Our ncert solutions for Exercise 3.3 Class 11 maths 3. Trigonometric Functions - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.

# Exercise 3.3 Class 11 maths 3. Trigonometric Functions - ncert solutions - Toppers Study

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## Exercise 3.3 Class 11 maths 3. Trigonometric Functions - ncert solutions - Toppers Study

CBSE board students who preparing for **class 11 ncert solutions maths and Mathematics** solved exercise **chapter 3. Trigonometric Functions** available and this helps in upcoming exams
2024-2025.

### You can Find Mathematics solution Class 11 Chapter 3. Trigonometric Functions

- All Chapter review quick revision notes for chapter 3. Trigonometric Functions Class 11
- NCERT Solutions And Textual questions Answers Class 11 Mathematics
- Extra NCERT Book questions Answers Class 11 Mathematics
- Importatnt key points with additional Assignment and questions bank solved.

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### Exercise 3.3 class 11 Mathematics Chapter 3. Trigonometric Functions

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- Exercise 3.3 Class 11 Maths 3. Trigonometric Functions - Ncert Solutions - Toppers Study
- Class 11 Ncert Solutions
- Solution Chapter 3. Trigonometric Functions Class 11
- Solutions Class 11
- Chapter 3. Trigonometric Functions Exercise 3.3 Class 11

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## 3. Trigonometric Functions

### | Exercise 3.3 |

## Exercise 3.3 Class 11 maths 3. Trigonometric Functions - ncert solutions - Toppers Study

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. sin ^{2 }6x - sin^{2}**

**4x = sin 2x sin 10x**

**Solution: **

Sin^{2} A - sin^{2}B = sin(A + B) sin(A - B)

LHS = sin^{2 }6x - sin^{2} 4x

= sin (6x + 4x) sin(6x - 4x)

= sin 10x sin 2x

= sin 2x sin 10x

LHS = RHS

**Q13. cos ^{2} 2x - cos^{2} 6x = sin 4x sin 8x**

**Solution: **

LHS = cos^{2} 2x - cos^{2} 6x

= (1 - sin^{2} 2x) - (1 - sin^{2} 6x)

= (1 - sin^{2} 2x - 1 + sin^{2} 6x)

= - sin^{2} 2x + sin^{2} 6x

= sin^{2} 6x - sin^{2} 2x

= sin (6x + 2x) sin(6x -2x)

= sin 10x sin 4x

**Q24. ****cos 4 x = 1 – 8 sin^{2} x cos^{2} x**

**Solution: **

LHS = cos 2(2x)

= cos 2A [ Let be A = 2x ]

= 1 - 2 sin^{2} A

= 1 - 2 sin^{2} 2x [Putting A = 2x ]

= 1 - 2 [sin 2x]^{2}

= 1 - 2 [2 sin x cos x ]^{2}

= 1 - 2 [4 sin^{2} x cos^{2} x]

= 1 - 8 sin^{2} x cos^{2} x

**LHS = RHS **

**Q25. cos 6 x = 32 cos^{6} x – 48cos^{4} x + 18 cos^{2} x – 1**

**Solution: **

LHS = cos 6*x * = cos 3(2x)

= cos 3A [Let be A = 2x]

= 4 cos^{3} A - 3 cos A

= 4 cos^{3} 2x - 3 cos 2x [Putting A = 2x]

= 4 [cos 2x]^{3} - 3 [cos 2x]

= 4 [2cos^{2} *x *– 1]^{3} - 3 [2cos^{2} *x *– 1]

= 4 [(2cos^{2} *x*)^{3} - 1^{3} - 3(2cos^{2} *x)*^{2} (1) + 3 (2cos^{2} *x)*(1)^{2} ] - 3 [2cos^{2} *x *– 1]

= 4 [8 cos^{6} x - 1 - 12 cos^{2 }x + 6 cos^{2} x] - 6 cos^{2} + 3

= 32 cos^{6} x - 4 - 48 cos^{2 }x + 24 cos^{2} x - 6 cos^{2} + 3

= 32 cos^{6} x - 48 cos^{2 }x + 18 cos^{2} x - 1

**LHS = RHS**

##### Other Pages of this Chapter: 3. Trigonometric Functions

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