3. Trigonometric Functions Mathematics Exercise - 3.3 class 11 Maths in English - CBSE Study
NCERT Solutions for Class 11 Mathematics are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important 3. Trigonometric Functions with detailed explanations and step-by-step answers for better exam preparation. Each Exercise 3.3 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 11 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics.
Class 11 English Medium Mathematics All Chapters:
3. Trigonometric Functions
3. 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
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