If sin(x) = 1/3 and sec(y) = 5/4, where x and y lie between 0 and π/2, evaluate sin(x + y). 11?

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If sin(x) = 1/3 and sec(y) = 5/4, where x and y are in the interval [0, π/2], how can we evaluate sin(x + y)?

Anonymous · Accepted Answer

I am not sure what the .11 means&hellip;so I will assume you want us to evaluate sin(x+y)use the sin addition formula:sin(x+y) = sin x cosy +cos x sin ysin x = 1/3we find cos x from sin^2x+Cos^2x=1(1/3)^2+ cos^2x = 1 or cos x = 0.94 (we use the positive root since we are told this is in the first quadrant)secy = 5/4 so cos y = 4/5 since cos y = 1/sec ythen, sin^2 y+ cos^2y=1sin^2y+(4/5)^2=1sin y = 3/5so sin (x+y) = (1/3)(4/5)+(0.94)(3/5)=0.83alternately, you can find that x = arc sin (1/3) = 18.4and that y = arc cos (4/5)=36.9so that sin(18.4+36.9)=0.83

Anonymous · Answer

sin x = 1/3 cos x = √[(1 – (1/3)²] = (√8)/3 = (2√2)/3 = ⅔√2 sec y = 5/4 cos y = 1/sec y = 4/5 sin y = √[(1 – (4/5)²] = 3/5 sin (x + y) = sin x cos y + cos x sin y sin (x + y) = (1/3) (4/5) + (⅔√2) (3/5) sin (x + y) = (4/15) + (2/5)√2)

Anonymous · Answer

sin(x) = 1/3 
cos^2(x) + sin^2(x) = 1
cos^2(x) = 1 – (1/3)^2
cos^2(x) = 8/9
cos(x)  = [2sqrt(2)]/3
sec(y) = 5/4 ==> cos(y) = 4/5
sin^2(y) = 1 – cos^2(y)
sin^2(y) = 1 – 16/25
sin^2(y) = 9/25
sin(y) = 3/5
All answers are positive between 0 and pi/2
sin(x + y) = sin(x).cos(y) + cos(x)sin(y)
= (1/3)(4/5) + [2sqrt(2)/3](3/5)
= 4/15 + 6sqrt(2)/15
= [4 + 6sqrt(2)]/15

Anonymous · Answer

Well&hellip; if sin(x) = 1/3, then x = arcsin(1/3) = 19.47 degreesif sec(y) = 5/4, then y = arcsec(5/4) = 36.87 degreesso sin(x + y) = sin(19.47 + 36.87) = sin(56.34) = 0.812

Anonymous

If sin(x) = 1/3 and sec(y) = 5/4, where x and y lie between 0 and π/2, evaluate sin(x + y). 11?

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