
The graph of the sine function lies exactly in which of the following region?
A) \[y = - 1{\text{ }}to{\text{ }}y = 1\]
B) \[y = - \infty {\text{ }}to{\text{ }}y = \infty {\text{ }}\]
C) \[y = 1{\text{ }}to{\text{ }}y = \infty {\text{ }}\]
D) \[y = - \infty {\text{ }}to{\text{ }}y = - 1\]
Answer
486.6k+ views
Hint: To find the region in which graph of sine function lies exactly we will draw a graph by plotting various sine angles. From there, the maximum and minimum values on the y – axis will be the region or range of the sine function.
Trigonometric formulas:
$ \sin \left( {90 + \theta } \right) = \cos \theta $
$ \sin \left( {180 + \theta } \right) = - \sin \theta $
Complete step by step solution:
The value of different angles of sine is given as:
sin 0° = 0
sin 90° = 1
sin 180° = sin (90 + 90°)
= cos 90° $ \left[ {\because \sin \left( {90 + \theta } \right) = \cos \theta } \right] $
= 0
sin 270° = sin (180 + 90°)
= sin 90° $ \left[ {\because \sin \left( {180 + \theta } \right) = - \sin \theta } \right] $
= -1
sin 360° = sin (180 + 180°)
= sin 180°
= 0
To get a graph for sine function we can plot the values of certain angles on the graph. The table for values of sine can be drawn as:
Plotting these values on graph we get:
From the graph, it can be seen that the minimum value at y –axis is -1 and the maximum value is 1. So, the graph of sine function will lie between the range \[y = - 1{\text{ }}to{\text{ }}y = 1\].
So, the correct answer is “Option A”.
Note: We take the sine angle greater than 180° with a negative sign because the value does not lie in the second quadrant and the sine function is positive only in the second quadrant where values lie between 90° to 180° .
We know that the value of sine function lies between the interval -1 to 1 and represented as:
$ - 1 \leqslant \sin x \leqslant 1 $
So, we could have written the range along y- axis directly
Trigonometric formulas:
$ \sin \left( {90 + \theta } \right) = \cos \theta $
$ \sin \left( {180 + \theta } \right) = - \sin \theta $
Complete step by step solution:
The value of different angles of sine is given as:
sin 0° = 0
sin 90° = 1
sin 180° = sin (90 + 90°)
= cos 90° $ \left[ {\because \sin \left( {90 + \theta } \right) = \cos \theta } \right] $
= 0
sin 270° = sin (180 + 90°)
= sin 90° $ \left[ {\because \sin \left( {180 + \theta } \right) = - \sin \theta } \right] $
= -1
sin 360° = sin (180 + 180°)
= sin 180°
= 0
To get a graph for sine function we can plot the values of certain angles on the graph. The table for values of sine can be drawn as:
Angles | Measure |
sin 0° | 0 |
sin 90° $ \to \left[ {\sin \left( {\dfrac{\pi }{2}} \right)} \right] $ | 1 |
sin 180° $ \to \left[ {\sin \left( \pi \right)} \right] $ | 0 |
sin 270° $ \to \left[ {\sin \left( {\dfrac{{3\pi }}{2}} \right)} \right] $ | -1 |
sin 360° $ \to \left[ {\sin \left( {2\pi } \right)} \right] $ | 0 |
Plotting these values on graph we get:

From the graph, it can be seen that the minimum value at y –axis is -1 and the maximum value is 1. So, the graph of sine function will lie between the range \[y = - 1{\text{ }}to{\text{ }}y = 1\].
So, the correct answer is “Option A”.
Note: We take the sine angle greater than 180° with a negative sign because the value does not lie in the second quadrant and the sine function is positive only in the second quadrant where values lie between 90° to 180° .
We know that the value of sine function lies between the interval -1 to 1 and represented as:
$ - 1 \leqslant \sin x \leqslant 1 $
So, we could have written the range along y- axis directly
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Trending doubts
Which one is a true fish A Jellyfish B Starfish C Dogfish class 11 biology CBSE

What is the difference between superposition and e class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

In which part of the body the blood is purified oxygenation class 11 biology CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

State the laws of reflection of light
