RD Chapter 7 Trigonometric Ratios of Compound Angles Ex 7.2 Solutions
Question - 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
Answer - 1 : -
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 – 5cos 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 13and -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 -9and 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 -3and 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– √2and 1 + √2respectively.
Question - 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
Answer - 2 : -
(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)
Question - 3 : - Show thatSin 100o – Sin 10o is positive.
Answer - 3 : -
Let f(x) = sin 100° – sin 10°
Dividing And multiplying by √(12 + 12)i.e. by √2,
f(x) = √2(1/√2 sin 100o – 1/√2 sin 10o)
f(x) = √2(cos π/4 sin (90+10)o – sinπ/4 sin 10o) (since, 1/√2 = cos π/4 and 1/√2 = sin π/4)
f(x) = √2(cos π/4 cos 10o – sin π/4sin 10o)
We know that cos A cos B – sin A sin B = cos (A + B)
f(x) = √2 cos (π/4 + 10o)
∴ 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).
Answer - 4 : -
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 aboveinequality still holds.
So by rearranging the above expression √(15+12+12√5)weget, 2√3 + √15
– 2√3 + √15 ≤ (2√3 + 3) sin x + 2√3 cos x ≤ 2√3 + √15
Hence proved.