**Miscellaneous Exercise**

Trigonometric Functions

**Question and Answers**

**Class 11 – Maths**

Trigonometric Functions

Class | Class 11 |

Subject | Mathematics |

Chapter Name | Trigonometric Functions |

Chapter No. | Chapter 3 |

Exercise | Miscellaneous Exercise |

Category | Class 11 Maths NCERT Solutions |

**Question (1) Prove that**

**Answer**

**Question (2) Prove that (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0**

**Answer**

LHS = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x

LHS = sin 3x sin x + sin2 x + cos 3x cos x – cos2 x

Taking out the common terms,

LHS = cos 3x cos x + sin 3x sin x – (cos^{2} x – sin^{2} x)

Using the formula

cos (A – B) = cos A cos B + sin A sin B

= cos (3x – x) – cos 2x

So, we get

= cos 2x – cos 2x

= 0

= RHS

**Question (3) Prove that**

**Answer **

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

L.H.S=(cos x + cos y)^{2}+(sin x – sin y)^{2
}By expanding using the formula, we get

LHS = cos^{2} x + cos^{2} y + 2 cos x cos y + sin^{2} x + sin^{2} y – 2 sin x sin y

Grouping the terms,

= (cos^{2} x + sin^{2} x) + (cos^{2} y + sin^{2} y) + 2 (cos x cos y – sin x sin y)

Using the formula cos (A + B) = (cos A cos B – sin A sin B)

= 1 + 1 + 2 cos (x + y)

By further calculation,

= 2 + 2 cos (x + y)

Taking 2 as common

= 2 [1 + cos (x + y)]
From the formula cos 2A = 2 cos^{2} A – 1

Therefore , LHS =RHS

Hence Proved

**Question (4) Prove that**

**Answer**

LHS = (cos x – cos y) ^{2} + (sin x – sin y) ^{2
}By expanding using the formula,

= cos^{2} x + cos^{2} y – 2 cos x cos y + sin^{2} x + sin^{2} y – 2 sin x sin y

Grouping the terms,

= (cos^{2} x + sin^{2} x) + (cos^{2} y + sin^{2} y) – 2 (cos x cos y + sin x sin y)

Using the formula cos (A – B) = cos A cos B + sin A sin B

= 1 + 1 – 2 [cos (x – y)]
By further calculation,

= 2 [1 – cos (x – y)]
From formula cos 2A = 1 – 2 sin^{2} A

LHS =RHS

Hence proved

**Question (5) Prove that sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x**

**Answer**

L.H.S. =sin x + sin 3x+sin 5x+sin 7x

LHS=(sin x + sin 5x)+(sin 3x+sin 7x)

LHS =(sin x + sin 5x)+(sin 3x+sin 7x)

L.H.S = 4 cos 2x sin 4x cos x = R.H.S

**Question (6) Prove that**

**Answer**

Therefore L.H.S = R.H.S

**Question (7) Prove that sin 3x + sin 2x – sin x = 4 sin x cos ( x /2) cos (3x /2)****Answer**

**Question (8) Find**

**Answer**

Here, x is in 2^{nd} quadrant.

π / 2 < x < π

Dividing by 2

π / 4 < x/2 < π /2

Hence x/2 lies in the 1^{st} quadrant

Therefore , sin x/2 , cos x/2 and tan x/2 are positive

Given that tan = -4/3

sec^{2}x=1+tan^{2}x

sec^{2}x=1+ (-4/3)^{2
}sec^{2}x=1+ (16/9)

sec^{2}x= 25/9

sec x = ± 5/3

As x is in 2^{nd} quadrant, sec x is negative.

sec x = – 5/3

Then cos x= -3/5

2Cos^{2} x/2 = cos x + 1

2Cos^{2} x/2 = 2/5

Cos x/2 = 1/ √5 = √5 / 5

sin^{2}x/ 2 + cos^{2}x/2 =1

Sin^{2} x/2 + ( 1/ √5 )^{2 }=1

Sin^{2} x/2 = 4/5

Sin x/2 = 2 / √5

Sin x/2 = 2 √5 / 5

Sin x/2, cos x/2, and tan x/2 are 2√5 / 5, √5 / 5, 2

**Question (9) Find **

**Answer**

π < x < 3 π / 2

^{2}x/2 = cos x + 1

^{2}x/2 = (-1/3) + 1

^{2}x/2 = 2/3

^{2}x+cos

^{2}x=1

The values of sin x/2 , cos x/2 and tan x/2 are √2/ √3 , -1/3 , – √2

**Question (10) Find**

**Answer**

x lies in 2^{nd} quadrant

π /2 < x < π

Dividing by 2

π / 4 < x/2 < π /2

X/2 lies in the 1st quadrant therefore sin x/2, cos x/2, and tan x/2 are positive.

sin x =1/4

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

cos^{2} x = 1 – sin^{2}x

cos^{2} x = 1 – (1/4)^{2
}cos^{2} x = 15/16

cos x= – √15 /4

2 sin^{2 }x/2 = 1 – cos x/ 2

2 sin^{2 }x/2 = 1 + √15 /4

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