Sin 2pi/5: Exact Value √(10 + 2√5)/4 and Proof

What Does Sin 2pi/5 Mean?

A radian is the angle that cuts off an arc equal in length to the radius. A full turn is $2\pi$ radians, so $\frac{2\pi}{5}$ radians is one fifth of a turn, about $1.2566$ rad.

What is sin 2pi/5 in degrees? Multiply by $\frac{180^\circ}{\pi}$: $\frac{2\pi}{5} \times \frac{180^\circ}{\pi} = 72^\circ$, so $\sin\left(\frac{2\pi}{5}\right) = \sin 72^\circ$ (the full method is in radians to degrees). The unit changes; the value $0.9511$ does not.

On the unit circle (radius $1$, centre at the origin), the sine of an angle is the $y$-coordinate of the point the angle reaches. In a right triangle with a $72^\circ$ angle, the same number is the opposite side divided by the hypotenuse.

A quadrant is one of the four regions cut out by the axes, numbered I to IV counterclockwise starting top right. Since $0 < \frac{2\pi}{5} < \frac{\pi}{2}$, the angle lies in Quadrant I, where $y$ is positive.

The reference angle is the acute angle between the terminal side and the $x$-axis. For $\frac{2\pi}{5}$ it is $\frac{2\pi}{5}$ itself, and $\frac{3\pi}{5}$, $\frac{7\pi}{5}$ and $\frac{8\pi}{5}$ share it, so their sines are $\pm 0.9511$.

The principal value of $\sin^{-1}$ is the single answer it gives, always between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. Because $\frac{2\pi}{5}$ is inside that range, $\sin^{-1}\left(\sin\frac{2\pi}{5}\right) = \frac{2\pi}{5}$.

Standard-Angle Sine Reference Table

The angle $\frac{2\pi}{5}$ belongs to the "pentagon family" of angles, $\frac{\pi}{10}, \frac{\pi}{5}, \frac{3\pi}{10}, \frac{2\pi}{5}$, whose sines all contain $\sqrt{5}$. The table lists them in radian order next to the classic angles.

Sine rises through the first quadrant, so $\sin\left(\frac{2\pi}{5}\right)$ must sit between $\sin\left(\frac{\pi}{3}\right) \approx 0.8660$ and sin pi/2, which is $1$. The two square roots for $\frac{\pi}{5}$ and $\frac{2\pi}{5}$ differ only by the sign in front of $2\sqrt{5}$, which is exactly where most copying errors happen.

Where Does Sin 2pi/5 Show Up, And Why Does It Matter?

The five corners of a regular pentagon drawn in a unit circle sit at angles $0, \frac{2\pi}{5}, \frac{4\pi}{5}, \frac{6\pi}{5}, \frac{8\pi}{5}$, so $\sin\left(\frac{2\pi}{5}\right)$ is the height of the first corner above the centre line. It also fixes the diagonal: in a regular pentagon with circumradius $R$, each diagonal is $2R\sin\left(\frac{2\pi}{5}\right) \approx 1.902R$.

Each side is $2R\sin\left(\frac{\pi}{5}\right) \approx 1.176R$, so diagonal divided by side equals $\frac{\sin(2\pi/5)}{\sin(\pi/5)} = 1.6180$, the golden ratio. That ratio is why the five-pointed pentagram is full of golden proportions, and why $\sqrt{5}$ keeps appearing in these sines.

Outside geometry, $\sin\left(\frac{2\pi}{5}\right)$ is one of the fixed constants in a five-point discrete Fourier transform, the signal-processing routine that splits a signal into frequencies. The factor it uses is $\cos\left(\frac{2\pi}{5}\right) - i\sin\left(\frac{2\pi}{5}\right) \approx 0.3090 - 0.9511i$.

Is Sin 2pi/5 A Fraction Or A Root?

It is a root, not a fraction. $\sin\left(\frac{2\pi}{5}\right)$ is an irrational number, so the honest "in fraction" answer is the nested radical $\frac{\sqrt{10 + 2\sqrt{5}}}{4}$, and any decimal such as $\frac{9511}{10000}$ is rounded.

Squaring and clearing radicals shows that $s = \sin\left(\frac{2\pi}{5}\right)$ satisfies

$$16s^4 - 20s^2 + 5 = 0.$$

A fraction root would have to be $\pm 1$ or $\pm 5$ divided by $1, 2, 4, 8$ or $16$, and none of those twenty candidates makes the left side zero. So no fraction equals $\sin\left(\frac{2\pi}{5}\right)$.

How Do You Find The Exact Value Of Sin 2pi/5?

The fastest exact route finds $\cos\left(\frac{2\pi}{5}\right)$ first from a five-fold angle trick, then uses $\sin^2 + \cos^2 = 1$. Two other routes confirm it.

Method 1: The Five-Fold Angle Equation

Let $\theta = \frac{\pi}{10}$, so $5\theta = \frac{\pi}{2}$ and $2\theta = \frac{\pi}{2} - 3\theta$. Taking sine of both sides:

$$\sin 2\theta = \sin\left(\frac{\pi}{2} - 3\theta\right) = \cos 3\theta$$

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

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

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

$$4\sin^2\theta + 2\sin\theta - 1 = 0$$

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

The positive root is kept because $\frac{\pi}{10}$ is in Quadrant I. By the co-function rule, $\cos\left(\frac{2\pi}{5}\right) = \sin\left(\frac{\pi}{10}\right) = \frac{\sqrt{5} - 1}{4} \approx 0.3090$.

Now use $\sin^2 x + \cos^2 x = 1$:

$$\sin^2\left(\frac{2\pi}{5}\right) = 1 - \left(\frac{\sqrt{5} - 1}{4}\right)^2 = 1 - \frac{6 - 2\sqrt{5}}{16} = \frac{10 + 2\sqrt{5}}{16}$$

$$\sin\left(\frac{2\pi}{5}\right) = \frac{\sqrt{10 + 2\sqrt{5}}}{4} \approx 0.9510565163$$

Method 2: Double The Angle $\frac{\pi}{5}$

The golden triangle (a 36-72-72 isosceles triangle) gives $\cos\left(\frac{\pi}{5}\right) = \frac{1 + \sqrt{5}}{4}$ and so $\sin\left(\frac{\pi}{5}\right) = \frac{\sqrt{10 - 2\sqrt{5}}}{4}$. The sin double angle formula $\sin 2x = 2\sin x\cos x$ then gives

$$\sin\left(\frac{2\pi}{5}\right) = 2 \times \frac{\sqrt{10 - 2\sqrt{5}}}{4} \times \frac{1 + \sqrt{5}}{4} = \frac{(1 + \sqrt{5})\sqrt{10 - 2\sqrt{5}}}{8}$$

Squaring this gives $\frac{(6 + 2\sqrt{5})(10 - 2\sqrt{5})}{64} = \frac{40 + 8\sqrt{5}}{64} = \frac{10 + 2\sqrt{5}}{16}$, the same square as Method 1. Both roots are positive, so the two forms are equal.

Method 3: Use The Co-Function Rule

The cofunction identities say $\sin x = \cos\left(\frac{\pi}{2} - x\right)$. So $\sin\left(\frac{2\pi}{5}\right) = \cos\left(\frac{\pi}{10}\right) = \cos 18^\circ$, and any cosine table entry for $18^\circ$ gives $0.9511$.

Method 4: Read The Unit Circle And Use Periodicity

The point at $\frac{2\pi}{5}$ on the unit circle is $(0.3090, 0.9511)$; its $y$-coordinate is the answer. Sine repeats every $2\pi$, so $\sin\left(\frac{12\pi}{5}\right) = \sin\left(\frac{2\pi}{5} + 2\pi\right)$ is also $0.9511$.

Method 5: Calculator And Taylor Series

In radian mode, a calculator returns $0.9510565163$ for $\sin(2\pi \div 5)$. Underneath sits a series such as $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} + \cdots$ with $x \approx 1.256637$.

The series needs five terms here because $1.2566$ is not a small angle; the small-angle shortcut $\sin x \approx x$ would overshoot by about $0.31$.

Why Does A Calculator Sometimes Say Sin 2pi/5 Is 0?

Google's own search suggestion for "what is sin 2pi/5" displays "= 0", and that is not a bug in the value. It is a reading of the typed text as $\frac{\sin(2\pi)}{5}$, and $\sin(2\pi) = 0$.

Brackets around the whole angle, plus the right mode, are the entire fix. A quick check: any answer below $0.8660$ cannot be right, because $\frac{2\pi}{5}$ is larger than $\frac{\pi}{3}$.

What Is Sin 2pi/5 In Terms Of Other Trig Functions?

Each ratio of $\frac{2\pi}{5}$ rewrites $\sin\left(\frac{2\pi}{5}\right)$, which helps when a problem supplies a different one. Here $\cos\left(\frac{2\pi}{5}\right) \approx 0.3090$ and $\tan\left(\frac{2\pi}{5}\right) = \sqrt{5 + 2\sqrt{5}} \approx 3.0777$.

The square-root rows use the positive root because the angle is in Quadrant I.

Examples Of Sin 2pi/5

Example 1

Evaluate $4\sin\left(\frac{2\pi}{5}\right)$.

$$4\sin\left(\frac{2\pi}{5}\right) = 4 \times \frac{\sqrt{10 + 2\sqrt{5}}}{4} = \sqrt{10 + 2\sqrt{5}} \approx 3.8042$$

Example 2

Find $\sin\left(\frac{2\pi}{5}\right)$ given $\sin\left(\frac{\pi}{5}\right) \approx 0.5878$.

Wrong attempt: the angle doubled, so double the sine.

$$2 \times 0.5878 = 1.1756$$

That breaks immediately. No sine can exceed $1$, so $1.1756$ is impossible.

Correct: sine is not proportional to the angle. Use $\sin 2x = 2\sin x\cos x$ with $\cos\left(\frac{\pi}{5}\right) \approx 0.8090$.

$$\sin\left(\frac{2\pi}{5}\right) = 2 \times 0.5878 \times 0.8090 \approx 0.9511$$

Example 3

Simplify $\frac{3\sin(2\pi/5)}{\sin(12\pi/5)}$.

Since $\frac{12\pi}{5} = \frac{2\pi}{5} + 2\pi$, the two sines are equal.

$$\frac{3\sin(2\pi/5)}{\sin(12\pi/5)} = \frac{3\sin(2\pi/5)}{\sin(2\pi/5)} = 3$$

Example 4

A regular pentagon is drawn in a circle of radius 10 cm. How long is each diagonal?

A diagonal spans two sides, a central angle of $\frac{4\pi}{5}$, so its length is $2R\sin\left(\frac{2\pi}{5}\right)$.

$$\text{diagonal} = 2 \times 10 \times 0.9510565 \approx 19.02 \text{ cm}$$

Example 5

Prove that $\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5} = \frac{5}{16}$.

Use $\sin(\pi - x) = \sin x$: $\sin\frac{4\pi}{5} = \sin\frac{\pi}{5}$ and $\sin\frac{3\pi}{5} = \sin\frac{2\pi}{5}$. The product becomes

$$\sin^2\frac{\pi}{5} \times \sin^2\frac{2\pi}{5} = \frac{10 - 2\sqrt{5}}{16} \times \frac{10 + 2\sqrt{5}}{16}$$

$$= \frac{100 - 20}{256} = \frac{80}{256} = \frac{5}{16}$$

Example 6

Find the exact value of $\tan\left(\frac{2\pi}{5}\right)$.

$$\tan\left(\frac{2\pi}{5}\right) = \frac{\sin(2\pi/5)}{\cos(2\pi/5)} = \frac{\sqrt{10 + 2\sqrt{5}}}{\sqrt{5} - 1}$$

Square it: $\frac{10 + 2\sqrt{5}}{6 - 2\sqrt{5}}$. Multiply top and bottom by $6 + 2\sqrt{5}$ to get $\frac{80 + 32\sqrt{5}}{16} = 5 + 2\sqrt{5}$, so $\tan\left(\frac{2\pi}{5}\right) = \sqrt{5 + 2\sqrt{5}} \approx 3.0777$.

Students who reach for a calculator at line one get $3.0777$ and stop; the squaring step is what produces the clean radical, and it is worth doing once by hand.

Where Students Trip Up On Sin 2pi/5?

Mistake 1: Mixing Up The Two Pentagon Radicals

Where it slips in: Copying exact values for $\frac{\pi}{5}$ and $\frac{2\pi}{5}$ from memory or from a page.

Don't do this: Write $\sin\left(\frac{2\pi}{5}\right) = \frac{\sqrt{10 - 2\sqrt{5}}}{4}$ (that is $\sin\frac{\pi}{5} = 0.5878$) or $\frac{\sqrt{10 - 2\sqrt{5}}}{2}$ (that is $1.1756$, more than $1$). Both versions circulate online.

The correct way: Bigger angle, plus sign: $\frac{\sqrt{10 + 2\sqrt{5}}}{4}$, then check the decimal lands between $0.8660$ and $1$.

The memorizer who stores four radicals as a list mixes them; the student who stores the one quadratic $4s^2 + 2s - 1 = 0$ can rebuild all of them in a minute.

Mistake 2: Treating $\frac{2\pi}{5}$ As Degrees

Where it slips in: A calculator left in degree mode, or a student who reads "2pi/5" as a small number of degrees.

Don't do this: Accept $\sin\left(\frac{2\pi}{5}\right) \approx 0.0219$, which is the sine of $1.2566^\circ$.

The correct way: A $\pi$ in the angle almost always means radians. Switch to radian mode, or convert to $72^\circ$ first.

Mistake 3: Getting The Principal Value Wrong

Where it slips in: Inverse questions such as $\sin^{-1}\left(\sin\frac{3\pi}{5}\right)$.

Don't do this: Answer $\frac{3\pi}{5}$ because "inverse cancels".

The correct way: $\sin^{-1}$ only returns angles from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. Since $\sin\frac{3\pi}{5} = \sin\frac{2\pi}{5}$, the answer is $\frac{2\pi}{5}$; the rusher who cancels the functions before checking the range loses this mark every time.

Where Can You Get Help With Trigonometry Values?

The quickest help comes from someone who can see whether your problem is the radian conversion, the radical or the calculator, because each needs a different fix. A trigonometry tutor can walk you from the unit circle to pentagon-family values, and a high school math tutor can connect them to the rest of your syllabus.

For a structured course, math classes online take trigonometry in order, while one-to-one math tutoring targets a single weak step.

Key Takeaways

• Sin 2pi/5 equals $\frac{\sqrt{10 + 2\sqrt{5}}}{4} \approx 0.9511$, the same as $\sin 72^\circ$.

• The exact value comes from $\cos\left(\frac{2\pi}{5}\right) = \frac{\sqrt{5} - 1}{4}$ (the five-fold angle equation) and $\sin^2 + \cos^2 = 1$.

• It is irrational; $\frac{\sqrt{10 - 2\sqrt{5}}}{4}$ belongs to $\frac{\pi}{5}$, not $\frac{2\pi}{5}$.

• Typed without brackets, "sin 2pi/5" is read as $\frac{\sin 2\pi}{5} = 0$; use brackets and radian mode.

• In a regular pentagon, $2R\sin\left(\frac{2\pi}{5}\right)$ is the diagonal, and diagonal over side is the golden ratio.

Practice These Before Moving On

1. Evaluate $5\sin\left(\frac{2\pi}{5}\right)$ to four decimal places.

2. Give $\sin\left(\frac{7\pi}{5}\right)$ and $\sin\left(\frac{8\pi}{5}\right)$ using the reference angle $\frac{2\pi}{5}$.

3. Write $\sin\left(\frac{2\pi}{5}\right)$ as a cosine of another first-quadrant angle in radians.

4. An 8 m rope is pegged to the ground at an angle of $\frac{2\pi}{5}$ and pulled straight to the top of a pole. How tall is the pole, to two decimal places?

5. Find $\sin^{-1}\left(\sin\frac{2\pi}{5}\right)$ and $\sin^{-1}\left(\sin\frac{3\pi}{5}\right)$.

Answers:
(1) $4.7553$.
(2) Both are $-0.9511$ (Quadrants III and IV).
(3) $\cos\left(\frac{\pi}{10}\right)$.
(4) $8 \times 0.9510565 \approx 7.61$ m.
(5) Both equal $\frac{2\pi}{5}$.

If question 5 caught you, reread Mistake 3 and check the range of $\sin^{-1}$ before cancelling anything.

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