# 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 BYJUâ€™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 â€“ 5 cos x
(ii) 12 cos x + 5 sin x + 4
(iii) 5 cos x + 3 sin (Ï€/6 – x) + 4
(iv) sin x â€“ cos x + 1

Solution:

We know that the maximum value of A cos Î± + B sin Î± + C is C + âˆš(A2Â +B2),

And the minimum value is C – âˆš(a2Â +B2).

(i) 12 sin x â€“ 5 cos x

Given: f(x) = 12 sin x â€“ 5 cos x

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

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

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

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

-13Â â‰¤Â 12 sinÂ x – 5 cosÂ x â‰¤ 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 – âˆš(122 + 52) â‰¤ 12 cos x + 5 sin x + 4 â‰¤ 4 + âˆš(122 + 52)

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

4 –âˆš169 â‰¤ 12 cos x + 5 sin x + 4 â‰¤ 4 + âˆš169

-9Â â‰¤Â 12 cos x + 5 sin x + 4 â‰¤ 17

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

(iii) 5 cos x + 3 sin (Ï€/6 – x) + 4Â

Given: f(x) = 5 cos x + 3 sin (Ï€/6 – x) + 4Â

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

f(x) = 5 cos x + 3 sin (Ï€/6 – x) + 4Â

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

= 5 cos x + 3/2 cos x – 3âˆš3/2 sin x + 4

= 13/2 cos x – 3âˆš3/2 sin x + 4

So, here A = 13/2, B = – 3âˆš3/2, C = 4

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

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

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

-3 â‰¤ 13/2 cos x – 3âˆš3/2 sin x + 4 â‰¤ 11

Hence, the maximum and minimum values of f(x) are -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 – âˆš[(-1)2 + 12] â‰¤ sin x â€“ cos x + 1 â‰¤ 1 + âˆš[(-1)2 + 12]

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

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

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

2. Reduce each of the following expressions to the Sine and Cosine of a single expression:

(i) âˆš3 sin x â€“ cos x

(ii) cos x â€“ sin x

(iii) 24 cos x + 7 sin x

Solution:

(i) âˆš3 sin 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)

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

f(x) = 2 sin (x – 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)

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 – 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)2Â + 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)Â 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)Â where, cos Î± = 24/25 and sin Î± = 7/25

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

f(x) = 25 cos (Î± – x)

3. Show that Sin 100oÂ â€“ Sin 10oÂ 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 100o â€“ 1/âˆš2 sin 10o)

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

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

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

f(x) = âˆš2 cos (Ï€/4 + 10o)

âˆ´Â f(x) = âˆš2 cos 55Â°

4. Prove that (2âˆš3 + 3) sin x + 2âˆš3 cos xÂ 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

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

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

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

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

We know that (12âˆš3 + 6 < 12âˆš5) because the value of âˆš5 – âˆš3 is more than 0.5

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

So by rearranging the above expression âˆš(15+12+12âˆš5)we get, 2âˆš3 + âˆš15

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

Hence proved.