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

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

[-√3, √3]

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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, k^{2} + 1^{2} = (sin x - sin y )^{2} + (cos x + cosy )^{2}

k^{2} + 1^{2} = (sin x)^{2} + (sin y)^{2} – 2(sin x)(sin y) + (cos x)^{2} + (cos y)^{2} + 2(cos x)(cos y)

k^{2} + 1^{2}= sin^{2} x + cos^{2} x + sin^{2} y + cos^{2} y + 2(cos x)(cos y) – 2(sin x)(sin y)

k^{2} + 1^{2} = 1 + 1 + 2(cos x)(cos y) – 2(sin x)(sin y) {Because, sin^{2} x + cos^{2} x = 1}

k^{2} + 1^{2}= 2 + 2[(cos x)(cos y) – (sin x)(sin y)] {Because, (cos x)(cos y) – (sin x)(sin y) = cos(x + y)}

k^{2} = 1 + 2(cos(x + y))

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

Therefore the maximum value that k^{2} can take is 1+2(1) = 3,

and the minimum value that k^{2} can take is 1+2(-1) = -1,

This is impossible, k^{2} is always non-negative.

The minimum value that k^{2} can take is 0.

Therefore (k^{2})_{max} = 3 and (k^{2})_{min} = 0.

To get the range of k, we concentrate on (k^{2})_{max}.

(k^{2})_{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" **

Choice D is the correct answer

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