 # RD Sharma Solutions for Class 11 Chapter 7 - Values of Trigonometric Functions at Sum or Difference of Angles Exercise 7.2

Exercise 7.2 of Chapter 7, deals with problems based on maximum and minimum values of trigonometrical expressions and few algorithms to express a given expression in the desired form. To make learning easy, experts at BYJU’S have created the RD Sharma Class 11 Solutions which are a valuable resource for students. The solutions here are prepared in easy steps so that it is very useful for students to remember the concepts and score well in their exams. The solutions of this exercise are made available in the pdf format which can be downloaded easily from the links given below and students can start practising offline.

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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.