Asked by Saq2
                If sin(x) = 1/3 and sec(y) = 29/21, where x and y lie between 0 and  𝜋/2, evaluate the expression.
cos(x + y)
            
        cos(x + y)
Answers
                    Answered by
            mathhelper
            
    So both angles are in quad I, that makes it quite simple.
sinx = 1/3,
so we have x^2 + y^2 = r^2
1 + x^2 = 9
x = √8 = 2√2 , (the x is not the same x as the angle)
then cosx = 2√2/3
secy = 29/21 or cosy = 21/29
y^2 + 21^2 = 29^2
y = √400 = 20 ---> siny = 20/29
cos(x+y) = cosxcosy - sinxsiny
= (2√2/3)(21/29) - (1/3)(20/29)
= (42√2 - 20)/87
    
sinx = 1/3,
so we have x^2 + y^2 = r^2
1 + x^2 = 9
x = √8 = 2√2 , (the x is not the same x as the angle)
then cosx = 2√2/3
secy = 29/21 or cosy = 21/29
y^2 + 21^2 = 29^2
y = √400 = 20 ---> siny = 20/29
cos(x+y) = cosxcosy - sinxsiny
= (2√2/3)(21/29) - (1/3)(20/29)
= (42√2 - 20)/87
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