α is a root of the equation (2 sin x – cos x) (1 + cos x) = sin 2 x β is a root of the equation 3 cos 2 x – 10 cos x + 3 = 0 and γ is a root of the equation 1 – sin 2x = cos x – sin x. 0 ≤ α , β , γ ≤ π /2
(i) cos α + cos β + cos γ can be equal to –
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i)
(2 sin x – cos x) (1 + cos x) = sin 2 x
⇒ (1 + cos x) [2 sin x – cos x – 1 + cos x] = 0
⇒ (1 + cos x) (2 sin x – 1) = 0
⇒ cos x = –1 or sin x = ½
so sin α = 1/2 [as 0 ≤ α ≤ π /2]
⇒ cos α = 
/2……….(1)
Next, 3 cos 2 x – 10 cos x + 3 = 0
⇒ (3 cos x – 1) (cos x – 3) = 0
⇒ cos x = 1/3 as cos x ≠ 3
So cos β = 1/3, sin β = 
………(2) and 1 – sin 2x = cos x – sin x
⇒ sin 2 x + cos 2 x – 2 sin x cos x = cos x – sin x
⇒ (cos x – sin x ) (cos x – sin x – 1) = 0
⇒ Either sin x = cos x ⇒ sin γ = cos γ = 1/ 
……… (3)
or cos x – sin x = 1 ⇒ cos x = 1, sin x = 0 ⇒ cos γ = 1, sin γ = 0……. (4)
So that cos α + cos β + cos γ can be equal to
+ 
+ 
or 
+ 
+ 1
i.e. 
or 
sin α + sin β + sin γ can be equal to
+ 
+ 
or 
+ 
+ 0
i.e., 
or 
and sin ( α – β ) is equal to
sin α cos β – cos α sin β
= 
× 
– 
× 
= 
.
(ii)
(2 sin x – cos x) (1 + cos x) = sin 2 x
⇒ (1 + cos x) [2 sin x – cos x – 1 + cos x] = 0
⇒ (1 + cos x) (2 sin x – 1) = 0
⇒ cos x = –1 or sin x = ½
so sin α = 1/2 [as 0 ≤ α ≤ π /2]
⇒ cos α = 
/2……….(1)
Next, 3 cos 2 x – 10 cos x + 3 = 0
⇒ (3 cos x – 1) (cos x – 3) = 0
⇒ cos x = 1/3 as cos x ≠ 3
So cos β = 1/3, sin β = 
………(2)
and 1 – sin 2x = cos x – sin x
⇒ sin 2 x + cos 2 x – 2 sin x cos x = cos x – sin x
⇒ (cos x – sin x ) (cos x – sin x – 1) = 0
⇒ Either sin x = cos x ⇒ sin γ = cos γ = 1/ 
…(3)
or cos x – sin x = 1 ⇒ cos x = 1, sin x = 0
⇒ cos γ = 1, sin γ = 0……. (4)
So that cos α + cos β + cos γ can be equal to
+ 
+ 
or 
+ 
+ 1
i.e. 
or 
sin α + sin β + sin γ can be equal to
+ 
+ 
or 
+ 
+ 0
i.e., 
or 
and sin ( α – β ) is equal to
sin α cos β – cos α sin β
= 
× 
– 
× 
= 
.
(iii)
(2 sin x – cos x) (1 + cos x) = sin 2 x
⇒ (1 + cos x) [2 sin x – cos x – 1 + cos x] = 0
⇒ (1 + cos x) (2 sin x – 1) = 0
⇒ cos x = –1 or sin x = ½
so sin α = 1/2 [as 0 ≤ α ≤ π /2]
⇒ cos α = 
/2……….(1)
Next, 3 cos 2 x – 10 cos x + 3 = 0
⇒ (3 cos x – 1) (cos x – 3) = 0
⇒ cos x = 1/3 as cos x ≠ 3
So cos β = 1/3, sin β = 
………(2)
and 1 – sin 2x = cos x – sin x
⇒ sin 2 x + cos 2 x – 2 sin x cos x = cos x – sin x
⇒ (cos x – sin x ) (cos x – sin x – 1) = 0
⇒ Either sin x = cos x ⇒ sin γ = cos γ = 1/ 
… (3)
or cos x – sin x = 1 ⇒ cos x = 1, sin x = 0
⇒ cos γ = 1, sin γ = 0……. (4)
So that cos α + cos β + cos γ can be equal to
+ 
+ 
or 
+ 
+ 1
i.e. 
or 
sin α + sin β + sin γ can be equal to
+ 
+ 
or 
+ 
+ 0
i.e., 
or 
and sin ( α – β ) is equal to
sin α cos β – cos α sin β
= 
× 
– 
× 
= 
.
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