If cosx=4/5, cosy=12/13 in Quad. IV, find sin(x+y)

Answer 1

So, since you know Cos x= 0.8, plug in to the calculator Cos inv of 0.8. Do the same for Cos Y= 12/13.

X=36.9 degrees
Y=22.6 degrees

I think that where you are in precalculus, you have to know these trig identities. You MUST memorize them as they are the basis of good foundations for advanced math.

Math dot com/tables/trig/identities.htm

So Sin(X + Y)= Sinx Cos y + Cosx Sin y
So, Sin x= sin 36.9=0.60
Cos y= 12/13=0.92
Cos x= 4/5=0.80
Sin y= sin 22.6=0.38

So, the last part to do is:

(0.60*0.92) + (0.80*0.38)
0.552 + 1.18
1.732

I hope this helps!

Answer 2

Oh. I realized that you must also pay attention to signs, which I failed to do. In the fourth Quadrant, sin is negative, while cos is positive. So, your new solution will be:

So Sin(X + Y)= Sinx Cos y + Cosx Sin y
So, Sin x= sin 36.9=-0.60
Cos y= 12/13=0.92
Cos x= 4/5=0.80
Sin y= sin 22.6=-0.38

So, the last part to do is:

(-0.60*0.92) + (0.80*-0.38)
-0.552 - 0.30
-0.852

I hope I have corrected my errors, and if not, someone please help me out, as it is quite early in the morning and I have not had coffee yet!

Answer 3

In these type of questions, they usually expect the "exact" value of the trig ratios.

construct 2 right-angled triangles
The first you should recognize as the 3-4-5 right-angled triangle
The second you should recognize as the 5-12-13 triangle

if cosx = 4/5 , and x is in IV, then sinx = -3/5
if cosy = 12/13, and y is in IV, then siny = -5/13

sin(x+y) = sinxcosy + cosxsiny
= (-3/5)(12/13) + (4/5)((-5/13)
= -36/65 - 20/65
= -56/65