What Is The Value Of Cos pi/12?
Cos pi/12 has the exact value $\cos\frac{\pi}{12} = \dfrac{\sqrt{6}+\sqrt{2}}{4}$, which is approximately $0.9659$. The angle $\frac{\pi}{12}$ radians is the same as $15^\circ$, because $15^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{12}$.
Both forms describe one point. In radians the angle is $\frac{\pi}{12}$; in degrees it is $15^\circ$. Whenever this article states the angle, keep both pictures in mind, since exams and calculators switch between them freely.
The value is a positive number just below $1$. That already makes sense: a $15^\circ$ angle is small, and the cosine of a small angle stays close to $1$.
How Do You Find Cos pi/12 Step By Step?
The angle $\frac{\pi}{12}$ is not one of the five values you memorise ($0$, $\frac{\pi}{6}$, $\frac{\pi}{4}$, $\frac{\pi}{3}$, $\frac{\pi}{2}$). The trick is to build it from two angles you do know. In degrees, $15^\circ = 45^\circ - 30^\circ$; in radians, $\frac{\pi}{12} = \frac{\pi}{4} - \frac{\pi}{6}$.
That opens the difference identity for cosine:
$$\cos(A - B) = \cos A \cos B + \sin A \sin B$$
Set $A = \frac{\pi}{4}$ and $B = \frac{\pi}{6}$:
$$\cos\frac{\pi}{12} = \cos\left(\frac{\pi}{4} - \frac{\pi}{6}\right) = \cos\frac{\pi}{4}\cos\frac{\pi}{6} + \sin\frac{\pi}{4}\sin\frac{\pi}{6}$$
Now substitute the special-angle values $\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$, $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$, and $\sin\frac{\pi}{6} = \frac{1}{2}$:
$$= \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2}$$
$$= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6}+\sqrt{2}}{4}$$
A quick decimal check: $\sqrt{6} \approx 2.449$ and $\sqrt{2} \approx 1.414$, so the sum is about $3.863$, and dividing by $4$ gives $0.9659$. That matches what a calculator returns for $\cos 15^\circ$.
Example 1: confirm the value is positive using the quadrant.
The angle $\frac{\pi}{12}$ (that is, $15^\circ$) lands in the first quadrant, between $0^\circ$ and $90^\circ$. Under the ASTC rule (All, Sine, Tangent, Cosine), every ratio is positive in the first quadrant, and the reference angle is the angle itself, $15^\circ$. So $\cos\frac{\pi}{12}$ is positive, with no sign to flip.
Final answer: $\cos\dfrac{\pi}{12} = \dfrac{\sqrt{6}+\sqrt{2}}{4} \approx 0.9659$.
Can You Derive Cos pi/12 With The Half-Angle Formula?
Yes, and it is a good cross-check. Since $15^\circ$ is half of $30^\circ$, the half-angle formula for cosine gives a second route to the same answer:
$$\cos\frac{\theta}{2} = \sqrt{\frac{1 + \cos\theta}{2}}$$
Take $\theta = 30^\circ = \frac{\pi}{6}$, so $\frac{\theta}{2} = 15^\circ = \frac{\pi}{12}$. Using $\cos 30^\circ = \frac{\sqrt{3}}{2}$:
$$\cos\frac{\pi}{12} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{3}}{4}} = \frac{\sqrt{2 + \sqrt{3}}}{2}$$
The positive root is correct because the angle is in the first quadrant. This surd looks different from $\frac{\sqrt{6}+\sqrt{2}}{4}$, but the two are equal. Squaring $\frac{\sqrt{6}+\sqrt{2}}{4}$ gives $\frac{8 + 2\sqrt{12}}{16} = \frac{2 + \sqrt{3}}{4}$, the same number under the root above. Both forms equal $0.9659$.
Where Does π/12 Sit On The Unit Circle?
On the unit circle, a point is written as $(\cos\theta, \sin\theta)$, so the cosine of an angle is the $x$-coordinate of the point where the angle's arm meets the circle. For $\frac{\pi}{12}$, that point is:
$$\left(\cos\frac{\pi}{12},\ \sin\frac{\pi}{12}\right) = \left(\frac{\sqrt{6}+\sqrt{2}}{4},\ \frac{\sqrt{6}-\sqrt{2}}{4}\right) \approx (0.9659,\ 0.2588)$$
The arm points barely above the horizontal, so its shadow on the $x$-axis is long (close to $1$) and its height on the $y$-axis is short (about $0.26$). That is the unit-circle reason $\cos 15^\circ$ is large while $\sin 15^\circ$ is small.
The same value also comes from a right triangle: in a right triangle with an angle of $15^\circ$, the cosine is the adjacent side over the hypotenuse. Both pictures give $\frac{\sqrt{6}+\sqrt{2}}{4}$, which is the point of anchoring the value two ways.
How Does Cos pi/12 Compare To The Other Special Angles?
Placing $\frac{\pi}{12}$ beside its neighbours shows the pattern: as the angle grows from $0^\circ$ to $90^\circ$, the cosine falls steadily from $1$ to $0$.
Table: cosine and sine of the first-quadrant special angles, in degrees and radians.
Angle | Radians | Cosine | Sine |
|---|---|---|---|
$0^\circ$ | $0$ | $1$ | $0$ |
$\frac{\pi}{12}$ | $\frac{\sqrt{6}+\sqrt{2}}{4} \approx 0.9659$ | ||
$\frac{\pi}{6}$ | $\frac{\sqrt{3}}{2} \approx 0.8660$ | $\frac{1}{2} = 0.5$ | |
$\frac{\pi}{4}$ | $\frac{\sqrt{2}}{2} \approx 0.7071$ | ||
$\frac{\pi}{3}$ | $\frac{1}{2} = 0.5$ | $\frac{\sqrt{3}}{2} \approx 0.8660$ | |
$75^\circ$ | $\frac{5\pi}{12}$ | $\frac{\sqrt{6}-\sqrt{2}}{4} \approx 0.2588$ | |
$\frac{\pi}{2}$ | $0$ | $1$ |
Read across the $15^\circ$ and $75^\circ$ rows and you see a mirror: $\cos 15^\circ$ and $\sin 75^\circ$ are the same number, and $\sin 15^\circ$ and $\cos 75^\circ$ match too. That mirror is the cofunction rule, and it is the next idea.
Why Is Cos pi/12 Equal To (√6 + √2)/4?
The exact value is not a coincidence of the formula. Three ideas make it inevitable.
The angle is built, not memorised. Because $15^\circ = 45^\circ - 30^\circ$, its cosine is forced by the cosines and sines of $45^\circ$ and $30^\circ$, which are fixed surds. Combine fixed surds and you get another fixed surd.
The quadrant fixes the sign. The angle lies in the first quadrant, where cosine is positive, so the answer must be a positive number, ruling out the negative root in the half-angle method.
The size is believable. A $15^\circ$ arm barely rises above the horizontal, so its $x$-coordinate stays near $1$, and $0.9659$ is exactly that.
There is also a cofunction identity at work: $\cos\theta = \sin(90^\circ - \theta)$. Setting $\theta = 15^\circ$ gives $\cos 15^\circ = \sin 75^\circ$, which is why the same surd appears in both the cosine of $15^\circ$ and the sine of $75^\circ$.
Who Discovered The Formula Behind Cos pi/12?
Long before anyone wrote $\cos(A - B)$, astronomers needed the length of a chord across a circle to predict where the Moon and planets would be. Those chord tables are the ancestors of every sine and cosine value we use, including this one.
Two later mathematicians carried the idea forward:
Claudius Ptolemy (c. 100 – c. 170 CE, Roman Egypt) collected and extended these chord tables in the Almagest, and his chord-of-a-difference rule is, in modern language, the cosine and sine difference identities used above.
Aryabhata (476 – 550 CE, India) tabulated the half-chord, the jya, which became our word "sine" through Arabic and Latin, and pushed these tables toward the functions we use now.
Where Is Cos pi/12 Used In The Real World?
Angles of about $15^\circ$, and the cosines that measure them, appear across many fields.
Surveying and construction: a shallow roof pitch or a graded slope is set to a small angle, and cosine converts that angle into the horizontal run a builder must mark out.
Optics and light: when a beam strikes a surface at a small angle, the cosine of that angle scales how much light lands per unit area, which matters for solar panels and lenses.
Signal processing: alternating currents and sound waves are sums of cosines, and a phase shift of a small angle changes each term by its cosine.
Navigation and astronomy: the same chord-and-angle reasoning Hipparchus used still converts an observed angle into a distance or a position.
One surd, $\frac{\sqrt{6}+\sqrt{2}}{4}$, quietly sits inside rooftops, lenses, audio, and star charts. That reach is why exact trigonometric values are worth knowing, not just approximating.
What Are The Most Common Mistakes With Cos pi/12?
These four errors account for most lost marks on angles like this one. They were the recurring confusions in the unit-circle and identity practice threads surfaced during research.
Leaving the calculator in radian mode.
Where it slips in:
A student types "cos(15)" expecting $15^\circ$ but the calculator is set to radians, so it returns $\cos(15\text{ rad}) \approx -0.7597$.
Don't do this:
Do not trust a cosine reading without checking the angle mode first.
The correct way:
Set the calculator to degree mode for $\cos 15^\circ$, or enter $\cos(\pi/12)$ in radian mode. Both give about $0.9659$.
Dropping the second term of the difference formula.
Where it slips in:
A student writes $\cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ$ and forgets the $+\sin 45^\circ \sin 30^\circ$ piece.
Don't do this:
Do not treat $\cos(A - B)$ as $\cos A \cos B$. Cosine does not distribute over subtraction.
The correct way:
Use the full identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$, keeping both products and the plus sign.
Confusing cos 15° with sin 15°.
Where it slips in:
A student reads the small number $0.2588$ off the unit circle and calls it the cosine, when $0.2588$ is the sine (the height) and the cosine is the long horizontal $0.9659$.
Don't do this:
Do not assign the $y$-coordinate to cosine. Cosine is the $x$-coordinate.
The correct way:
Match cosine to the horizontal reach and sine to the vertical height, so $\cos 15^\circ \approx 0.9659$ and $\sin 15^\circ \approx 0.2588$, and remember $\cos 15^\circ = \sin 75^\circ$.
Calling 0.966 the exact value.
Where it slips in:
A student rounds to $0.966$ and writes that as the final exact answer on a proof or an exact-value question.
Don't do this:
Do not present a decimal as exact when the question asks for an exact value.
The correct way:
Give the surd $\frac{\sqrt{6}+\sqrt{2}}{4}$ as the exact value, and offer $0.9659$ only as its decimal approximation.
Practice Problems On Cos pi/12
Work each one, then check the answer beside it.
Convert $\frac{\pi}{12}$ radians to degrees.
(Answer: $\frac{\pi}{12} \times \frac{180^\circ}{\pi} = 15^\circ$.)Use $\cos(45^\circ - 30^\circ)$ to state $\cos 15^\circ$.
(Answer: $\frac{\sqrt{6}+\sqrt{2}}{4}$.)Is $\cos\frac{\pi}{12}$ positive or negative, and why?
(Answer: positive, because $\frac{\pi}{12}$ is in the first quadrant.)Given $\cos 15^\circ = \frac{\sqrt{6}+\sqrt{2}}{4}$, write $\sin 75^\circ$ without new work.
(Answer: $\frac{\sqrt{6}+\sqrt{2}}{4}$, by the cofunction rule.)Derive $\cos\frac{\pi}{12}$ from the half-angle formula on $30^\circ$.
(Answer: $\frac{\sqrt{2+\sqrt{3}}}{2}$, which equals $\frac{\sqrt{6}+\sqrt{2}}{4}$.)Give $\cos\frac{\pi}{12}$ as a decimal to four places.
(Answer: $0.9659$.
Where Should You Go Next After Cos pi/12?
Cos pi/12 is one worked example of a much larger toolkit, and a few natural doors open from here.
Cos A minus B formula. The identity that made this value possible, with more worked cases.
Cos 15 degrees. The degree-form twin of this page, if you prefer working in degrees.
Trigonometric table. Every standard angle in one place, so you never have to rebuild a value under exam pressure.
If your child is building these foundations, a live Bhanzu trainer teaches exact values starting from the "why" (how the angle is built and where it sits on the unit circle) in the Bhanzu trigonometry program.
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