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3. Trigonometric Functions Mathematics class 11 in English Medium ncert book solutions Exercise 3.3
3. Trigonometric Functions Exercise 3.3 – Complete NCERT Book Solutions for Class 11 Mathematics (English Medium). Get all chapter explanations, extra questions, solved examples and additional practice questions for 3. Trigonometric Functions Exercise 3.3 to help you master concepts and score higher.
3. Trigonometric Functions Mathematics class 11 in English Medium ncert book solutions Exercise 3.3
NCERT Solutions for Class 11 Mathematics play an important role in helping students understand the concepts of the chapter 3. Trigonometric Functions clearly. This chapter includes the topic Exercise 3.3, which is essential from both academic and examination point of view. The solutions provided here are prepared strictly according to the latest NCERT syllabus and follow the guidelines of CBSE to ensure accuracy and relevance. Each question is explained in a simple and student-friendly manner so that learners can grasp the concepts without confusion. These NCERT Solutions are useful for regular study, homework help, and exam preparation. All textbook questions are solved step by step to improve problem-solving skills and conceptual clarity. Students of Class 11 studying Mathematics can use these solutions to revise important topics, understand difficult questions, and practise effectively before examinations. The chapter 3. Trigonometric Functions is explained in a structured way, making it easier for students to connect the theory with the topic Exercise 3.3. By studying these updated NCERT Solutions for Class 11 Mathematics, students can build a strong foundation, boost their confidence, and score better marks in school and board exams.
3. Trigonometric Functions
Exercise 3.3
Exercise 3.3
Q5. Find the value of
(i) sin 75° (ii) tan 15°
Solution:
(i) sin 75° = sin (45° + 30°) [∵ sin(x + y) = sin x cos y + cos x sin y ]
= sin 45° cos 30° + cos 45° sin 30°
(ii) tan 15°
Solution:
(ii) tan 15° = tan (45° - 30°)
Q10. sin(n + 1)x sin(n + 2)x + cos(n + 1)x cos(n + 2)x = cos x
Solution:
LHS = sin(n + 1)x sin(n + 2)x + cos(n + 1)x cos(n + 2)x
Or cos(n + 2)x cos(n + 1)x + sin(n + 2)x sin(n + 1)x
Let be the A = (n + 2)x, B = (n + 1)x
Now we have,
LHS = cos A cos B + sin A sin B
= cos ( A - B) [∵ sin(x + y) = sin x cos y + cos x sin y ]
= cos [(n + 2)x - (n + 1)x]
= cos [nx + 2x - (nx + x) ]
= cos [nx + 2x - nx - x ]
= cos x
LHS = RHS
Q12. sin2 6x - sin2 4x = sin 2x sin 10x
Solution:
Sin2 A - sin2B = sin(A + B) sin(A - B)
LHS = sin2 6x - sin2 4x
= sin (6x + 4x) sin(6x - 4x)
= sin 10x sin 2x
= sin 2x sin 10x
LHS = RHS
Q13. cos2 2x - cos2 6x = sin 4x sin 8x
Solution:
LHS = cos2 2x - cos2 6x
= (1 - sin2 2x) - (1 - sin2 6x)
= (1 - sin2 2x - 1 + sin2 6x)
= - sin2 2x + sin2 6x
= sin2 6x - sin2 2x
= sin (6x + 2x) sin(6x -2x)
= sin 10x sin 4x
Q24. cos 4x = 1 – 8 sin2 x cos2 x
Solution:
LHS = cos 2(2x)
= cos 2A [ Let be A = 2x ]
= 1 - 2 sin2 A
= 1 - 2 sin2 2x [Putting A = 2x ]
= 1 - 2 [sin 2x]2
= 1 - 2 [2 sin x cos x ]2
= 1 - 2 [4 sin2 x cos2 x]
= 1 - 8 sin2 x cos2 x
LHS = RHS
Q25. cos 6x = 32 cos6 x – 48cos4 x + 18 cos2 x – 1
Solution:
LHS = cos 6x = cos 3(2x)
= cos 3A [Let be A = 2x]
= 4 cos3 A - 3 cos A
= 4 cos3 2x - 3 cos 2x [Putting A = 2x]
= 4 [cos 2x]3 - 3 [cos 2x]
= 4 [2cos2 x – 1]3 - 3 [2cos2 x – 1]
= 4 [(2cos2 x)3 - 13 - 3(2cos2 x)2 (1) + 3 (2cos2 x)(1)2 ] - 3 [2cos2 x – 1]
= 4 [8 cos6 x - 1 - 12 cos2 x + 6 cos2 x] - 6 cos2 + 3
= 32 cos6 x - 4 - 48 cos2 x + 24 cos2 x - 6 cos2 + 3
= 32 cos6 x - 48 cos2 x + 18 cos2 x - 1
LHS = RHS
See other sub-topics of this chapter:
1. Exercise 3.1 2. Exercise 3.2 3. Exercise 3.3 4. Exercise 3.4 5. Miscellaneous Exercise on Chapter - 3