**Exercise 3.3 Trigonometric Functions**

**Question and Answers**

**Class 11 – Maths**

Class | Class 11 |

Subject | Mathematics |

Chapter Name | Trigonometric Functions |

Chapter No. | Chapter 3 |

Exercise | Exercise 3.3 |

Category | Class 11 Maths NCERT Solutions |

**Question (1) Prove that **

**Answer**

**Question (2) Prove that**

**Answer **

**Question 3 Prove that**

**Answer **

**Question 4 Prove that**

**Answer**

**Question 5 Find the value of
(i) Sin 75°**

**Answer**

Sin 75°= Sin ( 45°+ 30°)

Sin 75°= Sin 45° Cos 30° + Cos 45° Sin 30°

sin(x+y) = sin x cos y + cos x sin y

Therefore we have,

**Question (5) Find the value of
(ii) tan 15°**

**Answer
**tan 15°= tan ( 45°- 30° )

**Question (6) Prove the following:**

**Answer**

cos(x+y) = cos x cos y – sin x sin y

**Question (7 ) Prove that**

**Answer**

**Question (8) Prove that**

**Answer**

Observe that cos x repeats the same value after an interval

2 π and sin x repeats the same value after an interval 2 π

**Question 9**

**Answer**

We know that cot x repeats the same value after an interval 2π

**Question (10) Prove that sin (n+1) x sin ( n+2 ) x + cos (n+1) x cos (n+2) x = cos x**

**Answer**

cos(x-y) = cos x cos y + sin x sin y

L.H.S.= sin (n+1) x sin (n+2) x + cos (n+1) x cos (n+2) x

=cos [(n+1) x – ( n+2 ) x] = cos [ (n+1) x – (n+2) x]
=cos (-x)

=cos x

= R.H.S

**Question (11) Prove that
**

**Answer
**

**Question (12) Sin ^{2}6x – sin ^{2} 4x = Sin 2x sin 10x **

**Answer
**

**Question (13) cos ^{2} 2x – cos^{2} 6x = sin 4x sin 8x**

**Answer**

**Question (14) sin ^{2} x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x**

**Answer
**

**Question (15) cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)**

**Answer
**

**Question (16)**

**Answer**

**Question (17)**

**Answer**

**Question (18)**

**Answer**

**Question (19) Prove that**

**Answer**

**Question (20) Prove that**

**Answer**

**Question (21)**

**Answer**

**Question (22) cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1**

**Answer**

**Question (23)**

**Answer**

**Question (24) cos 4x = 1 – 8sin ^{2} x cos^{2} x**

**Answer **

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

And sin 2x=2sin x cos x

L.H.S.=cos 4x

=cos 2(2x)

=1-2sin^{2}2x

=1-2(2sin x cos x)^{2
}L.H.S =1- 8sin^{2}x cos^{2}x

= R.H.S.

Hence proved.

**Question (25) cos 6x = 32 cos ^{6} x – 48cos^{4} x + 18 cos^{2} x – 1**

**Answer **

cos 3A = 4cos^{3}A – 3cosA

cos 2x= 1 – 2 sin2x

cos 2*x* = 2 cos^{2} *x *– 1

L.H.S. = Cos 6x = Cos cos 3(2*x*)

= 4 cos^{3} 2*x* – 3 cos^{ }2*x
*= 4 [(2 cos

^{2}

*x*– 1)

^{3}– 3 (2 cos

^{2}

*x*– 1)

= 4 [(2 cos

^{2}

*x*)

^{3}– (1)

^{3}– 3 (2 cos

^{2}

*x*)

^{2}+ 3 (2 cos

^{2}

*x*)] – 6cos

^{2}

*x*+ 3

= 4 [8cos

^{6}

*x*– 1 – 12 cos

^{4}

*x*+ 6 cos

^{2}

*x*] – 6 cos

^{2}

*x*+ 3

= 32 cos

^{6}

*x*– 4 – 48 cos

^{4}

*x*+ 24 cos

^{2}

*x*– 6 cos

^{2}

*x*+ 3

= 32 cos

^{6}

*x*– 48 cos

^{4}

*x*+ 18 cos

^{2}

*x*– 1

= R.H.S

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