The value of tan 15 degrees is approximately 0.2679, and it can also be written in fraction form as:

tan 15° = (√3 − 1) / (√3 + 1)

When measured in radians, the value of tan 15 radians is about -0.85599.

In trigonometry, the tangent of an angle is one of the main six trigonometric functions. It is defined as the ratio of the opposite side to the adjacent side in a right-angled triangle.

Using the unit circle, tangent is defined as:

tan(θ) = y / x
or
tan(θ) = sin(θ) / cos(θ)

In this topic, we will explore how to calculate tan 15°, how it behaves in degrees and radians, and understand it better with simple examples.

Value Of Tan 15

The domain of a function is the set of all input values for which the function works. When we talk about the value of tan 15°, we are using 15 as the input. To find this value, 15 must fall within the domain of the tangent (tan) function. The tangent function is defined for all angles between –90° and 90° (excluding exactly –90° and 90°). Since 15° is between –90° and 90°, the value of tan 15° exists.

The range of a function is the set of all possible outputs (results) you can get. Since the tangent function is periodic and continues repeating, it doesn't have a highest or lowest value overall. But within one period, the outputs of tan x lie between –π/2 and π/2 (in radians). So, the value of tan 15° will fall within this range.

In short, tan 15° exists and lies within the allowed output values of the tangent function.

Value of tan 15 in Fraction

The value of tan 15 in fraction form is what we refer to when we express the value as a numerator and denominator. The value of tan 15 degrees is \(\frac{\sqrt{3} – 1}{\sqrt{3} + 1}\) in fraction form. We will see how this value was determined in the following sections. Continue to read more. Also, do check out these articles related to fractions, like and unlike fractions, proper fractions, mixed fractions and Equivalent fractions.

Value of tan 15 Degrees in Decimals

The value of tan 15 in fractions has already been shown. We are aware that we will obtain a number or decimal number if we divide the numerator of the fraction by its denominator. We’ll see the same thing here. We can obtain a decimal number by dividing the fractional denominator of the value of tan 15 by its numerator. Let’s have a look at the process step by step.

Value of tan 15 in fraction form = \(\frac{\sqrt{3} – 1}{\sqrt{3} + 1}\)

We will substitute the values of \(\sqrt{3}\) in the above fraction. We know that \(\sqrt{3}=1.732\).

Thus, the equation becomes.

\(= \frac{1.732 – 1}{1.732 + 1}\)

\(= \frac{0.732}{2.732}\)

\(= 0.267933557833\)

Thus, the value of sin 15 in decimal form is \(0.267933557833 ≈ 0.2679\)

Also, read more about decimals, Decimals in Daily Life, and decimal fractions here.

Value of tan15 in Radians

The standard unit of angular measurement used in many branches of mathematics is the radian, indicated by the symbol rad. It is the unit of angle in the International System of Units. In some calculations, the angles are required in radians because all other quantities are measured in corresponding units. Degrees can be converted into radians. The formula for converting degrees into radians is given as,

Radians = Degrees × \(\frac{{\pi}}{180^{\circ}}\).

Thus, in order to calculate the value of tan 15 in radians we need to multiply it by the fraction of \(\frac{{\pi}}{180^{\circ}}\). We will take the decimal value of tan 15 for the purpose of ease of calculation. The resulting value will be a decimal too.

Value of tan 15 in radians = value of tan 15 in decimals × \(\frac{{\pi}}{180^{\circ}}\).

Value of tan 15 in radians = \(0.2679\times\frac{{\pi}}{180^{\circ}}\).

Value of tan 15 in radians = \(0.6509\).

Thus, the value of tan 15 in radians is \(0.6509\).

To find the values of tan 15 degrees we make use of previously known values of sine function like tan 30 We then apply the identities of sin function to derive the value of tan 15.

Value of Tan 15 Degree with Respect to Sin and Cos Function

We need to first find the value of sin 15 and cos 15.

\((sin\frac{x}{2} + cos\frac{x}{2})^2 = sin^2\frac{x}{2} + cos^2\frac{x}{2} + 2sin\frac{x}{2} cos\frac{x}{2} = 1 + sinx\)

For all values of the angle x.

Therefore, taking square root on both the sides gives us,

\(sin\frac{x}{2} + cos\frac{x}{2} = \pm\sqrt{(1 + sinx)}\)

Now, let \(x = 30^{\circ}\) then, \(\frac{x}{2} = \frac{30^{\circ}}{2} = 15^{\circ}\) and from the above equation we get,

\(sin 15^{\circ} + cos 15^{\circ} = \pm\sqrt{(1 + sin30^{\circ})}\) …………… (i)

Similarly, for all values of the angle x we know that,

\((sin\frac{x}{2} – cos \frac{x}{2})^2 = sin^2\frac{x}{2} + cos^2\frac{x}{2} – 2 sin\frac{x}{2} cos\frac{x}{2} = 1 – sinx\)

Therefore, taking square root on both the sides gives us,

\(sin\frac{x}{2 – cos\frac{x}{2 = \pm\sqrt{(1 – sinx)}\)

Now, let \(x = 30^{\circ}\) then, \(\frac{x}{2} = \frac{30^{\circ}}{2} = 15^{\circ}\) and from the above equation we get,

\(sin 15^{\circ} – cos 15^{\circ} = \pm\sqrt{(1 – sin30^{\circ})}\) …………… (ii)

\(sin 15^{\circ} – cos 15^{\circ} = \sqrt{2}(\frac{1}{\sqrt{2}}sin 15˚ – \frac{1}{\sqrt{2}}cos 15˚)\)

Or, \(sin 15^{\circ} – cos 15^{\circ} = \sqrt{2}(cos 45^{\circ} sin 15˚ – sin 45^{\circ} cos 15^{\circ})\)

Or, \(sin 15^{\circ} – cos 15^{\circ} = \sqrt{2}sin (15˚ – 45˚)\)

Or, \(sin 15^{\circ} – cos 15^{\circ} = \sqrt{2}sin (- 30˚)\)

Or, \(sin 15^{\circ} – cos 15^{\circ} = -\sqrt{2}sin 30^{\circ}\)

Or, \(sin 15^{\circ} – cos 15^{\circ} = -\sqrt{2}∙\frac{1}{2}\)

Or, \(sin 15^{\circ} – cos 15^{\circ} = – \frac{\sqrt{2}}{2}\)

Thus, \(sin 15^{\circ} – cos 15^{\circ} < 0\)

Therefore, from (i) we get,

\(sin 15^{\circ} – cos 15^{\circ}= -\sqrt{(1 – sin 30^{\circ})}\) ………. (iii)

Now, adding (ii) and (iii) we get,

\(2 sin 15^{\circ} = \sqrt{1 + \frac{1}{2}} – \sqrt{1 – \frac{1}{2}}\)

\(2 sin 15^{\circ} = \frac{\sqrt{3}- 1}{\sqrt{2}}\)

\(\sin(15^{\circ}) = \frac{\sqrt3-1}{2\sqrt2}\)

Similarly,

Now, subtracting (ii) and (iii) we get,

\(2 cos 15^{\circ} = \sqrt{1 + \frac{1}{2}} + \sqrt{1 – \frac{1}{2}}\)

\(2 cos 15^{\circ} = \frac{\sqrt{3}+ 1}{\sqrt{2}}\)

Therefore, \(cos 15^{\circ} = \frac{\sqrt{3} + 1}{2\sqrt{2}}\)

Now, \(tan 15^{\circ} = \frac{sin 15^{\circ}}{cos 15^{\circ}}\)

\(= \frac{\frac{\sqrt{3} – 1}{2\sqrt{2}}}{\frac{\sqrt{3} + 1}{2\sqrt{2}}}\)

\(= \frac{\sqrt{3} – 1}{\sqrt{3} + 1}\)

Thus, \(tan 15^{\circ} = \frac{\sqrt{3} – 1}{\sqrt{3} + 1}\)

Tan \(15^{\circ}\) in Terms of Trigonometric Functions

There is a second alternative method to find the value of tan 15. Split 15 into two angles where the values of the six trigonometric functions are known.

tan(45−30)

Apply the difference of angles identity, given by:

\(tan(A-B)=\frac{tanA-tanB}{1+tanAtanB}\)

\(tan(45−30)=\frac{tan(45)-tan(30)}{1+tan(45)tan(30)}\)

\(tan(45−30)=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}\)

\(tan(45−30)=\frac{{\sqrt{3}}-1}{{\sqrt{3}}+1}\)

Tan 15 Degrees Using Unit Circle

Using the unit circle, we can calculate the value of tan 15 in radians and in degrees. As was already noted, the radius of a circle serves as the basis for the measurement of angles in radians. We are going to need a ruler, compass and protractor for this exercise.

To find out the value of tan 15 using unit circle, follow the steps below:

• Draw a circle of radius of 1 unit i.e unit magnitude.
• Draw the two axes and four quadrants as shown in the figure. Since 15 degrees lies in the first quadrant, we will only work in the first quadrant.
• Draw an arm of length ‘r’ in \(15^{\circ}\) anticlockwise angle with the positive x-axis using a protractor.
• The sin of 15 degrees equals the y-coordinate whereas cos of 5 degrees equals the x-coordinates of the point of intersection of unit circle and r. Whenever we plot an angle, its cosine value lies on the horizontal x-axis whereas as its sine value lies on the y-axis. Thus, \(x = 0.9659 is actually the value of cos 15.
• If we measure it, we will get the coordinates of the end point of our arm on the circle as \(\)(0.9659, 0.2588)\). Hence the value of \(sin 15^{\circ} = y = 0.2588\) and value of \(cos 15^{\circ} = x = 0.9659\).
• We know that, \(tan 15^{\circ} = \frac{sin 15^{\circ}}{cos 15^{\circ}}= 0.2679\)
• Now, we can convert the radians value to degrees value as we have already seen earlier.

The reason why we take the y-coordinate of our arm in this case lies in the concept of resolution of angle.

Tan Value Table

We are aware of the trigonometric ratio values for a few common angles, including \(0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}\), and \(90^{\circ}\). We may also need to employ the values of trigonometric ratios for nonstandard angles, such as \(sin 62^{\circ}, sin 47^{\circ} 45′, cos 83^{\circ}, cos 41^{\circ} 44′\), and \(tan 39^{\circ}\), when using the notion of trigonometric ratios to solve problems involving heights and distances. Trigonometric tables contain the approximations, accurate to four decimal places, of the natural sines, cosines, and tangents of all angles between \(0^{\circ}\) and \(90^{\circ}\).

Trigonometric tables consist of three parts.

(i) There is a column with the numbers 0 to 90 on the far left (in degrees).

(ii) The degree column is followed by ten columns with the headings

\(0′, 6′, 12′, 18′, 24′, 30′, 36′, 42′, 48′\) and \(54′\) or \(0.0^{\circ}, 0.1^{\circ}, 0.2^{\circ}, 0.3^{\circ}, 0.4^{\circ}, 0.5^{\circ}, 0.6^{\circ}, 0.7^{\circ}, 0.8^{\circ}\) and \(0.9^{\circ}\)

(iii) After that, on the right, there are five columns known as mean difference columns with the headings \(1′, 2′, 3′, 4′\) and \(5′\).

\(60′ = 60\text{ minutes } = 1^{\circ}\).

Tan value table is given below:

Tan Function Periodicity

A function f is periodic if there is a positive number \(p such that \(\)f(x + p) = f(x)\) for all \(x in the domain of \(\)f\). The smallest such \(p, if it exists, is called the period of \(\)f\).

For any real number x and any integer n,

\(tan x = tan (x + n·\pi) = tan (x + n·180^\circ)\) the sine and cosine functions are periodic functions with period \(360^\circ\) or \(2\pi\) radians, while the tangent function is a periodic function with period \(180^\circ\) or \(\pi\) radians.

Properties of the Value of Tan 15°

1. Exact Value (in Surd Form):
   The exact value of tan 15° is:
   tan 15° = 2 – √3
2. Approximate Decimal Value:
   The approximate value of tan 15°, rounded to four decimal places, is:
   tan 15° ≈ 0.2679
3. Irrational Number:
   Since √3 is an irrational number, the value 2 – √3 is also irrational.
   Therefore, tan 15° is an irrational number.
4. Lies in the First Quadrant:
   The angle 15° lies in the first quadrant, where all trigonometric functions are positive.
   So, tan 15° > 0.
5. Derived Using Angle Difference Formula:
   We can find tan 15° using the identity:
   tan(45° − 30°) = (tan 45° − tan 30°) / (1 + tan 45° × tan 30°)
   Substitute the values:
   tan 45° = 1 and tan 30° = 1/√3
   So:
   tan 15° = (1 − 1/√3) / (1 + 1 × 1/√3) = (2 − √3)
6. Graph Behavior Near 15°:
   On the graph of y = tan x, the value of tan 15° is close to the x-axis and increases as the angle moves toward 45°.

Applications of Value of Tan 15°

1. Trigonometry Problems:

• tan 15° is used in solving problems involving right-angled triangles where one of the angles is 15°.
• Helps find the height or distance using trigonometric ratios.

2. Construction and Architecture:

• In engineering designs, slopes or angles of 15° are common in ramps, roofs, and roads.
• tan 15° helps calculate rise and run of slopes.

3. Physics and Mechanics:

• tan 15° is useful when calculating forces, inclinations, or angles of motion on inclined planes at 15°.

4. Navigation and Surveying:

• In land surveying or navigation, if an angle of 15° is involved, tan 15° helps determine distances or elevations.

5. Angle Difference Formula:

• tan 15° = tan(45° – 30°) is used to derive or verify trigonometric identities and simplify complex expressions.

6. Mathematics Education:

• tan 15° serves as a standard example in teaching trigonometric formulas and how to handle irrational numbers.

Solved Examples

Now that we have learned all about tan15. Here are some solved examples.

Example 1: If \(tan 15^{\circ} = 2 – \sqrt{3}\), the value of \(tan 15^{\circ} cot 75^{\circ}+ tan 75^{\circ}cot75^{\circ}\) is

Solution:

\(\Large{tan15 ^{\circ} cot75 ^{\circ} +tan75 ^{\circ} cot15 ^{\circ}}\)

\(=\Large{tan15 ^{\circ} .cot(90 ^{\circ} -15 ^{\circ} )+tan(90 ^{\circ} -15 ^{\circ} ).cot15 ^{\circ}}\)

\(=\Large{tan^{2}15 ^{\circ} +cot^{2}15 ^{\circ} =tan^{2}15 ^{\circ} +cot^{2}15 ^{\circ}}\)

We know that,

\(\Large{cot(90 ^{\circ} – \theta )=tan\theta}\)

\(\Large {an(90 ^{\circ} – \theta )=cot\theta}\)

Put the value of \(tan15 ^{\circ} \)

\(\Large cot15 ^{\circ} =2+\sqrt{3}\)

Now let’s substitute this in the original equation

\(\Large tan^{2}15 ^{\circ} +cot^{2}15 ^{\circ}\)

\(=\Large{(2-\sqrt{3})^{2}+(2+\sqrt{3})^{2}}\)

\(=\Large{4+3-4\sqrt{3}+4+3+4\sqrt{3}}\)

\(=14\)

Example 2: Show that \(\tan15^{\circ}+\cot15^{\circ}=4\)

Solution:

\(tanx^{\circ}+cotx^{\circ}\)

\(=tanx^{\circ} + \frac{1}{tanx^{\circ}}\)

\(= \frac{tan^2x^{\circ}+1}{tanx^{\circ}}\)

\(= \frac{sec^2x^{\circ})}{tanx^{\circ}}\)

\(= \frac{1}{cos^2x^{\circ}}\frac{cosx^{\circ}}{sinx^{\circ}}\)

\(= \frac{1}{cosx^{\circ}sinx^{\circ}}\)

\(= \frac{2}{2cosx^{\circ}sinx^{\circ}}\)

\(tanx^{\circ}+cotx^{\circ}= \frac{2}{sin2x^{\circ}}\)

Put \(x=15\)

\(tan15^{\circ}+cot15^{\circ}= \frac{2}{sin(2\times15)^{\circ}}\)

\(tan15^{\circ}+cot15^{\circ}= \frac{2}{sin30^{\circ}}\)

\(tan15^{\circ}+cot15^{\circ}= \frac{2}{\frac{1}{2}}\)

\(tan15^{\circ}+cot15^{\circ}= 4\)

Thus, LHS = RHS.

Example 3: If the roots of the quadratic equation \(x^2 + px + q = 0\) are \(tan30^{\circ}\) and \(tan15^{\circ}\), respectively then the value of \(2 + q − p\) is ?

Solution:

If \(tan30^{\circ}\) and \(tan15^{\circ}\) are the roots of the quadratic equation, \(x^2 + px + q = 0 then the following equations are true.

\(tan 30^{\circ} ⋅ tan 15^{\circ} = q\)

\(tan45 = \frac{tan 30^{\circ}+tan15^{\circ}}{1+tan 30^{\circ}tan15^{\circ}}\)

\(1 = \frac{-p}{1+q}\)

Thus,

\(1 + q = -p\)

\( q − p = 1\)

\(2 + q − p = 3\)

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