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Question 36 : Given that cos x + cos y = 1, the range of sin x - sin y is

  1. [-1, 1]
  2. [-2, 2]
  3. [0, √3]
  4. [-√3, √3]

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Explanatory Answer

Let’s say sin x – sin y = k. So we intend to find the range of k.
We have cos x + cos y = 1

So, k2 + 12 = (sin x - sin y )2 + (cos x + cosy )2
k2 + 12 = (sin x)2 + (sin y)2 – 2(sin x)(sin y) + (cos x)2 + (cos y)2 + 2(cos x)(cos y)
k2 + 12= sin2 x + cos2 x + sin2 y + cos2 y + 2(cos x)(cos y) – 2(sin x)(sin y)
k2 + 12 = 1 + 1 + 2(cos x)(cos y) – 2(sin x)(sin y) {Because, sin2 x + cos2 x = 1}
k2 + 12= 2 + 2[(cos x)(cos y) – (sin x)(sin y)] {Because, (cos x)(cos y) – (sin x)(sin y) = cos(x + y)}
k2 = 1 + 2(cos(x + y))

The maximum value that a cos function can take is 1, and the minimum value is 0.

Therefore the maximum value that k2 can take is 1+2(1) = 3,
and the maximum value that k2 can take is 1+2(0) = 1,

Therefore (k2)max = 3 and (k2)min = 1.

To get the range of k, we concentrate on (k2)max.

(k2)max = 3, implies that the extreme values of k are:

k = +√3 or -√3
So, the values of k range between +√3 and -√3, both +√3 and -√3 inclusive.

k = sin x – sin y, Remember!!

Therefore, Range of sin x – sin y = [-√3, √3]

The question is "Given that cos x + cos y = 1, the range of sin x - sin y is"

Hence, the answer is [-√3, √3]

Choice D is the correct answer

 

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