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RD Chapter 7- Trigonometric Ratios of Compound Angles Ex-7.2 Interview Questions Answers

Question 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

Answer 1 :

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 – 5cos 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 13and -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 -9and 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 -3and 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– 2and 1 + 2respectively.

Question 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

Answer 2 :

(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)

Question 3 :

Show thatSin 100o – Sin 10o is positive.

Answer 3 :

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 – sinπ/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/4sin 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°

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

Answer 4 :

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 aboveinequality still holds.

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

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

Hence proved.


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RD Chapter 7- Trigonometric Ratios of Compound Angles Ex-7.2 Contributors

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