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Question

If x and y are the maximum and minimum values of sin θ + cos θ, find the value of x2y2


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Solution

If you know that the maximum and minimum value of asinθ + bcosθ are a2+b2, and -a2+b2,you can directly use this result to arrive at the answer.

We will try to arrive at the answer similar to the derivation of the above result.

Let E = asinθ + bcosθ, If we can express it as k sin (A + B), we can say the minimum and maximum values are -k and +k, where k is a +ve number.

we will expand ksin (A+B) and campare it with asinθ + bsinθ.

ksin(A+B) = asinθ + bcosθ

k[sinAcosB + cosAsinB) = asinθ + bcosθ 

If we take A = θ, then we get k cos B = a and Ksin B = b ___________(1)

cosB = ak and sinB = bk

cos2Bsin2Ba2+b2k2 = 1

k = a2+b2

From (1) we get = cosB = aa2+b2

sinB = ba2+b2

So we will divide and multiply by a2+b2 in asinθ + bcosθ.

 a2+b2 (aa2+b2sinθ+ba2+b2cosθ)

asinθ + bcosθa2+b2 (sinθx cosB + cosθsinB)
                                                                      = a2+b2 sin(θ + B)

  Maximum and minimum values are 

a2+b2 and - a2+b2

In our question a = 1 , b =1

The minimum and maximum values are 2 and 2

x2y2 = 2 + 2 = 4

Key steps concepts: writing the given expression in the form k sin (A+B)

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