Question

# If x and y are the maximum and minimum values of sin θ + cos θ, find the value of x2 + y2 __

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 cos2B + sin2B = a2+b2k2 = 1 ⇒ k = √a2+b2 From (1) we get = cosB = a√a2+b2 sinB = b√a2+b2 So we will divide and multiply by √a2+b2 in asinθ + bcosθ. ⇒ √a2+b2 (a√a2+b2sinθ+b√a2+b2cosθ) ⇒ asinθ + bcosθ = √a2+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 ⇒ x2 + y2 = 2 + 2 = 4 Key steps concepts: writing the given expression in the form k sin (A+B)

Suggest corrections