Toppers Study Toppers Study Toppers StudyDetailed NCERT Solutions for 11 Mathematics 3. Trigonometric Functions to simplify learning. Understand chapters clearly and practice with free solutions for better results.
Detailed NCERT Solutions for 11 Mathematics 3. Trigonometric Functions to simplify learning. Understand chapters clearly and practice with free solutions for better results.
Preparing for exams becomes easier with Exercise 3.3. Whether you are studying for board exams or mid-term exams, 11 Mathematics Chapter 3. Trigonometric Functions solutions provide quick revising points, well-structured answers, and additional practice material to help you score better.
ncert_solutionsExercise 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
Exercise 3.3 are created by experts to give step-by-step explanations. Around 60–70% of exam questions are based on NCERT concepts. Our 11 Mathematics Chapter 3. Trigonometric Functions solutions help you understand the core concepts and practice effectively.
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Every NCERT chapter ends with exercises, and solving them is crucial. Our Exercise 3.3 include complete solutions for 11 Mathematics Chapter 3. Trigonometric Functions exercises. With step-by-step answers, you gain clarity and confidence to attempt similar exam questions.
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Our Exercise 3.3 are useful for both board exams and mid-term exams. For 11 Mathematics, we provide notes, exercises, and important Q&A so that you can revise smartly and write perfect answers in exams.
In short, Exercise 3.3 for 11 Mathematics Chapter 3. Trigonometric Functions are a complete study package. With quick revising points, NCERT exercises, and additional important questions, you can prepare effectively for exams. Make these solutions your study companion and excel in your academic journey.
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