RD Sharma Solutions for Class 11 Chapter 7 - Values of Trigonometric Functions at Sum or Difference of Angles

In this chapter, we shall derive formulae which expresses the values of trigonometric functions at the sum or difference of two real numbers (or angles) in terms of the values of trigonometric functions at individual numbers (or angles). Experts at BYJU’S have formulated the RD Sharma Class 11 Solutions for Maths in the most lucid and easy manner. Solutions are developed using shortcut techniques to help students grasp the concepts faster and to make learning fun. The solutions to this chapter are available in the pdf format, which can be downloaded easily from the links provided below.

Chapter 7- Values of Trigonometric Functions at Sum or Difference of Angles contains two exercises and the RD Sharma Solutions present in this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.

  • Values of trigonometric functions at the sum or difference.
    • Cosine of the difference and sum of two numbers.
    • The sine of the difference and sum of two numbers.
    • Tangent of the difference and sum of two numbers.
  • Few theorems.
  • Maximum and minimum values of trigonometrical expressions.
  • To express the given expressions in the desired form.

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Exercise 7.1 Solutions

Exercise 7.2 Solutions

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EXERCISE 7.1 PAGE NO: 7.19

1. If sin A = 4/5 and cos B = 5/13, where 0 <A, B < π/2, find the values of the following:
(i) sin (A + B)
(ii) cos (A + B)
(iii) sin (A – B)
(iv) cos (A – B)

Solution:

Given:

sin A = 4/5 and cos B = 5/13

We know that cos A = √(1 – sin2 A) and sin B = √(1 – cos2 B), where 0 <A, B < π/2

So let us find the value of sin A and cos B

cos A = √(1 – sin2 A)

= √(1 – (4/5)2)

= √(1 – (16/25))

= √((25 – 16)/25)

= √(9/25)

= 3/5

sin B = √(1 – cos2 B)

= √(1 – (5/13)2)

= √(1 – (25/169))

= √(169 – 25)/169)

= √(144/169)

= 12/13

(i) sin (A + B)

We know that sin (A +B) = sin A cos B + cos A sin B

So,

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

= 4/5 × 5/13 + 3/5 × 12/13

= 20/65 + 36/65

= (20+36)/65

= 56/65

(ii) cos (A + B)

We know that cos (A +B) = cos A cos B – sin A sin B

So,

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

= 3/5 × 5/13 – 4/5 × 12/13

= 15/65 – 48/65

= -33/65

(iii) sin (A – B)

We know that sin (A – B) = sin A cos B – cos A sin B

So,

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

= 4/5 × 5/13 – 3/5 × 12/13

= 20/65 – 36/65

= -16/65

(iv) cos (A – B)

We know that cos (A -B) = cos A cos B + sin A sin B

So,

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

= 3/5 × 5/13 + 4/5 × 12/13

= 15/65 + 48/65

= 63/65

2. (a) If Sin A = 12/13 and sin B = 4/5, where π/2<A < π and 0 <B < π/2, find the following:
(i) sin (A + B) (ii) cos (A + B)

(b) If sin A = 3/5, cos B = –12/13, where A and B, both lie in second quadrant, find the value of sin (A +B).

Solution:

(a) Given:

Sin A = 12/13 and sin B = 4/5, where π/2<A < π and 0 <B < π/2

We know that cos A = – √(1 – sin2 A) and cos B = √(1 – sin2 B)

So let us find the value of cos A and cos B

cos A = – √(1 – sin2 A)

= – √(1 – (12/13)2)

= – √(1-144/169)

= -√((169-144)/169)

= – √(25/169)

= – 5/13

cos B = √(1 – sin2 B)

= √(1 – (4/5)2)

= √(1-16/25)

= √((25-16)/25)

=√(9/25)

= 3/5

(i) sin (A +B)

We know that sin (A + B) = sin A cos B + cos A sin B

So,

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

= 12/13 × 3/5 + (-5/13) × 4/5

= 36/65 – 20/65

= 16/65

(ii) cos (A + B)

We know that cos (A +B) = cos A cos B – sin A sin B

So,

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

= -5/13 × 3/5 – 12/13 × 4/5

= -15/65 – 48/65

= – 63/65

(b) Given:

sin A = 3/5, cos B = –12/13, where A and B, both lie in second quadrant.

We know that cos A = – √(1 – sin2 A) and sin B = √(1 – cos2 B)

So let us find the value of cos A and sin B

cos A = – √(1 – sin2 A)

= – √(1 – (3/5)2)

= – √(1- 9/25)

= – √((25-9)/25)

= – √(16/25)

= – 4/5

sin B = √(1 – cos2 B)

= √(1 – (-12/13)2)

= √(1 – 144/169)

= √((169-144)/169)

= √(25/169)

= 5/13

We need to find sin (A + B)

Since, sin (A + B) = sin A cos B + cos A sin B

= 3/5 × (-12/13) + (-4/5) × 5/13

= -36/65 – 20/65

= -56/65

3. If cos A = – 24/25 and cos B = 3/5, where π <A < 3π/2 and 3π/2 <B < 2π, find the following:
(i) sin (A + B) (ii) cos (A + B)

Solution:

Given:

cos A = – 24/25 and cos B = 3/5, where π <A < 3π/2 and 3π/2 <B < 2π

We know that A is in third quadrant, B is in fourth quadrant. So sine function is negative.

By using the formulas,

sin A = – √(1 – cos2 A) and sin B = -√(1 – cos2 B)

So let us find the value of sin A and sin B

sin A = – √(1 – cos2 A)

= – √(1-(-24/25)2)

= – √(1-576/625)

= – √((625-576)/625)

= – √(49/625)

= -7/25

sin B = -√(1 – cos2 B)

= – √(1-(3/5)2)

= – √(1-9/25)

= – √((25-9)/25)

= – √(16/25)

= – 4/5

(i) sin (A + B)

We know that sin (A + B) = sin A cos B + cos A sin B

So,

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

= -7/25 × 3/5 + (-24/25) × (-4/5)

= -21/125 + 96/125

= 75/125

= 3/5

(ii) cos (A + B)

We know that cos (A + B) = cos A cos B – sin A sin B

So,

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

= (-24/25) × 3/5 – (-7/25) × (-4/5)

= -72/125 – 28/125

= -100/125

= – 4/5

4. If tan A = 3/4, cos B = 9/41, where π<A < 3π/2 and 0 <B < π/2, find tan (A + B).

Solution:

Given:

tan A = 3/4 and cos B = 9/41, where π <A < 3π/2 and 0 <B < π/2

We know that, A is in third quadrant, B is in first quadrant.

So, tan function And sine function are positive.

By using the formula,

sin B = √(1 – cos2 B)

Let us find the value of sin B.

sin B = √(1 – cos2 B)

= √(1- (9/41)2)

= √(1- 81/1681)

= √((1681-81)/1681)

= √(1600/1681)

= 40/41

We know, tan B = sin B/cos B

= (40/41) / (9/41)

= 40/9

So, tan (A + B) = (tan A + tan B) / (1 – tan A tan B)

= (3/4 + 40/9) / (1 – 3/4 × 40/9)

= (187/36) / (1- 120/36)

= (187/36) / ((36-120)/36)

= (187/36) / (-84/36)

= -187/84

5. If sin A = 1/2, cos B = 12/13, where π/2<A < π and 3π/2 <B < 2π, find tan(A – B).

Solution:

Given:

sin A = 1/2, cos B = 12/13, where π/2<A < π and 3π/2 <B < 2π

We know that, A is in second quadrant, B is in fourth quadrant.

In the second quadrant, sine function is positive, cosine and tan functions are negative.

In the fourth quadrant, sine and tan functions are negative, cosine function is positive.

By using the formulas,

cos A = – √(1 – sin2 A) and sin B = -√(1 – cos2 B)

So let us find the value of cos A and sin B

cos A = – √(1 – sin2 A)

= – √(1 – (1/2)2)

= – √(1- 1/4)

= – √((4-1)/4)

= – √(3/4)

= -√3/2

sin B = -√(1 – cos2 B)

= – √(1-(12/13)2)

= – √(1- 144/169)

= – √((169-144)/169)

= – √(25/169)

= – 5/13

We know, tan A = sin A / cos A and tan B = sin B / cos B

tan A = (1/2)/( -√3/2) = -1/√3 and

tan B = (-5/13)/(12/13) = -5/12

So, tan (A – B) = (tan A – tan B) / (1 + tan A tan B)

= ((-1/√3) – (-5/12)) / (1 + (-1/√3) × (-5/12))

= ((-12+5√3)/12√3) / (1 + 5/12√3)

= ((-12+5√3)/12√3) / ((12√3 + 5)/12√3)

= (5√3 – 12) / (5 + 12√3)

6. If sin A = 1/2, cos B = √3/2, where π/2<A < π and 0 <B < π/2, find the following:
(i) tan (A + B) (ii) tan (A – B)

Solution:

Given:

Sin A = 1/2 and cos B = √3/2, where π/2 <A < π and 0 <B < π/2

We know that, A is in second quadrant, B is in first quadrant.

In the second quadrant, sine function is positive. cosine and tan functions are negative.

In first quadrant, all functions are positive.

By using the formulas,

cos A = – √(1 – sin2 A) and sin B = √(1 – cos2 B)

So let us find the value of cos A and sin B

cos A = – √(1 – sin2 A)

= – √(1 – (1/2)2)

= – √(1- 1/4)

= – √((4-1)/4)

= – √(3/4)

= -√3/2

sin B = √(1 – cos2 B)

= √(1-(√3/2)2)

= √(1- 3/4)

= √((4-3)/4)

= √(1/4)

= 1/2

We know, tan A = sin A / cos A and tan B = sin B / cos B

tan A = (1/2)/( -√3/2) = -1/√3 and

tan B = (1/2)/(√3/2) = 1/√3

(i) tan (A + B) = (tan A + tan B) / (1 – tan A tan B)

= (-1/√3 + 1/√3) / (1 – (-1/√3) × 1/√3)

= 0 / (1 + 1/3)

= 0

(ii) tan (A – B) = (tan A – tan B) / (1 + tan A tan B)

= ((-1/√3) – (1/√3)) / (1 + (-1/√3) × (1/√3))

= ((-2/√3) / (1 – 1/3)

= ((-2/√3) / (3-1)/3)

= ((-2/√3) / 2/3)

= – √3

7. Evaluate the following:
(i) sin 780 cos 180 – cos 780 sin 180 

(ii) cos 470 cos 130 – sin 470 sin 130
(iii) sin 360 cos 90 + cos 360 sin 90 

(iv) cos 800 cos 200 + sin 800 sin 200

Solution:

(i) sin 780 cos 180 – cos 780 sin 180

We know that sin (A – B) = sin A cos B – cos A sin B

sin 780 cos 180 – cos 780 sin 180 = sin(78 – 18) °

= sin 60°

= √3/2

(ii) cos 470 cos 130 – sin 470 sin 130

We know that cos A cos B – sin A sin B = cos (A + B)

cos 470 cos 130 – sin 470 sin 130 = cos (47 + 13) °

= cos 60°

= 1/2

(iii) sin 360 cos 90 + cos 360 sin 90

We know that sin (A +B) = sin A cos B + cos A sin B

sin 360 cos 90 + cos 360 sin 90 = sin (36 + 9) °

= sin 45°

= 1/√2

(iv) cos 800 cos 200 + sin 800 sin 200

We know that cos A cos B + sin A sin B = cos (A – B)

cos 800 cos 200 + sin 800 sin 200 = cos (80 – 20) °

= cos 60°

= ½

8. If cos A = –12/13 and cot B = 24/7, where A lies in the second quadrant and B in the third quadrant, find the values of the following:
(i) sin (A + B) (ii) cos (A + B) (iii) tan (A + B)

Solution:

Given:

cos A = -12/13 and cot B = 24/7

We know that, A lies in second quadrant, B in the third quadrant.

In the second quadrant sine function is positive.

In the third quadrant, both sine and cosine functions are negative.

By using the formulas,

sin A = √(1 – cos2 A), sin B = – 1/√(1 + cot2 B) and cos B = -√(1 – sin2 B),

So let us find the value of sin A and sin B

sin A = √(1 – cos2 A)

= √(1 – (-12/13)2)

= √(1 – 144/169)

= √((169-144)/169)

= √(25/169)

= 5/13

sin B = – 1/√(1 + cot2 B)

= – 1/√(1 + (24/7)2)

= – 1/√(1 + 576/49)

= -1/√((49+576)/49)

= -1/√(625/49)

= -1/(25/7)

= -7/25

cos B = -√(1 – sin2 B)

= -√(1-(-7/25)2)

= -√(1-(49/625))

= -√((625-49)/625)

= -√(576/625)

= -24/25

So, now let us find

(i) sin (A + B)

We know that sin (A + B) = sin A cos B + cos A sin B

So,

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

= 5/13 × (-24/25) + (-12/13) × (-7/25)

= -120/325 + 84/325

= -36/325

(ii) cos (A + B)

We know that cos (A + B) = cos A cos B – sin A sin B

So,

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

= -12/13 × (-24/25) – (5/13) × (-7/25)

= 288/325 + 35/325

= 323/325

(iii) tan (A + B)

We know that tan (A + B) = sin (A+B) / cos (A+B)

= (-36/325) / (323/325)

= -36/323

9. Prove that: cos 7π/12 + cos π/12 = sin 5π/12 – sin π/12

Solution:

We know that, 7π/12 = 105°, π/12 = 15°; 5π/12 = 75°

Let us consider LHS: cos 105° + cos 15°

cos (90° + 15°) + sin (90° – 75°)

-sin 15° + sin 75°

sin 75° – sin 15°

= RHS

∴ LHS = RHS

Hence proved.

10. Prove that: (tan A + tan B) / (tan A – tan B) = sin (A + B) / sin (A – B)  

Solution:

Let us consider LHS: (tan A + tan B) / (tan A – tan B)

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles image- 1

= RHS

∴ LHS = RHS

Hence proved.

11. Prove that:
(i) (cos 11o + sin 11o) / (cos 11o – sin 11o) = tan 56o  

(ii) (cos 9o + sin 9o) / (cos 9o – sin 9o) = tan 54o

(iii) (cos 8o – sin 8o) / (cos 8o + sin 8o) = tan 37o

Solution:

(i) (cos 11o + sin 11o) / (cos 11o – sin 11o) = tan 56o  

Let us consider LHS:

(cos 11o + sin 11o) / (cos 11o – sin 11o)

Now let us divide the numerator and denominator by cos 11o we get,

(cos 11o + sin 11o) / (cos 11o – sin 11o) = (1 + tan 11o) / (1 – tan 11o)

= (1 + tan 11o) / (1- 1×tan 11o)

= (tan 45o + tan 11o) / (1 – tan 45o × tan 11o)

We know that tan (A+B) = (tan A + tan B) / (1 – tan A tan B)

So,

(tan 45o + tan 11o) / (1 – tan 45o × tan 11o) = tan (45o + 11o)

= tan 56o

= RHS

∴ LHS = RHS

Hence proved.

(ii) (cos 9o + sin 9o) / (cos 9o – sin 9o) = tan 54o

Let us consider LHS:

(cos 9o + sin 9o) / (cos 9o – sin 9o)

Now let us divide the numerator and denominator by cos 9o we get,

(cos 9o + sin 9o) / (cos 9o – sin 9o) = (1 + tan 9o) / (1 – tan 9o)

= (1 + tan 9o) / (1 – 1 × tan 9o)

= (tan 45o + tan 9o) / (1 – tan 45o × tan 9o)

We know that tan (A+B) = (tan A + tan B) / (1 – tan A tan B)

So,

(tan 45o + tan 9o) / (1 – tan 45o × tan 9o) = tan (45o + 9o)

= tan 54o

= RHS

∴ LHS = RHS

Hence proved.

(iii) (cos 8o – sin 8o) / (cos 8o + sin 8o) = tan 37o

Let us consider LHS:

(cos 8o – sin 8o) / (cos 8o + sin 8o)

Now let us divide the numerator and denominator by cos 8o we get,

(cos 8o – sin 8o) / (cos 8o + sin 8o) = (1 – tan 8o) / (1 + tan 8o)

= (1 – tan 8o) / (1 + 1×tan 8o)

= (tan 45o – tan 8o) / (1 + tan 45o ×tan 8o)

We know that tan (A+B) = (tan A + tan B) / (1 – tan A tan B)

So,

(tan 45o – tan 8o) / (1 + tan 45o ×tan 8o) = tan (45o – 8o)

= tan 37o

= RHS

∴ LHS = RHS

Hence proved.

12. Prove that:

(i)

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles image- 2

(ii)

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles image- 3

(iii)

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles image- 4

Solution:

(i)

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles image- 5

= sin 90o

= 1

= RHS

∴ LHS = RHS

Hence proved.

(ii)

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles image- 6

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles image- 7

= sin 60o

= √3/2

= RHS

∴ LHS = RHS

Hence proved.

(iii)

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles image- 8

= sin 90o

= 1

= RHS

∴ LHS = RHS

Hence proved.

13. Prove that: (tan 69o + tan 66o) / (1 – tan 69o tan 66o) = -1

Solution:

Let us consider LHS:

(tan 69o + tan 66o) / (1 – tan 69o tan 66o)

We know that, tan (A + B) = (tan A + tan B) / (1 – tan A tan B)

Here, A = 69o and B = 66o

So,

(tan 69o + tan 66o) / (1 – tan 69o tan 66o) = tan (69 + 66)o

= tan 135o

= – tan 45o

= – 1

= RHS

∴ LHS = RHS

Hence proved.

14. (i) If tan A = 5/6 and tan B = 1/11, prove that A + B = π/4

(ii) If tan A = m/(m–1) and tan B = 1/(2m – 1), then prove that A – B = π/4

Solution:

(i) If tan A = 5/6 and tan B = 1/11, prove that A + B = π/4

Given:

tan A = 5/6 and tan B = 1/11

We know that, tan (A + B) = (tan A + tan B) / (1 – tan A tan B)

= [(5/6) + (1/11)] / [1 – (5/6) × (1/11)]

= (55+6) / (66-5)

= 61/61

= 1

= tan 45o or tan π/4

So, tan (A + B) = tan π/4

∴ (A + B) = π/4

Hence proved.

(ii) If tan A = m/(m–1) and tan B = 1/(2m – 1), then prove that A – B = π/4

Given:

tan A = m/(m–1) and tan B = 1/(2m – 1)

We know that, tan (A – B) = (tan A – tan B) / (1 + tan A tan B)

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles image- 9

= (2m2 – m – m + 1) / (2m2 – m – 2m + 1 + m)

= (2m2 – 2m + 1) / (2m2 – 2m + 1)

= 1

= tan 45o or tan π/4

So, tan (A – B) = tan π/4

∴ (A – B) = π/4

Hence proved.

15. prove that:
(i) cos2 π/4 – sin2 π/12 = √3/4

(ii) sin2 (n + 1) A – sin2nA = sin (2n + 1) A sin A

Solution:

(i) cos2 π/4 – sin2 π/12 = √3/4

Let us consider LHS:

cos2 π/4 – sin2 π/12

We know that, cos2A – sin2 B = cos (A + B) cos (A – B)

So,

cos2 π/4 – sin2 π/12 = cos (π/4 + π/12) cos (π/4 – π/12)

= cos 4π/12 cos 2π/12

= cos π/3 cos π/6

= 1/2 × √3/2

= √3/4

= RHS

∴ LHS = RHS

Hence proved.

(ii) sin2 (n + 1) A – sin2nA = sin (2n + 1) A sin A

Let us consider LHS:

sin2 (n + 1) A – sin2nA

We know that, sin2A – sin2 B = sin (A + B) sin (A – B)

Here, A = (n + 1) A and B = nA

So,

sin2 (n + 1) A – sin2n A = sin ((n + 1) A + nA) sin ((n + 1) A – nA)

= sin (nA +A + nA) sin (nA +A – nA)

= sin (2nA +A) sin (A)

= sin (2n + 1) A sin A

= RHS

∴ LHS = RHS

Hence proved.

16. Prove that:

(i)

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles image- 10

(ii)

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles image- 11

(iii)

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles image- 12

(iv) sin2 B = sin2 A + sin2 (A-B) – 2sin A cos B sin (A – B)

(v) cos2 A + cos2 B – 2 cos A cos B cos (A +B) = sin2 (A + B)

(vi)

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles image- 13

Solution:

(i)

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles image- 14

= tan A

= RHS

∴ LHS = RHS

Hence proved.

(ii)

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles image- 15

Let us consider LHS:

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles image- 16

= tan A – tan B + tan B – tan C + tan C – tan A

= 0

= RHS

∴ LHS = RHS

Hence proved.

(iii)

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles image- 17

= cotB – cotA + cot C – cotB + cotA – cot C

= 0

= RHS

∴ LHS = RHS

Hence proved.

(iv) sin2 B = sin2 A + sin2 (A-B) – 2sin A cos B sin (A – B)

Let us consider RHS:

sin2A + sin2 (A -B) – 2 sin A cos B sin (A – B)

sin2A + sin (A -B) [sin (A –B) – 2 sin A cos B]

We know that, sin (A –B) = sin A cos B – cos A sin B

So,

sin2A + sin (A -B) [sin A cos B – cos A sin B – 2 sin A cos B]

sin2A + sin (A -B) [-sin A cos B – cos A sin B]

sin2A – sin (A -B) [sin A cos B + cos A sin B]

We know that, sin (A +B) = sin A cos B + cos A sin B

So,

sin2A – sin (A – B) sin (A + B)

sin2 A – sin2 A + sin2 B

sin2 B

= LHS

∴ LHS = RHS

Hence proved.

(v) cos2 A + cos2 B – 2 cos A cos B cos (A + B) = sin2 (A + B)

Let us consider LHS:

cos2A + cos2B – 2 cos A cos B cos (A +B)

cos2A + 1 – sin2B – 2 cos A cos B cos (A +B)

1 + cos2A – sin2B – 2 cos A cos B cos (A +B)

We know that, cos2A – sin2B = cos (A +B) cos (A –B)

So,

1 + cos (A +B) cos (A –B) – 2 cos A cos B cos (A +B)

1 + cos (A +B) [cos (A –B) – 2 cos A cos B]

We know that, cos (A – B) = cos A cos B + sin A sin B.

So,

1 + cos (A +B) [cos A cos B + sin A sin B – 2 cos A cos B]

1 + cos (A +B) [-cos A cos B + sin A sin B]

1 – cos (A +B) [cos A cos B – sin A sin B]

We know that, cos (A +B) = cos A cos B – sin A sin B.

So,

1 – cos2 (A + B)

sin2 (A + B)

= RHS

∴ LHS = RHS

Hence proved.

(vi)

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles image- 18

∴ LHS = RHS

Hence proved.

17. Prove that:
(i) tan 8x – tan 6x – tan 2x = tan 8x tan 6x tan 2x

(ii) tan π/12 + tan π/6 + tan π/12 tan π/6 = 1

(iii) tan 36o + tan 9o + tan 36o tan 9o = 1

(iv) tan 13x – tan 9x – tan 4x = tan 13x tan 9x tan 4x

Solution:

(i) tan 8x – tan 6x – tan 2x = tan 8x tan 6x tan 2x

Let us consider LHS:

tan 8x – tan 6x – tan 2x

tan 8x = tan(6x + 2x)

We know that, tan (A + B) = (tan A + tan B) / (1 – tan A tan B)

So,

tan 8x = (tan 6x + tan 2x) / (1 – tan 6x tan 2x)

By cross-multiplying we get,

tan 8x (1 – tan 6x tan 2x) = tan 6x + tan 2x

tan 8x – tan 8x tan 6x tan2x = tan 6x + tan 2x

Upon rearranging we get,

tan 8x – tan 6x – tan 2x = tan 8x tan 6x tan 2x

= RHS

∴ LHS = RHS

Hence proved.

(ii) tan π/12 + tan π/6 + tan π/12 tan π/6 = 1

We know,

π/12 = 15° and π/6 = 30°

So, we have 15° + 30° = 45°

Tan (15° + 30°) = tan 45°

We know that, tan (A + B) = (tan A + tan B) / (1 – tan A tan B)

So,

(tan 15o + tan 30o) / (1 – tan 15o tan 30o) = 1

tan 15° + tan 30° = 1 – tan 15° tan 30°

Upon rearranging we get,

tan15° + tan30° + tan15° tan30° = 1

Hence proved.

(iii) tan 36o + tan 9o + tan 36o tan 9o = 1

We know 36° + 9° = 45°

So we have,

tan (36° + 9°) = tan 45°

We know that, tan (A + B) = (tan A + tan B) / (1 – tan A tan B)

So,

(tan 36o + tan 9o) / (1 – tan 36o tan 9o) = 1

tan 36° + tan 9° = 1 – tan 36° tan 9°

Upon rearranging we get,

tan 36° + tan 9° + tan 36° tan 9° = 1

Hence proved.

(iv) tan 13x – tan 9x – tan 4x = tan 13x tan 9x tan 4x

Let us consider LHS:

tan 13x – tan 9x – tan 4x

tan 13x = tan (9x + 4x)

We know that, tan (A + B) = (tan A + tan B) / (1 – tan A tan B)

So,

tan 13x = (tan 9x + tan 4x) / (1 – tan 9x tan 4x)

By cross-multiplying we get,

tan 13x (1 – tan 9x tan 4x) = tan 9x + tan 4x

tan 13x – tan 13x tan 9x tan 4x = tan 9x + tan 4x

Upon rearranging we get,

tan 13x – tan 9x – tan 4x = tan 13x tan 9x tan 4x

= RHS

∴ LHS = RHS

Hence proved.

EXERCISE 7.2 PAGE NO: 7.26

1. Find the maximum and minimum values of each of the following trigonometrical expressions:

(i) 12 sin x – 5 cos x
(ii) 12 cos x + 5 sin x + 4
(iii) 5 cos x + 3 sin (π/6 – x) + 4 
(iv) sin x – cos x + 1

Solution:

We know that the maximum value of A cos α + B sin α + C is C + √(A2 +B2),

And the minimum value is C – √(a2 +B2).


(i) 12 sin x – 5 cos x

Given: f(x) = 12 sin x – 5 cos x

Here, A = -5, B = 12 and C = 0

((-5)2 + 122) ≤ 12 sin x – 5 cos x ≤ ((-5)2 + 122)

(25+144) ≤ 12 sin x – 5 cos x ≤ (25+144)

169 ≤ 12 sin x – 5 cos x ≤ 169

-13 ≤ 12 sin x – 5 cos x ≤ 13

Hence, the maximum and minimum values of f(x) are 13 and -13 respectively.

(ii) 12 cos x + 5 sin x + 4

Given: f(x) = 12 cos x + 5 sin x + 4

Here, A = 12, B = 5 and C = 4

4 – (122 + 52) ≤ 12 cos x + 5 sin x + 4 ≤ 4 + (122 + 52)

4 – (144+25) ≤ 12 cos x + 5 sin x + 4 ≤ 4 + (144+25)

4 –169 ≤ 12 cos x + 5 sin x + 4 ≤ 4 + 169

-9 ≤ 12 cos x + 5 sin x + 4 ≤ 17

Hence, the maximum and minimum values of f(x) are -9 and 17 respectively.

(iii) 5 cos x + 3 sin (π/6 – x) + 4 

Given: f(x) = 5 cos x + 3 sin (π/6 – x) + 4 

We know that, sin (A – B) = sin A cos B – cos A sin B

f(x) = 5 cos x + 3 sin (π/6 – x) + 4 

= 5 cos x + 3 (sin π/6 cos x – cos π/6 sin x) + 4

= 5 cos x + 3/2 cos x – 33/2 sin x + 4

= 13/2 cos x – 33/2 sin x + 4

So, here A = 13/2, B = – 33/2, C = 4

4 – [(13/2)2 + (-33/2)2] ≤ 13/2 cos x – 33/2 sin x + 4 ≤ 4 + [(13/2)2 + (-33/2)2]

4 – [(169/4) + (27/4)] ≤ 13/2 cos x – 33/2 sin x + 4 ≤ 4 + [(169/4) + (27/4)]

4 – 7 ≤ 13/2 cos x – 33/2 sin x + 4 ≤ 4 + 7

-3 ≤ 13/2 cos x – 33/2 sin x + 4 ≤ 11

Hence, the maximum and minimum values of f(x) are -3 and 11 respectively.

(iv) sin x – cos x + 1

Given: f(x) = sin x – cos x + 1

So, here A = -1, B = 1 And c = 1

1 – [(-1)2 + 12] ≤ sin x – cos x + 1 ≤ 1 + [(-1)2 + 12]

1 – (1+1) ≤ sin x – cos x + 1 ≤ 1 + (1+1)

1 – 2 ≤ sin x – cos x + 1 ≤ 1 + 2

Hence, the maximum and minimum values of f(x) are 1 – 2 and 1 + 2 respectively.

2. Reduce each of the following expressions to the Sine and Cosine of a single expression:

(i) √3 sin x – cos x

(ii) cos x – sin x

(iii) 24 cos x + 7 sin x

Solution:

(i) √3 sin x – cos x

Let f(x) = √3 sin x – cos x

Dividing and multiplying by √((√3)2 + 12) i.e. by 2

f(x) = 2(√3/2 sin x – 1/2 cos x)

Sine expression:

f(x) = 2(cos π/6 sin x – sin π/6 cos x) (since, √3/2 = cos π/6 and 1/2 = sin π/6)

We know that, sin A cos B – cos A sin B = sin (A – B)

f(x) = 2 sin (x – π/6)

Again,

f(x) = 2(√3/2 sin x – 1/2 cos x)

Cosine expression:

f(x) = 2(sin π/3 sin x – cos π/3 cos x)

We know that, cos A cos B – sin A sin B = cos (A + B)

f(x) = -2 cos(π/3 + x)

(ii) cos x – sin x

Let f(x) = cos x – sin x

Dividing and multiplying by √(12 + 12) i.e. by √2,

f(x) = √2(1/√2 cos x – 1/√2 sin x)


Sine expression:

f(x) = √2(sin π/4 cos x – cos π/4 sin x) (since, 1/√2 = sin π/4 and 1/√2 = cos π/4)

We know that sin A cos B – cos A sin B = sin (A – B)

f(x) = √2 sin (π/4 – x)

Again,

f(x) = √2(1/√2 cos x – 1/√2 sin x)

Cosine expression:

f(x) = 2(cos π/4 cos x – sin π/4 sin x)

We know that cos A cos B – sin A sin B = cos (A + B)

f(x) = √2 cos (π/4 + x)

(iii) 24 cos x + 7 sin x

Let f(x) = 24 cos x + 7 sin x

Dividing and multiplying by √((√24)2 + 72) = √625 i.e. by 25,

f(x) = 25(24/25 cos x + 7/25 sin x)

Sine expression:

f(x) = 25(sin α cos x + cos α sin x) where, sin α = 24/25 and cos α = 7/25

We know that sin A cos B + cos A sin B = sin (A + B)

f(x) = 25 sin (α + x)

Cosine expression:

f(x) = 25(cos α cos x + sin α sin x) where, cos α = 24/25 and sin α = 7/25

We know that cos A cos B + sin A sin B = cos (A – B)

f(x) = 25 cos (α – x)

3. Show that Sin 100o – Sin 10o is positive.

Solution:

Let f(x) = sin 100° – sin 10°

Dividing And multiplying by √(12 + 12) i.e. by √2,

f(x) = √2(1/√2 sin 100o – 1/√2 sin 10o)

f(x) = √2(cos π/4 sin (90+10)o – sin π/4 sin 10o) (since, 1/√2 = cos π/4 and 1/√2 = sin π/4)

f(x) = √2(cos π/4 cos 10o – sin π/4 sin 10o)

We know that cos A cos B – sin A sin B = cos (A + B)

f(x) = √2 cos (π/4 + 10o)

∴ f(x) = √2 cos 55°

4. Prove that (2√3 + 3) sin x + 2√3 cos x lies between – (2√3 + √15) and (2√3 + √15).

Solution:

Let f(x) = (2√3 + 3) sin x + 2√3 cos x

Here, A = 2√3, B = 2√3 + 3 and C = 0

– √[(2√3)2 + (2√3 + 3)2] ≤ (2√3 + 3) sin x + 2√3 cos x ≤ √[(2√3)2 + (2√3 + 3)2]
– √[12+12+9+12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤ √[12+12+9+12√3]

– √[33+12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤ √[33+12√3]

– √[15+12+6+12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤ √[15+12+6+12√3]

We know that (12√3 + 6 < 12√5) because the value of √5 – √3 is more than 0.5

So if we replace, (12√3 + 6 with 12√5) the above inequality still holds.

So by rearranging the above expression √(15+12+12√5)we get, 2√3 + √15

– 2√3 + √15 ≤ (2√3 + 3) sin x + 2√3 cos x ≤ 2√3 + √15

Hence proved.

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