Cos 72 Degrees: Exact Value, Golden Ratio Proof And Examples

Where Does $\cos 72^\circ$ Show Up, And Why Does It Matter?

The number $72^\circ$ is one fifth of a full turn, so $\cos 72^\circ$ appears wherever something has five-fold symmetry. In a regular pentagon, each vertex sits $72^\circ$ around the centre from the next, and the ratio of diagonal to side, the golden ratio $\varphi \approx 1.618$, is exactly $\frac{1}{2\cos 72^\circ}$.

The same value balances forces. Five equal pulls spaced $72^\circ$ apart (a five-spoke bracket, a five-point star of cables) cancel exactly, because the horizontal parts add to $1 + 2\cos 72^\circ + 2\cos 144^\circ = 0$, which Example 6 checks.

Standard-Angle Cosine Reference Table

The table below lists the first-quadrant angles whose cosines have exact square-root forms, including the pentagon family ($18^\circ$, $36^\circ$, $54^\circ$, $72^\circ$) that most tables leave out. Every decimal was recomputed to 10 places and rounded.

Read down the last column and cosine falls from $1$ to $0$ as the angle opens. Notice that $\cos 36^\circ$ and $\cos 72^\circ$ differ only in one sign inside the numerator, $1+\sqrt{5}$ against $\sqrt{5}-1$.

What Does $\cos 72^\circ$ Mean On The Unit Circle?

On the unit circle, a circle of radius $1$ centred at the origin, the cosine of an angle is the $x$-coordinate of the point the rotating radius reaches. At $72^\circ$ that point is $(0.3090, 0.9511)$, so $\cos 72^\circ \approx 0.3090$.

A quadrant is one of the four regions the axes cut the plane into, numbered anticlockwise from the top right. Since $72^\circ$ lies between $0^\circ$ and $90^\circ$, it sits in Quadrant I, where both coordinates are positive, so $\cos 72^\circ$ is positive.

In a right triangle the same number is the ratio of the side adjacent to the $72^\circ$ angle to the hypotenuse. The triangle and the circle agree because the circle is just the triangle scaled so that the hypotenuse is $1$.

What Is Cos 72 Degrees In Radians?

A radian measures an angle by the arc it cuts on a radius-1 circle, and $180^\circ = \pi$ radians. So $72^\circ = 72 \times \frac{\pi}{180} = \frac{2\pi}{5} \approx 1.2566$, and $\cos \frac{2\pi}{5}$ is the same number, $\frac{\sqrt{5}-1}{4}$.

How Do You Find The Exact Value Of $\cos 72^\circ$?

Four routes reach the same answer, and each one teaches something different. The first is purely geometric, the second uses identities, the third leans on $\cos 36^\circ$, and the fourth comes from complex numbers.

Method 1: The Golden Triangle (36-72-72)

Draw an isosceles triangle $ABC$ with apex angle $\angle A = 36^\circ$, so the base angles are $\angle B = \angle C = 72^\circ$. Let $AB = AC = 1$ and $BC = x$.

1. Bisect $\angle B$ and let the bisector meet $AC$ at $D$. Now $\angle ABD = \angle DBC = 36^\circ$.

2. In triangle $BDC$, $\angle BDC = 180^\circ - 36^\circ - 72^\circ = 72^\circ$, so $BD = BC = x$.

3. In triangle $ABD$, $\angle A = \angle ABD = 36^\circ$, so $AD = BD = x$ and $DC = 1 - x$.

4. Triangles $ABC$ and $BDC$ have the same angles, so they are similar triangles:

$$\frac{BC}{AB} = \frac{DC}{BC}$$

$$\frac{x}{1} = \frac{1-x}{x}$$

$$x^2 + x - 1 = 0$$

5. The positive root is $x = \frac{\sqrt{5}-1}{2} \approx 0.6180$.

6. Drop the perpendicular from $A$ to $BC$. It bisects $BC$, so in the right triangle with angle $72^\circ$ at $B$, the adjacent side is $\frac{x}{2}$ and the hypotenuse is $1$:

$$\cos 72^\circ = \frac{x}{2} = \frac{\sqrt{5}-1}{4}$$

Method 2: Through $\sin 18^\circ$ And The Cofunction Identity

The cofunction identities say $\cos \theta = \sin(90^\circ - \theta)$, so $\cos 72^\circ = \sin 18^\circ$. That turns the question into one about $18^\circ$.

1. Let $\theta = 18^\circ$. Then $5\theta = 90^\circ$, so $2\theta = 90^\circ - 3\theta$.

2. Take sine of both sides: $\sin 2\theta = \cos 3\theta$.

3. Expand: $2\sin\theta\cos\theta = 4\cos^3\theta - 3\cos\theta$.

4. Divide by $\cos\theta$, which is not zero: $2\sin\theta = 4\cos^2\theta - 3 = 1 - 4\sin^2\theta$.

5. Rearrange: $4\sin^2\theta + 2\sin\theta - 1 = 0$.

6. Solve the quadratic in $s = \sin\theta$ and keep the positive root, since $\sin 18^\circ > 0$:

$$\sin 18^\circ = \frac{-2 + \sqrt{4 + 16}}{8} = \frac{\sqrt{5}-1}{4}$$

So $\cos 72^\circ = \sin 18^\circ = \frac{\sqrt{5}-1}{4}$. This is the "by hand" route the Flying Colours reader asked about, and it is the one exam boards expect when the question says "show that".

Method 3: The Double-Angle Formula From $\cos 36^\circ$

Method 1 also gives $\cos 36^\circ$ for free: in triangle $ABD$ the foot of the perpendicular from $D$ halves $AB$, so $\cos 36^\circ = \frac{1/2}{x} = \frac{1+\sqrt{5}}{4}$. The double angle formula $\cos 2\theta = 2\cos^2\theta - 1$ then finishes the job.

$$\cos 72^\circ = 2\left(\frac{1+\sqrt{5}}{4}\right)^2 - 1$$

$$= \frac{6 + 2\sqrt{5}}{8} - 1$$

$$= \frac{\sqrt{5}-1}{4}$$

Method 4: The Fifth Roots Of Unity

The five solutions of $z^5 = 1$ sit on the unit circle $72^\circ$ apart, and they add to zero. Adding their real parts gives $1 + 2\cos 72^\circ + 2\cos 144^\circ = 0$.

1. Let $c = \cos 72^\circ$. Then $\cos 144^\circ = 2c^2 - 1$.

2. Substitute: $1 + 2c + 4c^2 - 2 = 0$, so $4c^2 + 2c - 1 = 0$.

3. Positive root: $c = \frac{\sqrt{5}-1}{4}$.

The same quadratic $4c^2 + 2c - 1 = 0$ appears in Methods 2 and 4. That is not a coincidence: both are the algebraic fingerprint of a five-fold split of the circle.

Calculator Check

Set the calculator to degree mode and enter $\cos(72)$. It returns $0.3090169944\ldots$, which matches $\frac{\sqrt{5}-1}{4} = \frac{2.2360679\ldots - 1}{4}$.

Why Is $\cos 72^\circ$ Linked To The Golden Ratio?

Because doubling it gives the reciprocal of the golden ratio: $2\cos 72^\circ = \frac{\sqrt{5}-1}{2} = \frac{1}{\varphi} \approx 0.6180$. The side $x$ in Method 1 is exactly this number, so the golden triangle is the golden ratio drawn as a shape.

Two consequences are worth keeping:

• Decagon side. A regular decagon inscribed in a circle of radius $r$ has side $2r\sin 18^\circ = 2r\cos 72^\circ = \frac{r}{\varphi}$.

• Secant. $\sec 72^\circ = \frac{4}{\sqrt{5}-1} = 1 + \sqrt{5} = 2\varphi \approx 3.2361$.

Here is the part that took a while to sink in. The golden ratio usually arrives wrapped in rectangles and shells, and seeing it fall out of one cosine felt like a trick until the bisected triangle made it visible.

How Is $\cos 72^\circ$ Written Using Other Trig Functions?

The same value hides behind several angles and functions, which is why "cos 72 sin 18" is such a common search. The reference angle, the acute angle between an angle's terminal side and the $x$-axis, tells you which other angles share the size; the quadrant decides the sign.

On the graph of the cosine function, these are the points at height $0.3090$ or $-0.3090$ inside one period.

Examples Of Cos 72 Degrees

Example 1

Find the exact value of $\sec 72^\circ$.

$$\sec 72^\circ = \frac{1}{\cos 72^\circ} = \frac{4}{\sqrt{5}-1}$$

$$= \frac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)} = \frac{4(\sqrt{5}+1)}{4}$$

$$= 1 + \sqrt{5} \approx 3.2361$$

Example 2

Use $\cos 36^\circ = \frac{1+\sqrt{5}}{4}$ to find $\cos 72^\circ$.

First attempt. Since $72 = 2 \times 36$, write $\cos 72^\circ = 2\cos 36^\circ = \frac{1+\sqrt{5}}{2} \approx 1.618$.

That cannot be right. No cosine of a real angle is bigger than $1$, and the decimal $1.618$ is the golden ratio, not a cosine.

Correct route. Doubling the angle does not double the cosine. Use $\cos 2\theta = 2\cos^2\theta - 1$:

$$\cos 72^\circ = 2\left(\frac{1+\sqrt{5}}{4}\right)^2 - 1 = \frac{\sqrt{5}-1}{4} \approx 0.3090$$

Example 3

Show that $\cos 36^\circ \cos 72^\circ = \frac{1}{4}$ and $\cos 36^\circ - \cos 72^\circ = \frac{1}{2}$.

$$\cos 36^\circ \cos 72^\circ = \frac{(\sqrt{5}+1)(\sqrt{5}-1)}{16} = \frac{5-1}{16} = \frac{1}{4}$$

$$\cos 36^\circ - \cos 72^\circ = \frac{(1+\sqrt{5}) - (\sqrt{5}-1)}{4} = \frac{2}{4} = \frac{1}{2}$$

The first line is a difference of squares, and spotting that pairing is the step most solvers miss; they expand both surds term by term and lose a sign halfway.

Example 4

Evaluate $\cos^2 72^\circ - \sin^2 54^\circ$.

Since $\sin 54^\circ = \cos 36^\circ$, factor as a difference of squares:

$$\cos^2 72^\circ - \cos^2 36^\circ = (\cos 72^\circ - \cos 36^\circ)(\cos 72^\circ + \cos 36^\circ)$$

$$= \left(-\frac{1}{2}\right)\left(\frac{\sqrt{5}}{2}\right) = -\frac{\sqrt{5}}{4} \approx -0.5590$$

The sum $\cos^2 72^\circ + \sin^2 54^\circ$ comes out to $\frac{3-\sqrt{5}}{8} + \frac{3+\sqrt{5}}{8} = \frac{3}{4}$, a useful check.

Example 5

A regular pentagon has side $10$ cm. Find the length of a diagonal.

Each interior angle of a regular pentagon is $108^\circ$, and a diagonal faces that angle in a triangle of two sides. By the law of cosines, with $\cos 108^\circ = -\cos 72^\circ$:

$$d^2 = 10^2 + 10^2 - 2(10)(10)\cos 108^\circ$$

$$d^2 = 200(1 + \cos 72^\circ) = 200 \times \frac{3+\sqrt{5}}{4}$$

$$d = 5 + 5\sqrt{5} \approx 16.18 \text{ cm}$$

The diagonal is $\varphi$ times the side, as the golden-ratio section promised.

Example 6

Five cables pull on a ring with equal force $F$, spaced $72^\circ$ apart, one along the positive $x$-axis. Show the horizontal forces cancel.

The horizontal components are $F\cos 0^\circ$, $F\cos 72^\circ$, $F\cos 144^\circ$, $F\cos 216^\circ$ and $F\cos 288^\circ$. Pair the symmetric ones:

$$F\left(1 + 2\cos 72^\circ + 2\cos 144^\circ\right)$$

$$= F\left(1 + \frac{\sqrt{5}-1}{2} - \frac{1+\sqrt{5}}{2}\right) = F(1 - 1) = 0$$

Where Do Students Go Wrong With $\cos 72^\circ$?

Mistake 1: Typing $\cos 72$ In Radian Mode

Where it slips in: Searching or calculating "what is cos 72" on a device set to radians.

Don't do this: Accept $\cos 72 \approx -0.9673$. That is the cosine of $72$ radians, an angle of about $4125^\circ$.

The correct way: Switch to degree mode or enter $\cos\left(\frac{2\pi}{5}\right)$. A negative answer for a Quadrant I angle is the alarm bell; the rusher who never checks the sign is the one this catches.

Mistake 2: Swapping $\cos 72^\circ$ And $\cos 36^\circ$

Where it slips in: Recall under time pressure, because the two exact forms differ only by the sign in front of the $1$.

Don't do this: Write $\cos 72^\circ = \frac{1+\sqrt{5}}{4}$. That is $0.8090$, the value of $\cos 36^\circ$.

The correct way: Anchor on size: cosine shrinks as the angle grows, so the larger angle $72^\circ$ takes the smaller value $\frac{\sqrt{5}-1}{4}$. The memorizer who learned both surds as a pair without a size check is exactly the student who swaps them; asking "which is closer to 1?" fixes it in one step.

Mistake 3: Dropping The Factor Of 2

Where it slips in: The last line of the golden-triangle method, or after solving $x^2 + x - 1 = 0$.

Don't do this: Stop at $\frac{\sqrt{5}-1}{2} \approx 0.618$ and call it $\cos 72^\circ$.

The correct way: That number is the side $x$, which equals $2\cos 72^\circ$, so halve it to get $\frac{\sqrt{5}-1}{4}$. The giveaway is the decimal: $0.618$ is the cosine of about $51.8^\circ$, not $72^\circ$.

Where Can You Get Help With Trigonometry Values Like $\cos 72^\circ$?

Derivations like the golden triangle click fastest when someone watches you draw the bisector and asks why the small triangle is similar. A live trigonometry tutor can walk through all four methods at your pace.

For broader support, a high school math tutor connects these exact values to identities and equations, and math classes online give regular practice with feedback.

Key Takeaways

• Cos 72 degrees equals $\frac{\sqrt{5}-1}{4}$, which is $0.3090$ to four decimal places, and $72^\circ = \frac{2\pi}{5}$ radians.

• The golden 36-72-72 triangle gives it directly, and the sin $18^\circ$ identity, the double-angle formula and the fifth roots of unity confirm it.

• Doubling it gives $\frac{1}{\varphi} \approx 0.618$, the reciprocal of the golden ratio.

• The two traps are radian mode, which returns $-0.9673$, and swapping it with $\cos 36^\circ = \frac{1+\sqrt{5}}{4}$.

Practice These Before Moving On

1. Evaluate $2\cos 72^\circ + 1$ exactly.

2. Find $\cos 288^\circ$ and $\cos 108^\circ$ exactly.

3. A regular decagon is inscribed in a circle of radius $5$ cm. Find its side length.

4. Show that $4\cos 72^\circ + 1 = \sqrt{5}$.

Answers:

1. $\frac{1+\sqrt{5}}{2} \approx 1.618$, the golden ratio.

2. $\cos 288^\circ = \frac{\sqrt{5}-1}{4}$ and $\cos 108^\circ = -\frac{\sqrt{5}-1}{4}$.

3. Side $= 2 \times 5 \times \cos 72^\circ = \frac{5(\sqrt{5}-1)}{2} \approx 3.09$ cm.

4. $4 \times \frac{\sqrt{5}-1}{4} + 1 = \sqrt{5} - 1 + 1 = \sqrt{5}$.

If Question 3 stalls, go back to the decagon bullet in the golden-ratio section. Practice these to solidify your understanding.

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Read More

• Trigonometric table of standard angles

• Sum and difference formulas in trigonometry

• Angles in a pentagon

• Fibonacci numbers and the golden ratio

• Cos 60 degrees