Value of sin 15°

The function sine is one of the trigonometric functions applicable on right-angled triangles. It is primarily applicable on an acute angle in a triangle wherein the sine function is mathematically represented as sinθ, for acute angle θ in a right-angled triangle. The function sinθ is calculated as the ratio of perpendicular and hypotenuse. Mathematically,

sinθ = perpendicular/hypotenuse

For right triangle ABC

sinθ = AB/AC

Finding the value of sin 15° - Maths

The original value of sin15° = (√3 – 1) / 2√2. This value of sin15° can be obtained through different methods. To solve for the sine value, the following values of sine and cosine functions for some standard angles must be considered:

sin0° = 0
sin30° = ½
sin45° = 1/√2
sin60° = √3/2
sin90° = 1
cos0° = 1
cos30° = √3/2
cos45° = 1/√2
cos60° = ½
cos90° = 1

The following methods can be used to derive the value of sin15°:

Method 1

For finding the value of sin15° use this method. Consider the sine and cosine trigonometric functions of standard angles 60° and 45° i.e. sin45°, sin60°, cos45° and cos60°.

We know,

sin45° = 1/√2

sin60° = √3/2

cos45° = 1/√2

cos60° = ½

Proceed as below

sin15° = sin(60 – 45)…..(1)

Comparing the above equation with the formula

sin(A – B) = sinA . cosB – cosA . sinB…..(2)

Substituting the values of equation (1) in the equation (2), we get

sin(60 – 45) = (sin60 . cos45) – (cos60 . sin45)

sin15° = (√3/2 . 1/√2) – (½ . 1/√2)

sin15° = (√3 – 1) / 2√2

This can also be solved using trigonometric functions of standard angles 30° and 45°.

We know,

sin30° = ½

sin45° = 1/√2

cos30° = √3/2

cos45° = 1/√2

Substituting the above values in equation (2)

sin(45 – 30) = (sin45 . cos30) – (cos45 . sin30)

sin15° = (1/√2 . √3/2) – (1/√2 . ½)

sin15° = (√3 – 1) / 2√2

Method 2

This concept is based on half angle property of trigonometric function. Considering the trigonometric functions sine and cosine for the standard angle 30°, let

(sinA/2 + cosA/2)² = (sinA/2)² + (cosA/2)² + 2 . sinA/2 . cosA/2

(sinA/2 + cosA/2)² = 1 + 2 . sinA/2 . cosA/2…..(3)

As we known,

2 sinx. cosx = sin2x…..equation (4)

Using equation (4) to reduce equation (3), we get

(sinA/2 + cosA/2)² = 1 + sinA

On taking square root both sides,

(sinA/2 + cosA/2) = ±√ (1 + sinA) ….. (5)

Now, let A = 30°, then A/2 = 15°. putting the values of A and A/2 in equation (5), we get

sin15° + cos15° = ±√ (1 + sin30°)….. (6)

Similarly, obtaining the corresponding value for (sin15° + cos15°) which comes out to be

sin15° - cos15° = ±√ (1 - sin30°)….. (7)