Related Articles

# Class 11 RD Sharma Solutions – Chapter 7 Trigonometric Ratios of Compound Angles – Exercise 7.2

• Last Updated : 15 Dec, 2020

### (iv) sin x – cos x + 1

Solution:

As it is known the maximum value of A cos α + B sin α + C is C + √(A2 +B2),

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12.

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

And the minimum value is C – √(a2 + B2).

(i) 12sin x – 5cos 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

As 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.

### (iii) 24 cos x + 7 sin x

Solution:

(i) √3sin 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)

As 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)

As 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 that Sin 100° – Sin 10°] 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° – 1/√2 sin 10°)

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

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

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

f(x) = √2 cos (π/4 + 10°)

Therefore,

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).

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]

As we know that (12√3 + 6 < 12√5) because the value of √5 – √3 is more than 0.5

If we replace, (12√3 + 6 with 12√5) the above inequality still holds.

After 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.

My Personal Notes arrow_drop_up