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Sine function of an angle is a trigonometric function that is represented by sinα, where α is the acute angle in consideration. In a right-angled triangle, the sine of any angle is defined as the ratio of the length of the perpendicular to that of the hypotenuse. The hypotenuse is the opposite side to the angle in consideration (α). The graph of the sine function, also known as the sine wave, is a ‘sinusoidal’ curve. It is an up-down curve and repeats every 360°. In trigonometry, there are various identities involving sine function that are widely used for solving mathematical problems.
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Key Takeaways: Identities, Acute Angle, Triangle, Quadrant, Sine Function, Hypotenuse, Trigonometry, Trigonometric function
Sine Function
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In a right-angled triangle, the sine of any angle can be defined as the ratio of the length of the perpendicular to that of the hypotenuse. It provides the value of the sine function of the angle between the hypotenuse and the base of the same. It is useful to find out the unknown angles or sides of a right triangle. Generally, it is abbreviated as sinα where α is an acute angle between the base and the hypotenuse of the right triangle.
Sine Wave
Sine Function Formula
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In a right-angled triangle, the sine of an angle is equal to the ratio of length of the perpendicular to that of the hypotenuse.
Sine Function Formula
Suppose, ABC is a right-angled triangle
Where,
AC= base= b; AB= Perpendicular= a; BC= Hypotenuse=h and α is the acute angle.
Then, as per the sine formula sinα= perpendicular/hypotenuse =AB/BC= a/h.
So, sinα = a/h
Also Read:
Sine Function Table of values
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Here are the sine function values for some specific angles such as 0°, 30°,45°,60°,90° and so on are given in a tabular format-
| Sine Degrees | Sine Radians | Value of the sine function |
|---|---|---|
| Sin 0° | Sin 0 | 0 |
| Sin 30° | Sin π/6 | ½ |
| Sin 45° | Sin π/4 | 1/√2 |
| Sin 60° | Sin π/3 | √3/2 |
| Sin 90° | Sin π/2 | 1 |
| Sin 120° | Sin 2π/3 | √3/2 |
| Sin 150° | Sin 5π/6 | ½ |
| Sin 180° | Sin π | 0 |
| Sin 270° | Sin 3π/2 | -1 |
| Sin 360° | Sin 2π | 0 |
Sine Function Graphs
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Firstly, let us try to understand how the values of sine function vary on a unit circle and plot them graphically. As shown in the above picture,
Sin α = AB/ BC= AB/1= AB (as the radius of the unit circle is 1. Thus BC= 1). Thus, the value of the sin α varies with the length of AB. Now, the values of the sine function will vary in all four quadrants and if we plot those values graphically, we will get the sine graph. Here the horizontal axis represents the input variable α in radians and the vertical axis represents the value of the sine function itself. Putting the values of one complete cycle that is 0 rad to 2Π rad we will get the graph as following:
Sine Curve
Read More: Differentiation and Integration Formula
Sine Function Identities
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In trigonometry there are multiple identities involving sine function. Some of those sin functions are given below-
- Sinα = 1/cosα
- Sin-1α = arcsin α (where α lies between -1 and 1).
- Sin2α + cos2α= 1
- Sin (α+β)= sinαcosβ + sinβcosα
- Sin (α-β)= sinαcosβ - sinβcosα
- Sin 2α= 2 Sin αcos α
- d (sin x)/dx= cosx
- sinx dx = - cosx +c (where, c= integration constant).
Things to Remember
- In a right-angled triangle, the sine of any angle is defined as the ratio of the length of the perpendicular to that of the hypotenuse.
- Formula: sinα= perpendicular/hypotenuse
- f(x)=sinx is a periodic function and the sine function period is 2π.
- The domain of the sine function is (-∞,∞) while the range is (-1,1)
- The graph of the sine function, also known as sine wave, is a ‘sinusoidal’. It is an up-down curve and repeats every 360° or 2π.
Sample Questions
Ques: What is a sine function in trigonometry? (2 Marks)
Ans: In a right-angled triangle, the sine of any angle can be defined as the ratio of the length of the perpendicular to that of the hypotenuse.
Ques: State the sine function formula (2 Marks)
sinα= perpendicular/hypotenuse
Ques: Find the value of sin 135°. (2 Marks)
sin (a + b) = sin a cos b + sin b cos a and using the same we get,
sin(135°) = sin(90° + 45°)
= sin 90° cos 45° + sin 45° cos 90°
= (1 × 1/√2) + (1/√2 ×0)
= 1/√2
Therefore, sin 135°= 1/√2
Ques: Determine the value of the length of the perpendicular of a right-triangle if the value of sine of α is 0.6 and the length of the hypotenuse is 6 units. (2 Marks)
Ans: We know that sin x = Perpendicular/Hypotenuse
We have sin x = 0.6, Hypotenuse = 6 units
Therefore, 0.6 = Perpendicular/6
⇒ Perpendicular = 6 × 0.6 = 3.6
Hence, the length of the perpendicular is 3.6 units.
Ques: a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B) = 0 (3 Marks)
Ans: By using the sine rule we get,
a = k sin A, b = k sin B, c = k sin C
Let us consider LHS:
a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B)
Substituting the values of a, b, c from sine rule in above equation, we will obtain
a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B) = k sin A (sin B – sin C) + k sin B (sin C – sin A) + k sin C (sin A – sin B)
= k sin A sin B – k sin A sin C + k sin B sin C – k sin B sin A + k sin C sin A – k sin C sin B
Upon simplification, we get
= 0 = R.H.S (proved).
Ques: In a Δ ABC, if a = 5, b = 6 and C = 60o, show that its area is (15√3)/2 sq. units. (3 Marks)
Ans: Given:In a ΔABC, a = 5, b = 6 and C = 60o
By using the formula,
Area of ABC = 1/2 ab sin θ (where, a and b are the lengths of the sides of a triangle and θ is the angle between sides.)
So,
Area of ΔABC = 1/2 ab sin θ
= 1/2 × 5 × 6 × sin 60o
= 30/2 × √3/2
= (15√3)/2 sq. units
Ques: In a right-angle triangle ABC, angle A= 45°, angle B= 60° and angle c= 75°. Find out the ratio of the sides of the triangle. (5 Marks)
a:b:c= 1: 6:1+√3
\(\frac{a}{sin A}\)=\(\frac{b}{sin B}\)=\(\frac{c}{sin C}\)
Now, by substituting the values, we get
\(\frac{a}{sin 45}\)=\(\frac{b}{sin 60}\)=\(\frac{c}{sin 75}\)
\(\frac{a}{sin 45}\)=\(\frac{b}{sin 60}\)=\(\frac{c}{sin (30 +45)}\)
\(\frac{a}{sin 45}\)=\(\frac{b}{sin 60}\)=\(\frac{c}{sin30 cos45+sin45cos30)}\)
Now, by substituting the corresponding values, we get
\(\frac{a}{\frac{1}{\sqrt{2}}} = \frac{b}{\frac{\sqrt{3}}{2}} = \frac{c}{\frac{1}{2} \times \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}}\)
\(\frac{a}{\frac{1}{\sqrt{2}}} = \frac{b}{\frac{\sqrt{3}}{2}} = \frac{c}{\frac{1 + \sqrt{3}}{2\sqrt{2}}}\)
a:b:c = \(\frac{1}{\sqrt{2}} : \frac{\sqrt{3}}{2}:\frac{1 + \sqrt{3}}{2\sqrt{2}}\)
Therefore, the ratio of the sides of the triangle is 1: 6:1+√3
Ques: b sin B – c sin C = a sin (B – C) (5 Marks)
Ans: By using the sine rule we get,
So, c = k sin C
Similarly, a = k sin A
And b = k sin B
We know,
Now let us consider LHS:
b sin B – c sin C
Substituting the values of b and c in the above equation, we obtain
k sin B sin B – k sin C sin C = k (sin2 B – sin2 C) ……….(i)
We know,
Sin2 B – sin2 C = sin (B + C) sin (B – C),
Substituting the above values in equation (i), we get
k (sin2 B – sin2 C) = k (sin (B + C) sin (B – C)) [since, π = A + B + C ⇒ B + C = π –A]
The above equation becomes,
= k (sin (π –A) sin (B – C)) [since, sin (π – θ) = sin θ]
= k (sin (A) sin (B – C))
From sine rule, a = k sin A, thus, the above equation becomes,
= a sin (B – C)
= RHS (proved)
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