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

RD Sharma Solutions Class 11 Maths Chapter 7 – Free PDF Download

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles is provided here. In this chapter, we shall derive formulae that express 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). The subject experts at BYJU’S have formulated the RD Sharma Class 11 Solutions for Maths in the most lucid and easy manner. The solutions to this chapter are available in 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.  RD Sharma Solutions provide accurate answers to each question of these exercises as per the grasping abilities of students. 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.
• A few theorems.
• Maximum and minimum values of trigonometrical expressions.
• To express the given expressions in the desired form.

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

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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)

= RHS

∴ LHS = RHS

Hence proved.

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

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

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

Solution:

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

Let us consider LHS:

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

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

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

= (1 + tan 11^o) / (1- 1×tan 11^o)

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

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

So,

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

= tan 56^o

= RHS

∴ LHS = RHS

Hence proved.

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

Let us consider LHS:

(cos 9^o + sin 9^o) / (cos 9^o – sin 9^o)

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

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

= (1 + tan 9^o) / (1 – 1 × tan 9^o)

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

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

So,

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

= tan 54^o

= RHS

∴ LHS = RHS

Hence proved.

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

Let us consider LHS:

(cos 8^o – sin 8^o) / (cos 8^o + sin 8^o)

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

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

= (1 – tan 8^o) / (1 + 1×tan 8^o)

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

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

So,

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

= tan 37^o

= RHS

∴ LHS = RHS

Hence proved.

12. Prove that:

(i)

(ii)

(iii)

Solution:

(i)

= sin 90^o

= 1

= RHS

∴ LHS = RHS

Hence proved.

(ii)

= sin 60^o

= √3/2

= RHS

∴ LHS = RHS

Hence proved.

(iii)

= sin 90^o

= 1

= RHS

∴ LHS = RHS

Hence proved.

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

Solution:

Let us consider LHS:

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

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

Here, A = 69^o and B = 66^o

So,

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

= tan 135^o

= – tan 45^o

= – 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 45^o 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)

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

= (2m² – 2m + 1) / (2m² – 2m + 1)

= 1

= tan 45^o 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)

(ii)

(iii)

(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)

Solution:

(i)

= tan A

= RHS

∴ LHS = RHS

Hence proved.

(ii)

Let us consider LHS:

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

= 0

= RHS

∴ LHS = RHS

Hence proved.

(iii)

= 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)

∴ 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 36^o + tan 9^o + tan 36^o tan 9^o = 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 15^o + tan 30^o) / (1 – tan 15^o tan 30^o) = 1

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

Upon rearranging we get,

tan15° + tan30° + tan15° tan30° = 1

Hence proved.

(iii) tan 36^o + tan 9^o + tan 36^o tan 9^o = 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 36^o + tan 9^o) / (1 – tan 36^o tan 9^o) = 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 – 3√3/2 sin x + 4

= 13/2 cos x – 3√3/2 sin x + 4

So, here A = 13/2, B = – 3√3/2, C = 4

4 – √[(13/2)² + (-3√3/2)²] ≤ 13/2 cos x – 3√3/2 sin x + 4 ≤ 4 + √[(13/2)² + (-3√3/2)²]

4 – √[(169/4) + (27/4)] ≤ 13/2 cos x – 3√3/2 sin x + 4 ≤ 4 + √[(169/4) + (27/4)]

4 – 7 ≤ 13/2 cos x – 3√3/2 sin x + 4 ≤ 4 + 7

-3 ≤ 13/2 cos x – 3√3/2 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 100^o – Sin 10^o 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 100^o – 1/√2 sin 10^o)

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

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

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

f(x) = √2 cos (π/4 + 10^o)

∴ 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.