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 тАУ 5 cos x
(ii) 12 cos x + 5 sin x + 4
(iii) 5 cos x + 3 sin (╧А/6 тАУ x) + 4
(iv) sin x тАУ cos x + 1
Answer - 1 : -
We know that the maximum value of A cos ╬▒ + B sin ╬▒ +C is C + тИЪ(A2┬а+B2),
And the minimum value is C тАУ тИЪ(a2┬а+B2).
(i)┬а12 sin x тАУ 5 cos x
Given: f(x) = 12 sin x тАУ 5 cos x
Here, A = -5, B = 12 and C = 0
тАУтИЪ((-5)2┬а+ 122) тЙд 12 sin x тАУ 5cos x тЙд┬атИЪ((-5)2┬а+122)
тАУтИЪ(25+144) тЙд 12 sin x тАУ 5 cos x тЙд┬атИЪ(25+144)
тАУтИЪ169 тЙд 12 sin x тАУ 5 cos x тЙд┬атИЪ169
-13┬атЙд┬а12 sin┬аx тАУ 5 cos┬аx тЙд 13
Hence, the maximum and minimum values of f(x) are 13and -13 respectively.
(ii)┬а12 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 тАУ┬атИЪ(122┬а+ 52)тЙд 12 cos x + 5 sin x + 4 тЙд 4 +┬атИЪ(122┬а+ 52)
4 тАУ┬атИЪ(144+25) тЙд 12 cos x + 5 sin x + 4 тЙд4 +┬атИЪ(144+25)
4 тАУтИЪ169 тЙд 12 cos x + 5 sin x + 4 тЙд 4 +┬атИЪ169
-9┬атЙд┬а12 cos x + 5 sin x + 4 тЙд 17
Hence, the maximum and minimum values of f(x) are -9and 17 respectively.
(iii)┬а5 cos x + 3 sin (╧А/6 тАУ x) + 4┬а
Given: f(x) = 5 cos x + 3 sin (╧А/6 тАУ x) + 4┬а
We know that, sin (A тАУ B) = sin A cos B тАУ cos A sin B
f(x) = 5 cos x + 3 sin (╧А/6 тАУ x) + 4┬а
=┬а5 cos x + 3 (sin ╧А/6 cos x тАУ cos ╧А/6 sin x) + 4
= 5 cos x + 3/2 cos x тАУ 3тИЪ3/2 sin x+ 4
= 13/2 cos x тАУ 3тИЪ3/2 sin x + 4
So, here A = 13/2, B = тАУ 3тИЪ3/2, C =4
4 тАУ┬атИЪ[(13/2)2┬а+ (-3тИЪ3/2)2]тЙд 13/2 cos x тАУ 3тИЪ3/2 sin x + 4 тЙд 4 +┬атИЪ[(13/2)2┬а+(-3тИЪ3/2)2]
4 тАУ┬атИЪ[(169/4) + (27/4)] тЙд 13/2 cos x тАУ 3тИЪ3/2 sin x+ 4 тЙд 4 +┬атИЪ[(169/4) + (27/4)]
4 тАУ 7 тЙд 13/2 cos x тАУ 3тИЪ3/2 sin x + 4 тЙд 4 + 7
-3 тЙд 13/2 cos x тАУ 3тИЪ3/2 sin x + 4 тЙд 11
Hence, the maximum and minimum values of f(x) are -3and 11 respectively.
(iv)┬аsin x тАУ cos x + 1
Given: f(x) = sin x тАУ cos x + 1
So, here A = -1, B = 1 And c = 1
1 тАУ┬атИЪ[(-1)2┬а+ 12]тЙд sin x тАУ cos x + 1 тЙд 1 +┬атИЪ[(-1)2┬а+ 12]
1 тАУ┬атИЪ(1+1) тЙд sin x тАУ cos x + 1 тЙд 1+┬атИЪ(1+1)
1 тАУ┬атИЪ2 тЙд sin x тАУ cos x + 1 тЙд 1 +┬атИЪ2
Hence, the maximum and minimum values of f(x) are 1тАУ┬атИЪ2and 1 +┬атИЪ2respectively.
Question - 2 : - Reduce each of the following expressions to the Sine and Cosine of a single expression:
(i) тИЪ3 sin x тАУ cos x
(ii) cos x тАУ sin x
(iii) 24 cos x + 7 sin x
Answer - 2 : -
(i)┬атИЪ3 sin x тАУ cos x
Let f(x) = тИЪ3 sin x тАУ cos x
Dividing and multiplying by тИЪ((тИЪ3)2┬а+12) i.e. by 2
f(x) = 2(тИЪ3/2 sin x тАУ 1/2 cos x)
Sine expression:
f(x) = 2(cos ╧А/6 sin x тАУ sin ╧А/6 cos x) (since, тИЪ3/2 =cos ╧А/6 and 1/2 = sin ╧А/6)
We know that, sin A cos B тАУ cos A sin B = sin (A тАУ B)
f(x) = 2 sin (x тАУ ╧А/6)
Again,
f(x) = 2(тИЪ3/2 sin x тАУ 1/2 cos x)
Cosine expression:
f(x) = 2(sin ╧А/3 sin x тАУ cos ╧А/3 cos x)
We know that, cos A cos B тАУ sin A sin B = cos (A + B)
f(x) = -2 cos(╧А/3 + x)
(ii)┬аcos x тАУ sin x
Let f(x) = cos x тАУ sin x
Dividing and multiplying by тИЪ(12┬а+ 12)i.e. by тИЪ2,
f(x) = тИЪ2(1/тИЪ2 cos x тАУ 1/тИЪ2 sin x)
Sine expression:
f(x) = тИЪ2(sin ╧А/4 cos x тАУ cos ╧А/4 sin x) (since, 1/тИЪ2= sin ╧А/4 and 1/тИЪ2 = cos ╧А/4)
We know that sin A cos B тАУ cos A sin B = sin (A тАУ B)
f(x) = тИЪ2 sin (╧А/4 тАУ x)
Again,
f(x) = тИЪ2(1/тИЪ2 cos x тАУ 1/тИЪ2 sin x)
Cosine expression:
f(x) = 2(cos ╧А/4 cos x тАУ sin ╧А/4 sin x)
We know that cos A cos B тАУ sin A sin B = cos (A + B)
f(x) = тИЪ2 cos (╧А/4 + x)
(iii)┬а24 cos x + 7 sin x
Let f(x) = 24 cos x + 7 sin x
Dividing and multiplying by тИЪ((тИЪ24)2┬а+72) = тИЪ625 i.e. by 25,
f(x) = 25(24/25 cos x + 7/25 sin x)
Sine expression:
f(x) = 25(sin ╬▒ cos x + cos ╬▒ sin x)┬аwhere, sin ╬▒= 24/25 and cos ╬▒ = 7/25
We know that sin A cos B + cos A sin B = sin (A + B)
f(x) = 25 sin (╬▒ + x)
Cosine expression:
f(x) = 25(cos ╬▒ cos x + sin ╬▒ sin x)┬аwhere, cos ╬▒= 24/25 and sin ╬▒ = 7/25
We know that cos A cos B + sin A sin B = cos (A тАУ B)
f(x) = 25 cos (╬▒ тАУ x)
Question - 3 : - Show thatSin 100o┬атАУ Sin 10o┬аis positive.
Answer - 3 : -
Let f(x) = sin 100┬░ тАУ sin 10┬░
Dividing And multiplying by тИЪ(12┬а+ 12)i.e. by тИЪ2,
f(x) = тИЪ2(1/тИЪ2 sin 100o┬атАУ 1/тИЪ2 sin 10o)
f(x) = тИЪ2(cos ╧А/4 sin (90+10)o┬атАУ sin╧А/4 sin 10o) (since, 1/тИЪ2 = cos ╧А/4 and 1/тИЪ2 = sin ╧А/4)
f(x) = тИЪ2(cos ╧А/4 cos 10o┬атАУ sin ╧А/4sin 10o)
We know that cos A cos B тАУ sin A sin B = cos (A + B)
f(x) = тИЪ2 cos (╧А/4 + 10o)
тИ┤┬аf(x) = тИЪ2 cos 55┬░
Question - 4 : - Prove that (2тИЪ3 + 3) sin x + 2тИЪ3 cos x lies between тАУ (2тИЪ3 + тИЪ15) and (2тИЪ3 + тИЪ15).
Answer - 4 : -
Let f(x) = (2тИЪ3 + 3) sin x + 2тИЪ3 cos x
Here, A = 2тИЪ3, B = 2тИЪ3 + 3 and C = 0
тАУ тИЪ[(2тИЪ3)2┬а+ (2тИЪ3 + 3)2] тЙд(2тИЪ3 + 3) sin x + 2тИЪ3 cos x тЙд тИЪ[(2тИЪ3)2┬а+ (2тИЪ3 + 3)2]
тАУ тИЪ[12+12+9+12тИЪ3] тЙд (2тИЪ3 + 3) sin x + 2тИЪ3 cos x тЙдтИЪ[12+12+9+12тИЪ3]
тАУ тИЪ[33+12тИЪ3] тЙд (2тИЪ3 + 3) sin x + 2тИЪ3 cos x тЙдтИЪ[33+12тИЪ3]
тАУ тИЪ[15+12+6+12тИЪ3] тЙд (2тИЪ3 + 3) sin x + 2тИЪ3 cos x тЙдтИЪ[15+12+6+12тИЪ3]
We know that (12тИЪ3 + 6 < 12тИЪ5) because the value ofтИЪ5 тАУ тИЪ3 is more than 0.5
So if we replace, (12тИЪ3 + 6 with 12тИЪ5) the aboveinequality still holds.
So by rearranging the above expression тИЪ(15+12+12тИЪ5)weget, 2тИЪ3 + тИЪ15
тАУ 2тИЪ3 + тИЪ15 тЙд (2тИЪ3 + 3) sin x + 2тИЪ3 cos x тЙд 2тИЪ3 + тИЪ15
Hence proved.