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

Topper Study classes prepares ncert solutions on practical base problems and comes out with the best result that helps the students and teachers as well as tutors and so many ecademic coaching classes that they need in practical life. 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

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
2022-2023.

### 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
- Extra NCERT Book questions Answers
- Importatnt key points with additional Assignment and questions bank solved.

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

## 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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