \sin^2 10° + \sin 50°\sin 10°$.
10. Find $\tan 20° + \tan 80° - \tan 40°$.
11. Find $\cos^3 20° + \cos^3 140° + \cos^3 100°$.

Answers#