RD Sharma Class 11 Solutions Chapter 7 Exercise 7.2 (Updated For 2024)

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RD Sharma Class 11 Solutions Chapter 7 Exercise 7.2

RD Sharma Solutions Class 11 Maths Chapter 7 Exercise 7.2

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.

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