The value of sin 15 degrees is \(\frac{\sqrt{3} – 1}{2\sqrt{2}}\) or 0.25884. On the other hand, the value of sin 15 in radians is 0.6509. The area of mathematics known as trigonometry examines how the angles and sides of a right-angled triangle relate to one another.

The very first trigonometric ratio is the sine angle. The sine function takes an angle and returns the ratio of two sides of a right triangle. In this article, we will learn all about the value of sin 15 in degrees and in radians, and how it was determined with solved examples

## Value Of Sin 15

The domain is the possible range of values of a function for the input. When we talk about the value of sin 15, we are giving the input as 15. The value of sin 15 will exist only if it lies in the domain of the sine function. The domain of the sine function is x∈R. Thus, the value of sin 15 exists as 15∈R.

The output after providing an input value is given by the range of the function. So all values of sin will line between the given range. As f is a periodic function, its range is a bounded interval given by the max and min values of the function. The maximum output of sinx is 1, while its minimum is −1. In other words, we can say that the value of sin 15 will lie within the range i.e. −1 ≤ n ≤ 1.

## Value of Sin 15 in Fraction

When we represent the value of sin 15 in the form of numerator and denominator, we call it the value of sin 15 in fraction form. The value of sin 15 degrees is \(\frac{\sqrt{3} – 1}{2\sqrt{2}}\) 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 Sin 15 Degrees in Decimals

We have seen the value of sin 15 in the fractions above. We know that if we divide the numerator of the fraction by its denominator, we will get a number or decimal number. The same will happen here. If we divide the numerator of the value of sin 15 in fractional form with its denominator we will get a decimal number. Let’s see how we can do that step by step.

Value of sin 15 in fraction form = \(\frac{\sqrt{3} – 1}{2\sqrt{2}}\)

We will substitute the values of \(\sqrt{3}\) and \(\sqrt{2}\) in the above fraction. We know that \(\sqrt{3}=1.732\) and \(\sqrt{2}=1.414\)

Thus, the equation becomes.

\(= \frac{1.732 – 1}{2\times1.414}\)

\(= \frac{0.732}{2.828}\)

\(= 0.25884016973\)

Thus, the value of sin 15 in decimal form is \(0.25884016973 ≈ 0.25884

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

## Value of Sin 15 in Radians

Well, a Radian, simply put, is a unit of measure for angles that is based on the radius of a circle. It is alternate to degrees. In some calculations, the angles are required in radians because all other quantities are measured in corresponding units. The formula for converting degrees into radians is given as,

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

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

Value of sin 15 in radians = value of sin 15 in decimals × \(\frac{\pi}{180°}\).

Value of sin 15 in radians = \(0.25884\times\frac{\pi}{180°}\).

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

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

## How to Find the Value of Sin 15 Degree

To find the values of sin 15 degrees we make use of previously known values of sine functions like sin30 and sin45. We then apply the identities of sin function to derive the value of sin 15.

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,

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

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

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

## Sin 15 Degrees Using Unit Circle

We can find the value of sin 15 in radians and in degrees using the unit circle. As mentioned earlier, radians is unit of measure for angles that are based on the radius of a circle. We are going to need a ruler, compass and protractor for this exercise.

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

- Draw a circle of radius of 1 unit i.e unit magnitude.
- Draw the two axis and four quadrants as shown in the figure. Since we 15 degrees lies in first quadrant, we will only work in first quadrant.
- Draw an arm of length ‘r’ in 15° anticlockwise angle with the positive x-axis using a protractor.
- The sin of 15 degrees equals the y-coordinate of the point of intersection of unit circle and r.
- 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° = y = 0.2588.
- Now, we can convert the radians value to degrees value as we have already seen earlier.

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

## Sin 15° in Terms of Trigonometric Functions

All the remaining trigonometric functions can be used for writing sin 15. Let’s see how we can do that.

Function | Formula |

\(cos15\) | \( \pm\sqrt{(1 – cos^215^{\circ})} \) |

\(tan15\) | \(\pm\frac{tan 15^{\circ}}{\sqrt{(1 + tan^215^{\circ})}}\) |

\(cot15\) | \(\pm\frac{1}{\sqrt{(1 + cot^215^{\circ})}}\) |

\(sec15\) | \(\frac{1}{\cos15}\) |

\(cosec15\) | \(\frac{1}{\cosec15}\) |

## Alternative Method to Determine Sin 15 Values

There is one more method to determine the value of sin 15. It’s simpler but it requires more data. Let’s see it step by step.

**By using the values of sin 30, sin 45, cos 30 and cos 45:**

By using basic maths, we can write 15 as 45 – 30. Thus, we can write sin(15) as

\(sin(15) = sin(45 – 30)\)

By applying the identity of sine of difference of two angles we get,

\(\sin(\alpha -\beta ) =\sin \alpha \cos \beta -\cos \alpha \sin \beta \)

Here, \alpha = 45 and \beta = 30, we get;

\(\sin(45 – 30) =\sin45\cos30 -\cos45\sin30\)

Now, we substitute the values of sin 30, sin 45, cos 30 and cos 45.

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

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

## Sine Circle Graph

For ease of calculations, we can represent all the commonly asked values of sine function on a graph. It is as shown below. At every point, the x-coordinate represents the cosine value of that angle. On the other hand, the y-coordinate represents the cosine value of that angle.

## Solved Examples

Now that we have learned all about sine 15. Here are some solved examples.

**Solved Example 1: **What is the value of \(sin(75) – sin(15)\).

**Solution:**

\(sin(75) – sin(15) = sin(45 + 30) – sin(45 – 30)\)

We know that, \(sin(A + B) = sinA.cosB + cosA.sinB\) and,

\(sin(A – B) = sinA.cosB – cosA.sinB\)

\(sin(75) – sin(15)\)

\(= (sin45.cos30 + cos45.sin30) – (sin45.cos30 – cos45.sin30)\)

\(= 2cos45.sin30\)

\(= 2.\frac{1}{\sqrt{2}}.\frac{1}{2}\)

\(= \frac{1}{\sqrt{2}}\)

**Solved Example 2:** Show that \(\sin 45^{\circ}+\sin 15^{\circ}=\sin 75^{\circ}\)

**Solution:**

\(RHS = \sin 45^{\circ}+\sin 15^{\circ}\)

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

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

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

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

\(LHS = sin(75)\)

\(= (sin45.cos30 + cos45.sin30)\)

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

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

Thus, LHS = RHS.

Hence, \(\sin 45^{\circ}+\sin 15^{\circ}=\sin 75^{\circ}\)

**Solved Example 3:** Calculate the following expression: \(sin 15° + sin 75°\) = ?

**Solution:**

\(sin(75) + sin(15) = sin(45 + 30) + sin(45 – 30)\)

We know that, \(sin(A + B) = sinA.cosB + cosA.sinB\) and,

\(sin(A – B) = sinA.cosB – cosA.sinB\)

\(sin(75) + sin(15)\)

\(= (sin45.cos30 + cos45.sin30) + (sin45.cos30 – cos45.sin30)\)

\(= 2sin45.cos30\)

\(= 2.\frac{1}{\sqrt{2}}.\frac{\sqrt{3}}{2}\)

\(= \frac{\sqrt{6}}{2}\)

Hope this article on Value of Sine 15 was informative. Get some practice of the same on our free Testbook App. Download Now!

## Value of Sin 15 FAQs

**Q.1What is the value of sin 15 degrees?**

**Ans.1 **The value of sin 15 degrees is \(\frac{\sqrt{3} – 1}{2\sqrt{2}}\) or 0.25884.

**Q.2What is the value of sin 15 degrees in terms of tan 15°?**

**Ans.2 **We can write the value of sin 15 in terms of tan 15 by using standard trigonometric identities. The identity is \(sinx = −tanx\sqrt{1+tan2x}\). Thus, \(sin15 = −tan15\sqrt{1+tan(2\times15)}\). \(sin15 = −tan15\sqrt{1+tan30}\)

**Q.3How to find sin 15° in terms of other trigonometric functions?**

**Ans.3 **We can use the standard trigonometric identities to write sin 15 in terms of other trigonometric functions.

**Q.4Is sin 15 a rational number?**

**Ans.4 **Sin 15 involves \(\sqrt3\). Hence, the said numbers are irrational.

**Q.5How do you simplify sin 15?**

**Ans.5 **sin15 can be written as sin(45 – 30).