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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 â€“ cos x + 1

Solution:

As it is known the maximum value of A cos Î± + B sin Î± + C is C + âˆš(A2 +B2),

And the minimum value is C â€“ âˆš(a2 + B2).

(i) 12sin x â€“ 5cos x

Given:

f(x) = 12 sin x â€“ 5 cos x

Here, A = -5, B = 12 and C = 0

â€“âˆš((-5)2 + 122) â‰¤ 12 sin x â€“ 5 cos x â‰¤ âˆš((-5)2 + 122)

â€“âˆš(25+144) â‰¤ 12 sin x â€“ 5 cos x â‰¤ âˆš(25+144)

â€“âˆš169 â‰¤ 12 sin x â€“ 5 cos x â‰¤ âˆš169

â€“13 â‰¤ 12 sin x â€“ 5 cos x â‰¤ 13

Hence, the maximum and minimum values of f(x) are 13 and â€“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 â€“ âˆš(122 + 52) â‰¤ 12 cos x + 5 sin x + 4 â‰¤ 4 + âˆš(122 + 52)

4 â€“ âˆš(144+25) â‰¤ 12 cos x + 5 sin x + 4 â‰¤ 4 + âˆš(144+25)

4 â€“âˆš169 â‰¤ 12 cos x + 5 sin x + 4 â‰¤ 4 + âˆš169

â€“9 â‰¤ 12 cos x + 5 sin x + 4 â‰¤ 17

Hence, the maximum and minimum values of f(x) are â€“9 and 17 respectively.

(iii) 5 cos x + 3 sin (Ï€/6 â€“ x) + 4

Given:

f(x) = 5 cos x + 3 sin (Ï€/6 â€“ x) + 4

As we know that, sin (A â€“ B) = sin A cos B â€“ cos A sin B

f(x) = 5 cos x + 3 sin (Ï€/6 â€“ x) + 4

= 5 cos x + 3 (sin Ï€/6 cos x â€“ cos Ï€/6 sin x) + 4

= 5 cos x + 3/2 cos x â€“ 3âˆš3/2 sin x + 4

= 13/2 cos x â€“ 3âˆš3/2 sin x + 4

So, here A = 13/2, B = â€“ 3âˆš3/2, C = 4

4 â€“ âˆš[(13/2)2 + (-3âˆš3/2)2] â‰¤ 13/2 cos x â€“ 3âˆš3/2 sin x + 4 â‰¤ 4 + âˆš[(13/2)2 + (-3âˆš3/2)2]

4 â€“ âˆš[(169/4) + (27/4)] â‰¤ 13/2 cos x â€“ 3âˆš3/2 sin x + 4 â‰¤ 4 + âˆš[(169/4) + (27/4)]

4 â€“ 7 â‰¤ 13/2 cos x â€“ 3âˆš3/2 sin x + 4 â‰¤ 4 + 7

â€“3 â‰¤ 13/2 cos x â€“ 3âˆš3/2 sin x + 4 â‰¤ 11

Hence, the maximum and minimum values of f(x) are â€“3 and 11 respectively.

(iv) sin x â€“ cos x + 1

Given:

f(x) = sin x â€“ cos x + 1

So, here A = -1, B = 1 And c = 1

1 â€“ âˆš[(-1)2 + 12] â‰¤ sin x â€“ cos x + 1 â‰¤ 1 + âˆš[(-1)2 + 12]

1 â€“ âˆš(1+1) â‰¤ sin x â€“ cos x + 1 â‰¤ 1 + âˆš(1+1)

1 â€“ âˆš2 â‰¤ sin x â€“ cos x + 1 â‰¤ 1 + âˆš2

Hence, the maximum and minimum values of f(x) are 1 â€“ âˆš2 and 1 + âˆš2 respectively.

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

Solution:

(i) âˆš3sin x â€“ cos x

Let f(x) = âˆš3 sin x â€“ cos x

Dividing and multiplying by âˆš((âˆš3)2 + 12) i.e. by 2

f(x) = 2(âˆš3/2 sin x â€“ 1/2 cos x)

Sine expression:

f(x) = 2(cos Ï€/6 sin x â€“ sin Ï€/6 cos x) (since, âˆš3/2 = cos Ï€/6 and 1/2 = sin Ï€/6)

As we know that, sin A cos B â€“ cos A sin B = sin (A â€“ B)

f(x) = 2 sin (x â€“ Ï€/6)

Again,

f(x) = 2(âˆš3/2 sin x â€“ 1/2 cos x)

Cosine expression:

f(x) = 2(sin Ï€/3 sin x â€“ cos Ï€/3 cos x)

As we know that, cos A cos B â€“ sin A sin B = cos (A + B)

f(x) = -2 cos(Ï€/3 + x)

(ii) cos x â€“ sin x

Let f(x) = cos x â€“ sin x

Dividing and multiplying by âˆš(12 + 12) i.e. by âˆš2,

f(x) = âˆš2(1/âˆš2 cos x â€“ 1/âˆš2 sin x)

Sine expression:

f(x) = âˆš2(sin Ï€/4 cos x â€“ cos Ï€/4 sin x) (since, 1/âˆš2 = sin Ï€/4 and 1/âˆš2 = cos Ï€/4)

We know that sin A cos B â€“ cos A sin B = sin (A â€“ B)

f(x) = âˆš2 sin (Ï€/4 â€“ x)

Again,

f(x) = âˆš2(1/âˆš2 cos x â€“ 1/âˆš2 sin x)

Cosine expression:

f(x) = 2(cos Ï€/4 cos x â€“ sin Ï€/4 sin x)

We know that cos A cos B â€“ sin A sin B = cos (A + B)

f(x) = âˆš2 cos (Ï€/4 + x)

(iii) 24 cos x + 7 sin x

Let f(x) = 24 cos x + 7 sin x

Dividing and multiplying by âˆš((âˆš24)2 + 72) = âˆš625 i.e. by 25,

f(x) = 25(24/25 cos x + 7/25 sin x)

Sine expression:

f(x) = 25(sin Î± cos x + cos Î± sin x) where, sin Î± = 24/25 and cos Î± = 7/25

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

f(x) = 25 sin (Î± + x)

Cosine expression:

f(x) = 25(cos Î± cos x + sin Î± sin x) where, cos Î± = 24/25 and sin Î± = 7/25

We know that cos A cos B + sin A sin B = cos (A â€“ B)

f(x) = 25 cos (Î± â€“ x)

### Question 3: Show that Sin 100Â° â€“ Sin 10Â°] is positive.

Solution:

Let f(x) = sin 100Â° â€“ sin 10Â°

Dividing And multiplying by âˆš(12 + 12) i.e. by âˆš2,

f(x) = âˆš2(1/âˆš2 sin 100Â° â€“ 1/âˆš2 sin 10Â°)

f(x) = âˆš2(cos Ï€/4 sin (90+10)Â° â€“ sin Ï€/4 sin 10Â°) (since, 1/âˆš2 = cos Ï€/4 and 1/âˆš2 = sin Ï€/4)

f(x) = âˆš2(cos Ï€/4 cos 10Â° â€“ sin Ï€/4 sin 10Â°)

We know that cos A cos B â€“ sin A sin B = cos (A + B)

f(x) = âˆš2 cos (Ï€/4 + 10Â°)

Therefore,

f(x) = âˆš2 cos 55Â°

### Question 4: Prove that (2âˆš3 + 3) sin x + 2âˆš3 cos x lies between â€“ (2âˆš3 + âˆš15) and (2âˆš3 + âˆš15).

Solution:

Let f(x) = (2âˆš3 + 3) sin x + 2âˆš3 cos x

Here, A = 2âˆš3, B = 2âˆš3 + 3 and C = 0

â€“ âˆš[(2âˆš3)2 + (2âˆš3 + 3)2] â‰¤ (2âˆš3 + 3) sin x + 2âˆš3 cos x â‰¤ âˆš[(2âˆš3)2 + (2âˆš3 + 3)2]

â€“ âˆš[12+12+9+12âˆš3] â‰¤ (2âˆš3 + 3) sin x + 2âˆš3 cos x â‰¤ âˆš[12+12+9+12âˆš3]

â€“ âˆš[33+12âˆš3] â‰¤ (2âˆš3 + 3) sin x + 2âˆš3 cos x â‰¤ âˆš[33+12âˆš3]

â€“ âˆš[15+12+6+12âˆš3] â‰¤ (2âˆš3 + 3) sin x + 2âˆš3 cos x â‰¤ âˆš[15+12+6+12âˆš3]

As we know that (12âˆš3 + 6 < 12âˆš5) because the value of âˆš5 â€“ âˆš3 is more than 0.5

If we replace, (12âˆš3 + 6 with 12âˆš5) the above inequality still holds.

After rearranging the above expression:

âˆš(15+12+12âˆš5)we get, 2âˆš3 + âˆš15

â€“ 2âˆš3 + âˆš15 â‰¤ (2âˆš3 + 3) sin x + 2âˆš3 cos x â‰¤ 2âˆš3 + âˆš15

Hence, proved.

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