if sin x = 1/3 and sec y = 5/3 where x and y lie between 0 and π/2, evaluate sin(x + y).?
if sin x = 1/3 and sec y = 5/3 where x and y lie between 0 and π/2, evaluate sin(x + y).
in college cannot figure this one out none of us can, calc center is closed at 10pm at night, and this is due by 11:59
Update:mike we have gotten that answer and the system says it is incorrect, idk if there is another way to rewrite it or something?
Update 3:oh nvm we solved it we just typed in the quation for the answer rather than the decimal answer of .954424 etc.
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sin x = 1/3, cos x = sqrt(8)/3
cos y = 3/5 sin y = 4/5
sin(x+y) = sinxcosy+cosxsiny
= 1/5 + 4sqrt(8)/15
= [3+8sqrt(2)]/15
Sin(x+y)=sin(x)cos(y)+sin(y)cos(x)
Sin(x+y)= (1/3)(3/5)+(4/5)(rt8/3)
Dont have a calculator but theres the work.
Rt stands for square root btw, and you get the cos(x) and the sin(y) by pythagorean theorem