• +91 9971497814
  • info@interviewmaterial.com

RD Chapter 7- Trigonometric Ratios of Compound Angles Ex-7.1 Interview Questions Answers

Question 1 :
If sin A = 4/5 and cos B = 5/13, where 0
(i) sin (A + B)
(ii) cos (A + B)
(iii) sin (A – B)
(iv) cos (A – B)

Answer 1 :

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
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

Question 2 :
(a) If Sin A = 12/13 and sin B = 4/5, where π/2
(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).

Answer 2 :

(a) Given:
Sin A = 12/13 and sin B = 4/5, where 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

Question 3 :
If cos A = – 24/25 and cos B = 3/5, where π
(i) sin (A + B) (ii) cos (A + B)

Answer 3 :

Given:
cos A = – 24/25 and cos B = 3/5, where π
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

Question 4 :
If tan A = 3/4, cos B = 9/41, where π

Answer 4 :

Given:
tan A = 3/4 and cos B = 9/41, where π
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

Question 5 :
If sin A = 1/2, cos B = 12/13, where π/2

Answer 5 :

Given:
sin A = 1/2, cos B = 12/13, where π/2
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)

Question 6 :
If sin A = 1/2, cos B = √3/2, where π/2
(i) tan (A + B) (ii) tan (A – B)

Answer 6 :

Given:
Sin A = 1/2 and cos B = √3/2, where 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

Question 7 :

Evaluatethe following:
(i) sin 780 cos180 – cos 780 sin 180 

(ii) cos470 cos 130 – sin 470 sin 130
(iii) sin 360 cos90 + cos 360 sin 90 

(iv) cos800 cos 200 + sin 800 sin 200

Answer 7 :

(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 – cos780 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 – sin470 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 + cos360 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 + sin800 sin 200 = cos (80 – 20) °

= cos 60°

= ½

Question 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)

Answer 8 :

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

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

Answer 9 :

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.

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

Answer 10 : Let us consider LHS: (tan A + tan B) / (tan A – tan B)

= RHS
∴ LHS = RHS
Hence proved.


Selected

 

RD Chapter 7- Trigonometric Ratios of Compound Angles Ex-7.1 Contributors

krishan

Share your email for latest updates

Name:
Email:

Our partners