Cos 36 Degrees: Exact Value, Golden Ratio Proof And Examples

What Does Cos 36 Degrees Mean?

In a right triangle, cosine is the adjacent side divided by the hypotenuse, so $\cos 36^\circ$ is the fraction of the hypotenuse that lies next to a $36^\circ$ angle. The cosine function extends that ratio to every angle.

On the unit circle (radius 1, centred at the origin), the point at $36^\circ$ is $(\cos 36^\circ, \sin 36^\circ) \approx (0.8090, 0.5878)$. The angle lies in quadrant I, the upper-right region of the plane where both coordinates are positive, so the cosine is positive.

In radians, $36^\circ = \frac{\pi}{5}$, one fifth of a half turn. That fifth is the clue: five copies of $36^\circ$ make $180^\circ$, and that single fact is what pins down the exact value.

Standard-Angle Cosine Reference Table

Thirty-six degrees joins the familiar special angles because it comes from the regular pentagon, not from the square or the equilateral triangle. Here it is alongside its neighbours.

Cosine falls as the angle grows, and $36^\circ$ sits between $30^\circ$ and $45^\circ$ just as $0.8090$ sits between $0.8660$ and $0.7071$. The cos 30 degrees page covers the next special angle down.

Where Does Cos 36 Degrees Show Up, And Why Does It Matter?

In a regular pentagon, a diagonal is $2\cos 36^\circ = \varphi \approx 1.618$ times as long as a side (Pentagon). Each interior angle is $108^\circ$, and the triangle formed by two sides and one diagonal has base angles of $36^\circ$, which is where the factor comes from.

The thin rhombus in a Penrose tiling has corner angles of $36^\circ$ and $144^\circ$ (Penrose tiling). These non-repeating tilings later helped explain quasicrystals, so $\cos 36^\circ$ turns up in materials science as well as in geometry homework.

How Do You Find Cos 36 Degrees Without A Calculator?

Three routes give $\frac{1 + \sqrt{5}}{4}$ by hand: an equation from the triple-angle formula, a picture from the pentagon, and a shortcut through $\sin 18^\circ$. A fourth, the sum and product of $\cos 36^\circ$ and $\cos 72^\circ$, gives the same quadratic.

Method 1: The Triple-Angle Equation

Let $\theta = 36^\circ$. Then $5\theta = 180^\circ$, so $2\theta = 180^\circ - 3\theta$, and supplementary angles share a sine:

$$\sin 2\theta = \sin 3\theta$$

Expand both sides, using the sin 3x formula on the right:

$$2\sin\theta\cos\theta = 3\sin\theta - 4\sin^3\theta$$

Divide by $\sin\theta$, which is not zero, and replace $\sin^2\theta$ with $1 - \cos^2\theta$:

$$2\cos\theta = 3 - 4(1 - \cos^2\theta) = 4\cos^2\theta - 1$$

$$4\cos^2\theta - 2\cos\theta - 1 = 0$$

The quadratic formula (see roots of a quadratic equation) gives:

$$\cos\theta = \frac{2 \pm \sqrt{4 + 16}}{8} = \frac{1 \pm \sqrt{5}}{4}$$

Since $36^\circ$ is in quadrant I, keep the positive root: $\cos 36^\circ = \frac{1 + \sqrt{5}}{4}$.

Method 2: The Regular Pentagon And The Golden Triangle

In a regular pentagon with side 1, call the diagonal $d$. Similar isosceles triangles inside the pentagon give $d^2 = d + 1$, whose positive root is the golden ratio $\varphi = \frac{1 + \sqrt{5}}{2}$.

Now take the triangle made of two sides and one diagonal: legs 1 and 1, base $\varphi$, base angles $36^\circ$. Dropping a perpendicular from the apex cuts the base in half, giving a right triangle with hypotenuse 1 and adjacent side $\frac{\varphi}{2}$:

$$\cos 36^\circ = \frac{\varphi / 2}{1} = \frac{1 + \sqrt{5}}{4}$$

Method 3: Double Angle From Sin 18 Degrees

If you already know $\sin 18^\circ = \frac{\sqrt{5} - 1}{4}$, the double angle formula $\cos 2A = 1 - 2\sin^2 A$ finishes the job with $A = 18^\circ$:

$$\cos 36^\circ = 1 - 2\left(\frac{\sqrt{5} - 1}{4}\right)^2 = 1 - \frac{6 - 2\sqrt{5}}{8}$$

$$\cos 36^\circ = \frac{2 + 2\sqrt{5}}{8} = \frac{1 + \sqrt{5}}{4}$$

Method 4: Sum And Product With Cos 72 Degrees

The five vertices of a regular pentagon centred at the origin balance out, so their $x$-coordinates add to zero:

$$\cos 0^\circ + \cos 72^\circ + \cos 144^\circ + \cos 216^\circ + \cos 288^\circ = 0$$

Pairing equal terms and using $\cos 144^\circ = -\cos 36^\circ$ gives $1 + 2\cos 72^\circ - 2\cos 36^\circ = 0$, so $\cos 36^\circ - \cos 72^\circ = \frac{1}{2}$. Combining that difference with the product $\cos 36^\circ \times \cos 72^\circ = \frac{1}{4}$ (proved in Example 3) gives $4c^2 - 2c - 1 = 0$ again, the same quadratic as Method 1.

Why Is Cos 36 Degrees Half The Golden Ratio?

Because the golden ratio is the pentagon's diagonal-to-side ratio, and that diagonal is $2\cos 36^\circ$ times the side. Halving both sides of $2\cos 36^\circ = \varphi$ gives $\cos 36^\circ = \frac{\varphi}{2}$.

A neat consequence follows from $\varphi = 2\cos 36^\circ$ and $\frac{1}{\varphi} = \varphi - 1$:

$$\sec 36^\circ = \frac{1}{\cos 36^\circ} = \frac{2}{\varphi} = 2(\varphi - 1) = \sqrt{5} - 1 \approx 1.2361$$

Can Cos 36 Degrees Be Written As A Fraction?

Not as a fraction of whole numbers, because $\sqrt{5}$ is irrational. The exact value is the radical fraction $\frac{1 + \sqrt{5}}{4}$, sometimes written $\frac{\sqrt{5} + 1}{4}$ (the same number with the terms swapped).

For estimates, $\frac{809}{1000} = 0.809$ is within $0.00002$. The radical form is the one to give whenever a question asks for the exact value.

How Is Cos 36 Degrees Related To Sin 54, Sin 36 And Cos 72?

The cofunction identities and the Pythagorean identity link $\cos 36^\circ$ to a whole family of pentagon values:

The $\cos 108^\circ$ row explains the negative root in Method 1: $108^\circ$ also satisfies $\sin 2\theta = \sin 3\theta$, so the quadratic quietly answers for both angles.

Examples Of Cos 36 Degrees

Example 1

Simplify $4\cos 36^\circ - 1$.

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

$$= \sqrt{5} \approx 2.2361$$

Example 2

Solve $4c^2 - 2c - 1 = 0$ to find $c = \cos 36^\circ$.

Wrong attempt. The roots are $\frac{1 + \sqrt{5}}{4}$ and $\frac{1 - \sqrt{5}}{4}$, and a student picks $\frac{1 - \sqrt{5}}{4} \approx -0.3090$ because it "looks smaller, like a cosine should."

That breaks at once: $36^\circ$ is in quadrant I, where cosine is positive, so a negative answer is impossible.

Correct. Keep $\frac{1 + \sqrt{5}}{4} \approx 0.8090$. The other root is $\cos 108^\circ$, a second angle that satisfies the same equation.

Example 3

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

Multiply and divide by $\sin 36^\circ$, then apply $2\sin x\cos x = \sin 2x$ twice:

$$\sin 36^\circ\cos 36^\circ\cos 72^\circ = \frac{1}{2}\sin 72^\circ\cos 72^\circ = \frac{1}{4}\sin 144^\circ$$

Since $\sin 144^\circ = \sin 36^\circ$, dividing by $\sin 36^\circ$ leaves $\frac{1}{4}$. The cos A cos B product-to-sum route reaches the same answer.

Example 4

Evaluate $\cos 36^\circ - \cos 72^\circ$.

$$\frac{1 + \sqrt{5}}{4} - \frac{\sqrt{5} - 1}{4} = \frac{2}{4}$$

$$= \frac{1}{2}$$

The cos A minus cos B identity confirms it: $-2\sin 54^\circ\sin(-18^\circ) = 2\sin 54^\circ\sin 18^\circ = 2 \times \frac{1 + \sqrt{5}}{4} \times \frac{\sqrt{5} - 1}{4} = \frac{1}{2}$.

Example 5

A regular pentagon has sides of $5$ cm. How long is each diagonal?

$$d = 2 \times 5 \times \cos 36^\circ = 10 \times 0.8090170$$

$$d \approx 8.09 \text{ cm}$$

The habit that saves marks here is drawing the $108^\circ$ triangle first; students who skip the sketch often use $\cos 72^\circ$ and get $3.09$ cm, shorter than a side.

Example 6

Simplify $\cos(36^\circ - A)\cos(36^\circ + A) + \cos(54^\circ + A)\cos(54^\circ - A)$.

Use $\cos(x - A)\cos(x + A) = \cos^2 x - \sin^2 A$ on each product:

$$(\cos^2 36^\circ - \sin^2 A) + (\cos^2 54^\circ - \sin^2 A)$$

Since $\cos 54^\circ = \sin 36^\circ$, the first terms add to 1:

$$\cos^2 36^\circ + \sin^2 36^\circ - 2\sin^2 A = 1 - 2\sin^2 A = \cos 2A$$

Final answer: $\cos 2A$.

Where Do Students Trip Up On Cos 36 Degrees?

Mistake 1: Swapping The Plus And Minus Signs

Where it slips in: Writing the pentagon values from memory, where $\cos 36^\circ$ and $\cos 72^\circ$ differ only in one sign.

Don't do this: Writing $\cos 36^\circ = \frac{\sqrt{5} - 1}{4}$, which is $\cos 72^\circ \approx 0.3090$.

The correct way: The smaller angle has the larger cosine, so $\cos 36^\circ$ takes the plus sign: $\frac{\sqrt{5} + 1}{4} \approx 0.8090$.

The memorizer who stores two nearly identical radicals mixes them up under time pressure. Students who instead remember "$\cos 36^\circ$ is half of $\varphi$, and $\varphi$ is bigger than 1" never pick the minus sign.

Mistake 2: Getting -0.128 From A Calculator

Where it slips in: A calculator, or a search box, left in radian mode, which reads 36 as 36 radians.

Don't do this: Accepting $\cos 36 \approx -0.1280$, or $\cos 36 - \cos 72 \approx 0.8393$ in place of the exact $\frac{1}{2}$.

The correct way: Switch to degree mode, or enter $\cos\left(\frac{\pi}{5}\right)$ in radian mode; both return $0.8090$.

Mistake 3: Dropping The Half

Where it slips in: The golden-ratio link, said quickly as "cos 36 is the golden ratio."

Don't do this: Writing $\cos 36^\circ = \frac{1 + \sqrt{5}}{2} \approx 1.618$.

The correct way: A cosine can never exceed 1, so the value is $\frac{\varphi}{2} = \frac{1 + \sqrt{5}}{4}$. The rusher who checks "is it between 0 and 1?" catches this in a second.

Where Can You Get Help With Trigonometry Values?

Pentagon values like $\cos 36^\circ$ combine geometry, identities and a quadratic, which is exactly where a guided explanation saves time. A trigonometry tutor can walk through the triple-angle derivation until it feels natural.

A high school math tutor ties these values to the algebra behind them, and math classes online offer regular practice with exact values.

Key Takeaways

• Cos 36 degrees equals $\frac{1 + \sqrt{5}}{4} \approx 0.8090$, which is half the golden ratio.

• In radians it is $\cos\frac{\pi}{5}$, and five copies of $36^\circ$ make $180^\circ$.

• The triple-angle equation $4c^2 - 2c - 1 = 0$ gives the value, with the negative root belonging to $\cos 108^\circ$.

• In a regular pentagon, a diagonal is $2\cos 36^\circ = \varphi$ times a side.

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

Practice These Before Moving On

1. Find the exact value of $\cos^2 36^\circ$.

2. Show that $\sec 36^\circ = \sqrt{5} - 1$.

3. A regular pentagon has diagonals of $10$ cm. Find the side to two decimal places.

4. Use the cofunction identity to write the exact value of $\sin 54^\circ$.

Answers:

1. $\left(\frac{1 + \sqrt{5}}{4}\right)^2 = \frac{6 + 2\sqrt{5}}{16} = \frac{3 + \sqrt{5}}{8}$.

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

3. $\frac{10}{2\cos 36^\circ} = \frac{10}{\varphi} \approx 6.18$ cm.

4. $\sin 54^\circ = \cos 36^\circ = \frac{1 + \sqrt{5}}{4}$.

If Question 2 stalls, rationalise the denominator by multiplying top and bottom by $\sqrt{5} - 1$, as in Method 3's algebra.

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