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# HW Help!

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

Sep 1, 2019
edited by Guest  Sep 1, 2019

#1
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Not sure if my answer is correct.

If sin(x)=2/3

Therefore,

X=sin^-1(2/3)

=41.81 degrees.(Approx.)

If sec(y)=5/4

Therefore, y=sec^-1(5/4)=36.84 degrees.(Approx.)

So we know now that

X=41.81 degrees

Y=36.841

Using addition formula for the sin.

here:

Sin(A+B)=Sin(A)*Cos(B)+Sin(B)*cos(A)

Let X=A , B=Y

Let me know if I did any mistake.

Sep 1, 2019
#2
+2

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

Thanks guest,

It is really good to see you give the question a go.

Unfortunately, your answer cannot be fully correct because sine is a ratio.  It is not going to be in degrees or radians.

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x and y are both acute angles.

Consider a right angled triangle with one acue angle equal to x

opp=2

hyp=3

sinx=2/3        cosx=sqrt5 / 3

Consider a right angled triangle with one acue angle equal to x

cosy=4/5

hyp=5

so   opp=sqrt(16) = 4

$$sin(x+y)=sinx cosy+cosxsiny\\ sin(x+y)=\frac{2}{3} \cdot\frac{4}{5}+\frac{\sqrt5}{3}\cdot\frac{4}{5}\\ sin(x+y)=\frac{20-4\sqrt5}{15}\\ sin(x+y)=\frac{4}{15}(5-\sqrt5)$$

4/15(5-sqrt(5)) approx = 0.7370485393333894    This is a ratio, it is not going to be in degrees or radians.

Sep 1, 2019
#3
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sin (x)  = 2/3      ⇒  cos(x)   =  √[3^2 - 2^2]  / 3  =  √/3

sec (y)  =  5/4  ⇒  cos (y)   =  4/5  ⇒  sin(y)   =  3/5

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

(2/3)(4/5)  + (√5)/3 (3/5)  =

[ 8   + 3√5 ]

_________

15   Sep 1, 2019
#4
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Since Chris has answered after me I assume I made a silly mistake somewhere.

Sep 2, 2019