Cos 80 Degrees: Value 0.1736, Methods and Examples

What Does Cos 80 Degrees Mean?

In a right triangle, the cosine of an angle is the side adjacent to it divided by the hypotenuse. For an $80^\circ$ angle, the adjacent side is short compared with the hypotenuse, so the ratio is small: about $0.1736$.

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 angle's radius reaches. At $80^\circ$ that point is about $(0.1736, 0.9848)$, so $\cos 80^\circ \approx 0.1736$.

A quadrant is one of the four regions the axes cut the plane into, numbered I to IV anticlockwise from the positive $x$-axis. An angle of $80^\circ$ lies in Quadrant I, where both coordinates are positive, so $\cos 80^\circ$ is positive.

The reference angle of an angle is the acute angle its radius makes with the $x$-axis. For $80^\circ$ the reference angle is $80^\circ$ itself, which is why no sign change or angle shift is needed here.

Standard-Angle Cosine Reference Table

The angles between $0^\circ$ and $90^\circ$ split into two groups. Some have exact surd values; the rest, $80^\circ$ included, are given to four decimal places.

Read down the last column and cosine falls from $1$ to $0$. The drop between $80^\circ$ and $90^\circ$ is steep, which is why $\cos 80^\circ$ is already small even though $80^\circ$ is not yet a right angle.

The full set of values for other angles sits in the trigonometric table.

Where Does Cos 80 Degrees Show Up, And Why Does It Matter?

Cos 80 degrees measures how much of a nearly vertical length points sideways. A 6 m ladder leaning at $80^\circ$ to the ground has its foot only $6 \times 0.1736 \approx 1.04$ m from the wall, a standard heights and distances set-up.

On a globe, a circle of latitude at $80^\circ$ N has a circumference of $\cos 80^\circ \approx 17.4$ of the equator's (taking Earth as a sphere). A force pulling at $80^\circ$ to the horizontal also passes only about $17.4$ of its size along the ground, so a $50$ N pull gives roughly $8.68$ N of horizontal push.

How Do You Find The Value Of Cos 80 Degrees?

The value comes from four routes, and they agree to every digit a calculator shows. Pick the one that matches what the question gives you.

Method 1: The Cofunction Identity

The cofunction identity says $\cos\theta = \sin(90^\circ - \theta)$. Apply it with $\theta = 80^\circ$:

$$\cos 80^\circ = \sin(90^\circ - 80^\circ)$$

$$\cos 80^\circ = \sin 10^\circ \approx 0.1736$$

In the right triangle, the side adjacent to $80^\circ$ is the side opposite the other acute angle, $10^\circ$. The two ratios are the same fraction read from different corners; the cofunction identities page covers all six pairs.

Method 2: A Short Series (No Calculator)

Convert $10^\circ$ to radians first: $10^\circ = \frac{\pi}{18} \approx 0.174533$. For small angles in radians, $\sin x \approx x - \frac{x^3}{6}$, so:

$$\sin 10^\circ \approx 0.174533 - \frac{0.174533^3}{6}$$

$$\sin 10^\circ \approx 0.174533 - 0.000886 = 0.173647$$

That matches the true value $0.1736482$ to within about $0.0000013$.

Using only the first term, $\sin 10^\circ \approx 0.1745$, overshoots by about $0.0009$ (roughly $0.5$), which is fine for an estimate because $10^\circ$ is a small angle. The same shortcut fails badly on $80^\circ$ itself: $1 - \frac{x^2}{2}$ with $x = \frac{4\pi}{9}$ gives $0.0252$, nowhere near $0.1736$.

Method 3: Identities That Reach $80^\circ$

Several identities rebuild $\cos 80^\circ$ from other angles. Each gives the same $0.1736$, and each is a useful check.

• Double angle: $\cos 80^\circ = 2\cos^2 40^\circ - 1 = 2(0.76604)^2 - 1 \approx 0.1736$

• Difference of cosines: $\cos 80^\circ = \cos 20^\circ - \cos 40^\circ \approx 0.9397 - 0.7660 = 0.1737$ (the last digit drifts only because the inputs were rounded)

• Angle sum: $\cos 80^\circ = \cos(60^\circ + 20^\circ) = \frac{1}{2}\cos 20^\circ - \frac{\sqrt{3}}{2}\sin 20^\circ$

The values of cos 20 degrees and cos 40 degrees have the same problem as $80^\circ$: none of them has a real surd form. So these identities connect the values to each other; they do not turn any of them into a surd.

Method 4: A Calculator

Set the calculator to degree mode and enter $\cos(80)$. The screen shows $0.1736481777$, and rounding to four decimal places gives $0.1736$.

Is There An Exact Value Of Cos 80 Degrees?

Yes, but not in the form most people mean. The value is exact as $\sin 10^\circ$, and it is pinned down exactly by a cubic equation, but it cannot be written with whole numbers and real square or cube roots alone.

The cubic comes from the triple-angle identity $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$, the formula behind cos 3x. Put $\theta = 80^\circ$, so $3\theta = 240^\circ$ and $\cos 240^\circ = -\frac{1}{2}$:

$$4c^3 - 3c = -\frac{1}{2}$$

$$8c^3 - 6c + 1 = 0, \quad c = \cos 80^\circ$$

This cubic has three real roots: $\cos 80^\circ \approx 0.1736$, $\cos 40^\circ \approx 0.7660$ and $\cos 160^\circ \approx -0.9397$. None of the rational candidates ($\pm 1, \pm\frac{1}{2}, \pm\frac{1}{4}, \pm\frac{1}{8}$) is a root, so the cubic cannot be factored over the rationals.

A cubic like this, with three real roots and no rational factor, is the case mathematicians call casus irreducibilis. The cubic formula still solves it, but only by taking cube roots of complex numbers along the way; the real answer cannot be reached with real radicals.

The same fact explains why $80^\circ$ cannot be drawn exactly with compass and straightedge. Only whole-degree angles that are multiples of $3^\circ$ have real square-root forms, and the French mathematician Pierre Wantzel proved in 1837 that a general angle cannot be trisected with those tools.

The honest answer surprises most people, teachers included: the "exact value" of $\cos 80^\circ$ is a well-defined number, but its tidiest exact names are $\sin 10^\circ$, $\cos\frac{4\pi}{9}$ and $\frac{1}{2}\left(e^{i4\pi/9} + e^{-i4\pi/9}\right)$, not a surd.

Can Cos 80 Degrees Be Written As A Fraction?

No. The value is irrational, because a rational root of $8c^3 - 6c + 1 = 0$ would have to be one of the candidates above, and none works. A fraction such as $\frac{217}{1250} = 0.1736$ is only an approximation to four decimal places.

How Do You Write Cos 80 Degrees In Terms Of Other Trig Functions?

The same number can be named through every other trigonometric function. All of these give $0.1736$ when evaluated.

Examples Of Cos 80 Degrees

Example 1

Evaluate $\dfrac{\cos 80^\circ}{\sin 10^\circ}$.

By the cofunction identity, $\cos 80^\circ = \sin 10^\circ$.

$$\frac{\cos 80^\circ}{\sin 10^\circ} = \frac{\sin 10^\circ}{\sin 10^\circ} = 1$$

No calculator is needed; the numerator and denominator are the same number.

Example 2

Find the exact value of $\cos 80^\circ \cos 20^\circ + \sin 80^\circ \sin 20^\circ$.

Wrong path first. The product-of-cosines pattern suggests the angle-sum formula, so a student writes $\cos(80^\circ + 20^\circ) = \cos 100^\circ \approx -0.1736$.

That cannot be right. All four values $\cos 80^\circ$, $\cos 20^\circ$, $\sin 80^\circ$ and $\sin 20^\circ$ are positive, so the sum of their two products must be positive.

The fix is the sign. The angle-sum formula has a minus, $\cos(A + B) = \cos A\cos B - \sin A\sin B$, while a plus belongs to the cos(A − B) formula:

$$\cos 80^\circ \cos 20^\circ + \sin 80^\circ \sin 20^\circ = \cos(80^\circ - 20^\circ)$$

$$= \cos 60^\circ = \frac{1}{2}$$

Check: $0.1736 \times 0.9397 + 0.9848 \times 0.3420 \approx 0.1631 + 0.3368 = 0.4999$, which is $0.5$ once rounding in the inputs is allowed for.

Example 3

True or false: the value of $\sin 80^\circ - \cos 80^\circ$ is negative.

$$\sin 80^\circ - \cos 80^\circ \approx 0.9848 - 0.1736 = 0.8112$$

The statement is false. Between $45^\circ$ and $90^\circ$, sine is always larger than cosine, so the difference is positive without any calculation.

Example 4

A 6 m ladder leans against a wall and makes an angle of $80^\circ$ with the ground. How far is its foot from the wall?

The ladder is the hypotenuse and the ground distance is the side adjacent to $80^\circ$.

$$\text{distance} = 6 \times \cos 80^\circ$$

$$\text{distance} \approx 6 \times 0.1736 = 1.04 \text{ m}$$

The usual first move here is to multiply by $\sin 80^\circ$ instead; that gives $5.91$ m, which is the height the ladder reaches on the wall, not the gap at the bottom.

Example 5

Show that $\cos 20^\circ - \cos 40^\circ = \cos 80^\circ$.

Use the sum-to-product formulas: $\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$.

$$\cos 20^\circ - \cos 40^\circ = -2\sin 30^\circ \sin(-10^\circ)$$

$$= -2 \times \frac{1}{2} \times (-\sin 10^\circ)$$

$$= \sin 10^\circ = \cos 80^\circ$$

Example 6

Prove that $\cos 20^\circ \cos 40^\circ \cos 80^\circ = \dfrac{1}{8}$.

Let $P = \cos 20^\circ \cos 40^\circ \cos 80^\circ$ and multiply both sides by $8\sin 20^\circ$. Each step uses $2\sin\theta\cos\theta = \sin 2\theta$:

$$8\sin 20^\circ \times P = 4(2\sin 20^\circ \cos 20^\circ)\cos 40^\circ \cos 80^\circ$$

$$= 2(2\sin 40^\circ \cos 40^\circ)\cos 80^\circ$$

$$= 2\sin 80^\circ \cos 80^\circ = \sin 160^\circ$$

Since $\sin 160^\circ = \sin(180^\circ - 20^\circ) = \sin 20^\circ$:

$$8\sin 20^\circ \times P = \sin 20^\circ, \quad \text{so } P = \frac{1}{8}$$

Check: $0.9397 \times 0.7660 \times 0.1736 \approx 0.125$.

Where Do Students Trip Up On Cos 80 Degrees?

Mistake 1: A Calculator In Radian Mode

Where it slips in: Typing $\cos(80)$ on a calculator or phone app that is set to radians.

Don't do this: Copying the screen value $-0.1104$. That is the cosine of $80$ radians (about $263.7^\circ$ after full turns are removed, a Quadrant III angle).

The correct way: Switch to degree mode, or enter $\cos\left(\frac{4\pi}{9}\right)$ in radian mode, and expect a positive answer near $0.17$. The student who rushes and copies $-0.1104$ has skipped the one check that catches it: every angle between $0^\circ$ and $90^\circ$ has a positive cosine.

Mistake 2: Swapping Cos 80 And Sin 80

Where it slips in: Problems that need both values, such as Example 4 or the identity table.

Don't do this: Writing $\cos 80^\circ \approx 0.9848$. That number is $\sin 80^\circ$ (and also $\cos 10^\circ$).

The correct way: Anchor on the shape of the curve: cosine starts at $1$ and falls to $0$ at $90^\circ$, so at $80^\circ$ it must be small. The two numbers that get swapped are $0.9848$ and $0.1736$ because both feel like "the value for 80", and the picture of the unit circle settles it in one glance.

Mistake 3: Inventing A Surd

Where it slips in: Students who memorised the exact values for $30^\circ$, $45^\circ$ and $60^\circ$ expect every angle to have one.

Don't do this: Forcing a form such as $\cos 80^\circ = \frac{\sqrt{3} - 1}{4}$. Check it: that expression equals $0.1830$, not $0.1736$.

The correct way: State the value as $\sin 10^\circ \approx 0.1736$, and test any proposed surd against the decimal before trusting it. The memorizer is usually the one caught here, because a pattern that works for special angles does not extend to angles that are not multiples of $3^\circ$.

Where Can You Get Help With Trigonometry Values Like Cos 80 Degrees?

Non-special angles such as $80^\circ$ are mostly a calculator and identity skill, and the identities in this page are where students most often want a second explanation. A trigonometry tutor, a high school math tutor or live math classes online can walk through the cofunction, sum-to-product and triple-angle steps with you.

Key Takeaways

• Cos 80 degrees is about $0.1736$ and equals $\sin 10^\circ$ exactly, by the cofunction identity.

• In radians, $80^\circ = \frac{4\pi}{9}$, and the angle sits in Quadrant I, so the value is positive.

• No real surd gives $\cos 80^\circ$; it is a root of $8x^3 - 6x + 1 = 0$, a casus irreducibilis cubic.

• Identities such as $\cos 20^\circ - \cos 40^\circ = \cos 80^\circ$ and $\cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{8}$ use the value without needing its decimal.

• A negative calculator answer ($-0.1104$) means radian mode, not a new value.

Practice These Before Moving On

1. Evaluate $3\cos 80^\circ$ to four decimal places.

2. Find $\cos 440^\circ$ and $\cos(-80^\circ)$ without a calculator, using $\cos 80^\circ \approx 0.1736$.

3. Show that $\cos 40^\circ + \cos 80^\circ + \cos 160^\circ + \cos 240^\circ = -\frac{1}{2}$.

4. A right triangle has a hypotenuse of $12$ cm and an $80^\circ$ angle. Find the side adjacent to that angle.

5. Which is larger, $\cos 80^\circ$ or $\tan 10^\circ$?

Answers:

1. $3 \times 0.1736481777 \approx 0.5209$.

2. Both equal $0.1736$: $440^\circ = 360^\circ + 80^\circ$, and cosine is even.

3. $\cos 40^\circ + \cos 80^\circ = 2\cos 60^\circ \cos 20^\circ = \cos 20^\circ$, and $\cos 160^\circ = -\cos 20^\circ$, so the total is $\cos 240^\circ = -\frac{1}{2}$.

4. $12 \times \cos 80^\circ \approx 2.08$ cm.

5. $\tan 10^\circ \approx 0.1763$ is larger than $\cos 80^\circ \approx 0.1736$.

If question 3 stalls, go back to Example 5 and the sum-to-product step.

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