SOLUTION: Consider angles x and y such that 0 <_ y <_ x <_ pi/2 and sin(x+y) = 0.9 while sin(x-y) = 0.6. What is the value of (sin x + cos x)(sin y + cos y)?

Question 1187797: Consider angles x and y such that 0 <_ y <_ x <_ pi/2 and sin(x+y) = 0.9 while sin(x-y) = 0.6. What is the value of (sin x + cos x)(sin y + cos y)?
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
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Consider angles x and y such that 0 <_ y <_ x <_ pi/2 and sin(x+y) = 0.9 while sin(x-y) = 0.6.
What is the value of (sin x + cos x)(sin y + cos y)?
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            Actually, it is a joke problem of Trigonometry, and (as it should be for any joke problem),
            it has a simple unexpected solution.

  (sin x + cos x)(sin y + cos y) = sin(x)*sin(y) + cos(x)*sin(y) + sin(x)*cos(y) + cos(x)*cos(y) = 


        (regroup the terms)


= ( cos(x)*sin(y) + sin(x)*cos(y) ) + ( cos(x)*cos(y) + sin(x)*sin(y) ) = sin(x+y) + cos(x-y).



Now calculate  cos(x-y) =  =  =  =  = 0.8.


    Notice that the angle (x-y)  is in QI,  where  cosine is positive,

    so we select and use the positive value of the square root . . . 



At this point, we just have everything to continue and complete our calculations


    (sin x + cos x)(sin y + cos y) = sin(x+y) + cos(x-y) = 0.9 + 0.8 = 1.7.     ANSWER

Solved, answered, explained and completed.

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