Plus One Math's Solution Ex 3.3 Chapter 3 Trigonometric Functions


The solutions for the
 Trigonometric Functions are not readily available. While many schools may have a ready-made solution for this set in their school textbook, some might not. This is where our solution will be useful. In this article, you will find detailed solutions provided by us for the above set.Trigonometric Functions (Key Concept Reference) describes some basic and advanced uses of trigonometric functions, including identities, graph transformations, inverse functions, solutions of triangles, and polar coordinates.

Ncert Plus one Maths chapter-wise textbook solution for chapter 3 Trigonometric Functions Exercise 3.3. It contains detailed solutions for each question which have prepared by expert teachers to make each answer easily understand the students. they are well arranged solutions so that students would be able to understand easily.

BoardSCERT, Kerala
Text BookNCERT Based
ClassPlus One 
SubjectMath's Textbook Solution
ChapterChapter 3
ExerciseEx 3.3
Chapter NameTrigonometric Functions
CategoryPlus One Kerala

Kerala Syllabus Plus One Math's Textbook Solution Chapter  3 Trigonometric Functions Exercises 3.3


Chapter  3 Trigonometric Functions Textbook Solution



Kerala plus One maths NCERT textbooks, we provide complete solutions for the exercise and answers provided at the end of each chapter. We also cover the entire syllabus given by the Board of secondary education, Kerala state.

Chapter  3 Trigonometric Functions Exercise   3.3

 Question 1:

Prove that


L.H.S. = 


Prove that 

L.H.S. = 


Prove that 

L.H.S. =


Prove that 

L.H.S =


Find the value of:

(i) sin 75°

(ii) tan 15°

(i) sin 75° = sin (45° + 30°)

= sin 45° cos 30° + cos 45° sin 30°

[sin (x + y) = sin x cos y + cos x sin y]

(ii) tan 15° = tan (45° – 30°)


Prove that:


Prove that: 

It is known that 

∴L.H.S. =


 

L.H.S. = 


Prove that

It is known that.

∴L.H.S. = 


Prove that sin2 6x – sin2 4x = sin 2x sin 10x

It is known that 

∴L.H.S. = sin26x – sin24x

= (sin 6x + sin 4x) (sin 6x – sin 4x

= (2 sin 5x cos x) (2 cos 5x sin x)

= (2 sin 5x cos 5x) (2 sin x cos x)

= sin 10x sin 2x

= R.H.S.


Prove that cos2 2x – cos2 6x = sin 4x sin 8x

It is known that 

∴L.H.S. = cos2 2x – cos2 6x

= (cos 2x + cos 6x) (cos 2– 6x)

= [2 cos 4x cos 2x] [–2 sin 4(–sin 2x)]

= (2 sin 4x cos 4x) (2 sin 2x cos 2x)

= sin 8x sin 4x

= R.H.S.


Prove that sin 2x + 2sin 4x + sin 6x = 4cos2x sin 4x

L.H.S. = sin 2x + 2 sin 4x + sin 6x

= [sin 2x + sin 6x] + 2 sin 4x

= 2 sin 4x cos (– 2x) + 2 sin 4x

= 2 sin 4x cos 2x + 2 sin 4x

= 2 sin 4x (cos 2x + 1)

= 2 sin 4x (2 cos2x â€“ 1 + 1)

= 2 sin 4x (2 cos2x)

= 4cos2x sin 4x

= R.H.S.


Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

L.H.S = cot 4x (sin 5x + sin 3x)

= 2 cos 4x cos x

R.H.S. = cot x (sin 5x â€“ sin 3x)

= 2 cos 4x. cos x

L.H.S. = R.H.S.


Prove that 

It is known that

∴L.H.S. =


Prove that 

It is known that

∴L.H.S. =


Prove that 

It is known that

∴L.H.S. = 


Prove that 

L.H.S. =


Prove that cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x

= cot x cot 2x – cot 3x (cot 2x + cot x)

= cot x cot 2x – cot (2x) (cot 2x + cot x)

= cot x cot 2– (cot 2cot x – 1)

= 1 = R.H.S.


Prove that 

It is known that.

∴L.H.S. = tan 4x = tan 2(2x)


Prove that cos 4x = 1 – 8sinx cosx

L.H.S. = cos 4x

= cos 2(2x)

= 1 – 2 sin2 2x [cos 2A = 1 – 2 sin2A]

= 1 – 2(2 sin x cos x)2 [sin2A = 2sin A cosA]

= 1 – 8 sin2x cos2x

= R.H.S.


Prove that: cos 6x = 32 cos6x – 48 cos4x + 18 cos2x – 1

L.H.S. = cos 6x

= cos 3(2x)

= 4 cos3 2x – 3 cos2x [cos 3A = 4 cos3A – 3 cosA]

= 4 [(2 cos2– 1)3 – 3 (2 cos2x – 1) [cos 2x = 2 cos2– 1]

= 4 [(2 cos2x)3 – (1)3 – 3 (2 cos2x)2 + 3 (2 cos2x)] – 6cos2x + 3

= 4 [8cos6x – 1 – 12 cos4x + 6 cos2x] – 6 cos2x + 3

= 32 cos6x – 4 – 48 cos4x + 24 cos2x – 6 cos2x + 3

= 32 cos6– 48 cos4x + 18 cos2x – 1

= R.H.S.



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Chapter 3 Trigonometric Functions EX 3.3 Solution


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