Question 925334: Evaluate:
sin(x + y) if sin x =3/5, sec x > 0, cos y = − 2√(5)/5, and tan y < 0
sin(x + y) =
Found 2 solutions by ewatrrr, lwsshak3:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! sin x =3/5, cos x = 4/5, cos y = − 2√(5)/5, sin y = √5/5 
.....
sin(x + y) = sin x cosy + cos x sin y
Plug and Play
sin(x + y) = 
Answer by lwsshak3(11628)  (Show Source):
You can put this solution on YOUR website! Evaluate:
sin(x + y) if sin x =3/5, sec x > 0, cos y = − 2√(5)/5, and tan y < 0
sin(x + y) =
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sinx=3/5
secx>0
reference angle x is in quadrant I where sin>0, cos>0
cosx=4/5 (working with (3-4-5) reference right triangle in quadrant I)
...
cosy=-2√5/5
tany<0
reference angle y is in quadrant II where sin>0, cos<0
siny=√(1-cos^2(y))=√(1-20/25)=√(5/25)=√5/5
..
sin(x+y)=sinxcosy+cosxsiny=3/5*-2√5/5+4/5*√5/5=-6√5/25+4√5/25=-2√5/25
..
Check:
sinx=3/5
x≈36.87˚
cosy=-2√5/5
y≈153.43˚
x+y≈190.30˚
sin(x+y)≈sin(190.30)≈-0.1788
exact value as computed=-2√5/25≈-0.1788
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