Class 11 NCERT Mathematics Solutions - Chapter 3 Trigonometric Function - Exercise 3.2

Chapter 3 Trigonometric Functions, of the Class 11 NCERT Mathematics textbook deals with Trigonometric Functions which are a fundamental area in trigonometry. In Exercise 3.2, students explore the various problems related to trigonometric functions including their properties and applications. Understanding these exercises is crucial for grasping the concepts of angles, periodicity, and the relationship between trigonometric functions. The Mastery of this chapter is essential for solving complex mathematical problems and is foundational for advanced studies in the mathematics and related fields.

Trigonometric Function

The Trigonometric functions are fundamental mathematical functions that relate angles to the ratios of the side lengths in right-angled triangles. They are essential for modeling periodic phenomena and solving various real-world problems involving angles and distances. In this chapter, students study the definitions of trigonometric functions, their properties and their graphical representations. This exercise helps solidify understanding by applying these concepts to solve practical problems.

Class 11 NCERT Mathematics Solutions - Exercise 3.2

Question. Find the values of the other five trigonometric functions in Exercises 1 to 5.

1. cos x = –1/2, x lies in the third quadrant.

Solution:

Since cos x = (-1/2) 

We have sec x = 1/cos x = -2

Now sin² x + cos² x = 1

i.e., sin² x = 1 – cos² x 

or sin² x = 1 - (1/4) = (3/4) 

Hence sin x = ±(√3/2) 

Since x lies in third quadrant, sin x is negative. Therefore

sin x = (–√3/2) 

Which also gives

cosec x = 1/sin x = (–2/√3) 

Further, we have

tan x = sin x/cos x

         = (–√3/2)/(-1/2)

         = √3 

and cot x = 1/tanx = (1/√3) 

2. sin x = 3/5, x lies in the second quadrant.

Solution:

Since sin x = (3/5) 

we have cosec x = 1/sin x = (5/3) 

Now sin² x + cos² x = 1

i.e., cos² x = 1 – sin² x

or cos²  x = 1 - (3/5) 

              = 1 - (9/25) 

              = (16/25) 

Hence cos x = ±(4/5) 

Since x lies in second quadrant, cos  x is negative. 

Therefore

cos x = (–4/5) 

which also gives

sec x = 1/cos x= (-5/4) 

Further, we have

tan x =  sin x/cos x

         =  (3/5)/(-4/5) 

         =  (-3/4) 

and cot x = 1/tan x = (-4/3)

3. cot x = 3/4, x lies in the third quadrant.

Solution:

Since cot x = (3/4) 

we have tan x = 1 / cot x = (4/3) 

Now sec² x = 1 + tan² x

                  = 1 + (16/9)

                  = (25/9) 

Hence sec x = ±(5/3) 

Since x lies in third quadrant, sec x  will be negative. Therefore

sec x = (-5/3) 

which also gives

cos x = (-3/5) 

Further, we have

sin x = tan x  * cos x

        = (4/3) x (-3/5)

        = (-4/5) 

and cosec x = 1/sin x

                   = (-5/4) 

4. sec x = 13/5, x lies in the fourth quadrant.

Solution:

Since sec x = (13/5)

we have cos x = 1/secx = (5/13) 

Now sin² x + cos² x = 1

i.e., sin² x = 1 – cos² x

or sin² x = 1 - (5/13)²

              = 1 - (25/169)

              = 144/169

Hence sin x = ±(12/13)

Since x lies in forth quadrant, sin x is negative. 

Therefore

sin x = (–12/13)

which also gives

cosec x = 1/sin x = (-13/12)

Further, we have

tan x =  sin x/cos x

         =  (-12/13) / (5/13)

         =  (-12/5)

and cot x = 1/tan x = (-5/12) 

5. tan x = –5/12, x lies in the second quadrant.

Solution:

Since tan x = (-5/12)

we have cot x = 1/tan x = (-12/5)

Now sec² x = 1 + tan² x

                  = 1 + (25/144)

                  = 169/144

Hence sec x = ±(13/12)

Since x lies in second quadrant, sec x will be negative. Therefore

sec x = (-13/12)

which also gives

cos x = 1/sec x = (-12/13)

Further, we have

sin x = tan x  * cos x

        = (-5/12) x (-12/13) 

        = (5/13)

and cosec x = 1/sin x = (13/5) 

Question. Find the values of the trigonometric functions in Exercises 6 to 10.

6. sin(765°) 

Solution:

We known that the values of sin x repeats after an interval of 2π or 360^∘. 

So, sin(765°)

= sin(720° + 45°) { taking nearest multiple of 360 }

= sin(2 × 360° + 45°) 

= sin(45°) 

= 1/√2

Hence, sin(765°) = 1/√2

7. cosec(–1410°)

Solution:

We known that the values of cosec  x  repeats after an interval of 2π or 360^∘. 

So, cosec(-1410°) 

= - cosec(1410°) 

= - cosec(1440° - 30°)    { taking nearest multiple of 360 }

= - cosec(4 × 360° - 30°) 

= cosec(30°) 

= 2

Hence, cosec(–1410°) = 2.

8. tan(19π/3)

Solution:

We known that the values of tan x  repeats after an interval of π or 180^∘.

So, tan(19π/3) 

 = tan(18π/3 + π/3) { breaking into nearest integer } 

 = tan(6π + π/3)

 = tan(π/3) 

 = tan(60°) 

 = √3

Hence, tan(19π/3) = √3. 

9. sin(–11π/3)

Solution:

We known that the values of sin x  repeats after an interval of 2π or 360^∘.

So, sin(-11π/3)

= -sin(11π/3) 

= -sin(12π/3 - π/3) { breaking nearest multiple of 2π divisible by 3}

= -sin(4π - π/3)

= -sin(-π/3) 

= -[-sin(π/3)] 

= sin(π/3) 

= sin(60°)

= √3/2

Hence, sin(-11π/3) = √3/2.

10. cot(–15π/4) 

Solution:

We known that the values of cot x repeats after an interval of π or 180°. 

So, cot(-15π/4)

= -cot(15π/4)

= -cot(16π/4 - π/4) { breaking into nearest multiple of π  divisible by 4 }

= -cot(4π - π/4)

= -cot(-π/4)

= -[-cot(π/4)] 

= cot(45°)

= 1

Hence, cot(-15π/4) = 1.

Conclusion

Chapter 3 of the Class 11 NCERT Mathematics textbook, "Trigonometric Functions," introduces the basics of trigonometric functions, including their definitions and properties. Exercise 3.2 involves solving problems related to these functions and their applications. Detailed solutions are provided to help students understand and apply trigonometric concepts effectively.