
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
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:
The following methods can be used to derive the value of sin15°:
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
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)
On adding equations (6) and (7), we get
(sin15° + cos15°) + (sin15° - cos15°) = √ (1 + sin30°) + √ (1 - sin30°)
2 . sin15° = √ (1 + sin30°) + √ (1 - sin30°)
2 . sin15° = √(1 + ½) + √(1 – ½)
2 . sin15° = √(3 -1)/ √2
sin15° = √(3 -1)/ 2√2
which is the required value.
JEE Main marks vs rank vs percentile
JEE Advanced Eligibility Criteria
JEE Advanced Chemistry Syllabus
JEE Advanced Registration Dates
Derivation Of Lens Maker Formula
Unit Of Pressure Velocity Uses of Plane Mirror
Wave Theory of Light
Unit of Density Unit of Light Unit of Force Unit of Magnetic Field Unit of wavelength Unit of Viscosity Uses of Electroplating Young's Modulus
What is the Scattering of Light
Lenz Law Space Wave Propagation Schrodinger Wave Equation Relation between Fahrenheit and Celsius Refractive Index Potentiometer Working Pascal Law Oscillatory Motion Optical Instruments Newton's Laws of Motion - First Law Modulation and Demodulation Magnetic Flux Lens Formula and Magnification Kaleidoscope Faradays Law Epsilon Naught Value Energy Bands Electrostatics Electroscope AC Generator Unit of Current Lithosphere Bending Equation Derivation Difference Between Pound and Kilogram Semiconductor Devices OTEC - Ocean Thermal Energy Conversion Hall Effect Rectilinear Propagation of Light Difference Between Ammeter and Voltmeter Coefficient of Linear Expansion Ampere’s Law Cyclone and Thunderstorm Save The Environment From Pollution Particle Nature of Light Types of DC Motor Uses Of Transistor Derivation of Phase Rule Unit of Humidity