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# 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),

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.

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